b-1$ we can move only $b$ rungs downward. If we end up at $b-k0$ that $f^{k}(0)=0$. Since $f(m)=m+a$ or $f(m)=m-b$, it follows that $k$ can be written as $k=r+s$, where $r a-s b=0$. Since $a$ and $b$ are relatively prime, it follows that $k \geq a+b$.
Let us now prove that $f^{a+b}(0)=0$. In this case $a+b=r+s$ and hence $f^{a+b}(0)=(a+b-s) a-s b=(a+b)(a-s)$. Since $a+b \mid f^{a+b}(0)$ and $f^{a+b}(0) \in S$, it follows that $f^{a+b}(0)=0$. Thus for $(a, b)=1$ it follows that $k=a+b$. For other $a$ and $b$ we have $k=\frac{a+b}{(a, b)}$.
19. Let $d_{1}, d_{2}, d_{3}, d_{4}$ be the distances of the point $P$ to the tetrahedron. Let $d$ be the height of the regular tetrahedron. Let $x_{i}=d_{i} / d$. Clearly, $x_{1}+$ $x_{2}+x_{3}+x_{4}=1$, and given this condition, the parameters vary freely as we vary $P$ within the tetrahedron. The four tetrahedra have volumes $x_{1}^{3}, x_{2}^{3}, x_{3}^{3}$, and $x_{4}^{3}$, and the four parallelepipeds have volumes of $6 x_{2} x_{3} x_{4}$, $6 x_{1} x_{3} x_{4}, 6 x_{1} x_{2} x_{4}$, and $6 x_{1} x_{2} x_{3}$. Hence, using $x_{1}+x_{2}+x_{3}+x_{4}=1$ and setting $g(x)=x^{2}(1-x)$, we directly verify that
$$
\begin{aligned}
f(P) & =f\left(x_{1}, x_{2}, x_{3}, x_{4}\right)=1-\sum_{i=1}^{4} x_{i}^{3}-6 \sum_{1 \leq i0$ we have $g\left(x_{i}+x_{j}\right)+g(0) \geq$ $g\left(x_{i}\right)+g\left(x_{j}\right)$. Equality holds only when $x_{i}+x_{j}=2 / 3$.
Assuming without loss of generality $x_{1} \geq x_{2} \geq x_{3} \geq x_{4}$, we have $g\left(x_{1}\right)+$ $g\left(x_{2}\right)+g\left(x_{3}\right)+g\left(x_{4}\right)k$ and hence there exist two numbers $x, y \in S(n)$ such that $k \mid x-y$.
Let us show that $w=|x-y|$ is the desired number. By definition $k \mid w$. We also have
$$
w<1.2 \cdot 10^{n-1} \leq 1.2 \cdot\left(2^{3} \sqrt{2}\right)^{n-1} \leq 1.2 \cdot k^{3} \sqrt{k} \leq k^{4}
$$
Finally, since $x, y \in S(n)$, it follows that $w=|x-y|$ can be written using only the digits $\{0,1,8,9\}$. This completes the proof.
21. We must solve the congruence $\left(1+2^{p}+2^{n-p}\right) N \equiv 1\left(\bmod 2^{n}\right)$. Since $(1+$ $2^{p}+2^{n-p}$ ) and $2^{n}$ are coprime, there clearly exists a unique $N$ satisfying this equation and $0b$ we have in binary representation
$$
2^{a}-2^{b}=\underbrace{11 \ldots 11}_{a-b \text { times }} \underbrace{00 \ldots 00}_{b \text { times }}
$$
the binary representation of $N$ is calculated as follows:
$$
N=\left\{\begin{array}{cc}
\underbrace{11 \ldots 11}_{p \text { times }} \underbrace{11 \ldots 11}_{p \text { times }} \underbrace{00 \ldots 00}_{p \text { times }} \ldots \underbrace{11 \ldots 11}_{p \text { times }} \underbrace{00 \ldots 00}_{p-1 \text { times }} 1, & 2 \nmid \frac{n}{p} \\
\underbrace{11 \ldots 11}_{p-1 \text { times }} \underbrace{00 \ldots 00}_{p+1} \underbrace{11 \ldots 11}_{p \text { times }} \underbrace{00 \ldots 00}_{p \text { times }} \ldots \underbrace{11 \ldots 11}_{p \text { times }} \underbrace{00 \ldots 00}_{p-1 \text { times }} 1, & 2 \left\lvert\, \frac{n}{p}\right.
\end{array}\right.
$$
22. We can assume without loss of generality that each connection is serviced by only one airline and the problem reduces to finding two disjoint monochromatic cycles of the same color and of odd length on a complete graph of 10 points colored by two colors. We use the following two standard lemmas:
Lemma 1. Given a complete graph on six points whose edges are colored with two colors there exists a monochromatic triangle.
$\operatorname{Proof}$. Let us denote the vertices by $c_{1}, c_{2}, c_{3}, c_{4}, c_{5}, c_{6}$. By the pigeonhole principle at least three vertices out of $c_{1}$, say $c_{2}, c_{3}, c_{4}$, are of the same color, let us call it red. Assuming that at least one of the edges connecting points $c_{2}, c_{3}, c_{4}$ is red, the connected points along with $c_{1}$ form a red triangle. Otherwise, edges connecting $c_{2}, c_{3}, c_{4}$ are all of the opposite color, let us call it blue, and hence in all cases we have a monochromatic triangle.
Lemma 2. Given a complete graph on five points whose edges are colored two colors there exists a monochromatic triangle or a monochromatic cycle of length five.
Proof. Let us denote the vertices by $c_{1}, c_{2}, c_{3}, c_{4}, c_{5}$. Assume that out of a point $c_{i}$ three vertices are of the same color. We can then proceed as in Lemma 1 to obtain a monochromatic triangle. Otherwise, each
point is connected to other points with exactly two red and two blue vertices. Hence, we obtain monochromatic cycles starting from a single point and moving along the edges of the same color. Since each cycle must be of length at least three (i.e., we cannot have more than one cycle of one color), it follows that for both red and blue we must have one cycle of length five of that color.
We now apply the lemmas. Let us denote the vertices by $c_{1}, c_{2}, \ldots, c_{10}$. We apply Lemma 1 to vertices $c_{1}, \ldots, c_{6}$ to obtain a monochromatic triangle. Out of the seven remaining vertices we select 6 and again apply Lemma 1 to obtain another monochromatic triangle. If they are of the same color, we are done. Otherwise, out of the nine edges connecting the two triangles of opposite color at least 5 are of the same color, we can assume blue w.l.o.g., and hence a vertex of a red triangle must contain at least two blue edges whose endpoints are connected with a blue edge. Hence there exist two triangles of different colors joined at a vertex. These take up five points. Applying Lemma 2 on the five remaining points, we obtain a monochromatic cycle of odd length that is of the same color as one of the two joined triangles and disjoint from both of them.
23. Let us assume $n>1$. Obviously $n$ is odd. Let $p \geq 3$ be the smallest prime divisor of $n$. In this case $(p-1, n)=1$. Since $2^{n}+1 \mid 2^{2 n}-1$, we have that $p \mid 2^{2 n}-1$. Thus it follows from Fermat's little theorem and elementary number theory that $p \mid\left(2^{2 n}-1,2^{p-1}-1\right)=2^{(2 n, p-1)}-1$. Since $(2 n, p-1) \leq 2$, it follows that $p \mid 3$ and hence $p=3$.
Let us assume now that $n$ is of the form $n=3^{k} d$, where $2,3 \nmid d$. We first prove that $k=1$.
Lemma. If $2^{m}-1$ is divisible by $3^{r}$, then $m$ is divisible by $3^{r-1}$.
Proof. This is the lemma from (SL97-14) with $p=3, a=2^{2}, k=m$, $\alpha=1$, and $\beta=r$.
Since $3^{2 k}$ divides $n^{2} \mid 2^{2 n}-1$, we can apply the lemma to $m=2 n$ and $r=2 k$ to conclude that $3^{2 k-1} \mid n=3^{k} d$. Hence $k=1$.
Finally, let us assume $d>1$ and let $q$ be the smallest prime factor of $d$. Obviously $q$ is odd, $q \geq 5$, and $(n, q-1) \in\{1,3\}$. We then have $q \mid 2^{2 n}-1$ and $q \mid 2^{q-1}-1$. Consequently, $q \mid 2^{(2 n, q-1)}-1=2^{2(n, q-1)}-1$, which divides $2^{6}-1=63=3^{2} \cdot 7$, so we must have $q=7$. However, in that case we obtain $7|n| 2^{n}+1$, which is a contradiction, since powers of two can only be congruent to 1,2 and 4 modulo 7 . It thus follows that $d=1$ and $n=3$. Hence $n>1 \Rightarrow n=3$.
It is easily verified that $n=1$ and $n=3$ are indeed solutions. Hence these are the only solutions.
24. Let us denote $A=b+c+d, B=a+c+d, C=a+b+d, D=a+b+c$. Since $a b+b c+c d+d a=1$ the numbers $A, B, C, D$ are all positive. By trivially applying the AM-GM inequality we have:
$$
a^{2}+b^{2}+c^{2}+d^{2} \geq a b+b c+c d+d a=1 .
$$
We will prove the inequality assuming only that $A, B, C, D$ are positive and $a^{2}+b^{2}+c^{2}+d^{2} \geq 1$. In this case we may assume without loss of generality that $a \geq b \geq c \geq d \geq 0$. Hence $a^{3} \geq b^{3} \geq c^{3} \geq d^{3} \geq 0$ and $\frac{1}{A} \geq \frac{1}{B} \geq \frac{1}{C} \geq \frac{1}{D}>0$. Using the Chebyshev and Cauchy inequalities we obtain:
$$
\begin{aligned}
& \frac{a^{3}}{A}+\frac{b^{3}}{B}+\frac{c^{3}}{C}+\frac{d^{3}}{D} \\
& \geq \frac{1}{4}\left(a^{3}+b^{3}+c^{3}+d^{3}\right)\left(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}+\frac{1}{D}\right) \\
& \geq \frac{1}{16}\left(a^{2}+b^{2}+c^{2}+d^{2}\right)(a+b+c+d)\left(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}+\frac{1}{D}\right) \\
& =\frac{1}{48}\left(a^{2}+b^{2}+c^{2}+d^{2}\right)(A+B+C+D)\left(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}+\frac{1}{D}\right) \geq \frac{1}{3}
\end{aligned}
$$
This completes the proof.
25. Plugging in $x=1$ we get $f(f(y))=f(1) / y$ and hence $f\left(y_{1}\right)=f\left(y_{2}\right)$ implies $y_{1}=y_{2}$ i.e. that the function is bijective. Plugging in $y=1$ gives us $f(x f(1))=f(x) \Rightarrow x f(1)=x \Rightarrow f(1)=1$. Hence $f(f(y))=1 / y$. Plugging in $y=f(z)$ implies $1 / f(z)=f(1 / z)$. Finally setting $y=f(1 / t)$ into the original equation gives us $f(x t)=f(x) / f(1 / t)=f(x) f(t)$. Conversely, any functional equation on $\mathbb{Q}^{+}$satisfying (i) $f(x t)=f(x) f(t)$ and (ii) $f(f(x))=\frac{1}{x}$ for all $x, t \in \mathbb{Q}^{+}$also satisfies the original functional equation: $f(x f(y))=f(x) f(f(y))=\frac{f(x)}{y}$. Hence it suffices to find a function satisfying (i) and (ii).
We note that all elements $q \in \mathbb{Q}^{+}$are of the form $q=\prod_{i=1}^{n} p_{i}^{a_{i}}$ where $p_{i}$ are prime and $a_{i} \in \mathbb{Z}$. The criterion ( $a$ ) implies $f(q)=f\left(\prod_{i=1}^{n} p_{i}^{a_{i}}\right)=$ $\prod_{i=1}^{n} f\left(p_{i}\right)^{a_{i}}$. Thus it is sufficient to define the function on all primes. For the function to satisfy $(b)$ it is necessary and sufficient for it to satisfy $f(f(p))=\frac{1}{p}$ for all primes $p$. Let $q_{i}$ denote the $i$-th smallest prime. We define our function $f$ as follows:
$$
f\left(q_{2 k-1}\right)=q_{2 k}, \quad f\left(q_{2 k}\right)=\frac{1}{q_{2 k-1}}, \quad k \in \mathbb{N}
$$
Such a function clearly satisfies $(b)$ and along with the additional condition $f(x t)=f(x) f(t)$ it is well defined for all elements of $\mathbb{Q}^{+}$and it satisfies the original functional equation.
26. We note that $|P(x) / x| \rightarrow \infty$. Hence, there exists an integer number $M$ such that $M>\left|q_{1}\right|$ and $|P(x)| \leq|x| \Rightarrow|x|\left|q_{i}\right| \geq M$ and this ultimately contradicts $\left|q_{1}\right|0\right)$ both belonging to the set $\{j T+1, j T+2, \ldots, j T+T\}$ such that $q_{m_{j}}=q_{m_{j}+k_{j}}$. Since $k_{j}\sin ^{3} 30^{\circ}=\frac{1}{8} .
\end{aligned}
$$
On the other hand, since $\angle M A C+\angle M B A+\angle M C B<180^{\circ}-3 \cdot 30^{\circ}=90^{\circ}$, Jensen's inequality applied on the concave function $\ln \sin x(x \in[0, \pi])$ gives us $\sin \angle M A C \sin \angle M B A \sin \angle M C B<\sin ^{3} 30^{\circ}$, contradicting (*).
Second solution. Denote the intersections of $P A, P B, P C$ with $B C, C A$, $A B$ by $A_{1}, B_{1}, C_{1}$, respectively. Suppose that each of the angles $\angle P A B$, $\angle P B C, \angle P C A$ is greater than $30^{\circ}$ and denote $P A=2 x, P B=2 y, P C=$ $2 z$. Then $P C_{1}>x, P A_{1}>y, P B_{1}>z$. On the other hand, we know that
$$
\frac{P C_{1}}{P C+P C_{1}}+\frac{P A_{1}}{P A+P A_{1}}+\frac{P B_{1}}{P B+P B_{1}}=\frac{S_{A B P}}{S_{A B C}}+\frac{S_{P B C}}{S_{A B C}}+\frac{S_{A P C}}{S_{A B C}}=1 .
$$
Since the function $\frac{t}{p+t}$ is increasing, we obtain $\frac{x}{2 z+x}+\frac{y}{2 x+y}+\frac{z}{2 y+z}<1$. But on the contrary, Cauchy-Schwartz inequality (or alternatively Jensen's inequality) yields
$$
\frac{x}{2 z+x}+\frac{y}{2 x+y}+\frac{z}{2 y+z} \geq \frac{(x+y+z)^{2}}{x(2 z+x)+y(2 x+y)+z(2 y+z)}=1 .
$$
5. Let $P_{1}$ be the point on the side $B C$ such that $\angle B F P_{1}=\beta / 2$. Then $\angle B P_{1} F=180^{\circ}-3 \beta / 2$, and the sine law gives us $\frac{B F}{B P_{1}}=\frac{\sin (3 \beta / 2)}{\sin (\beta / 2)}=$ $3-4 \sin ^{2}(\beta / 2)=1+2 \cos \beta$.
Now we calculate $\frac{B F}{B P}$. We have $\angle B I F=120^{\circ}-\beta / 2, \angle B F I=60^{\circ}$ and $\angle B I C=120^{\circ}, \angle B C I=\gamma / 2=60^{\circ}-\beta / 2$. By the sine law,
$$
B F=B I \frac{\sin \left(120^{\circ}-\beta / 2\right)}{\sin 60^{\circ}}, \quad B P=\frac{1}{3} B C=B I \frac{\sin 120^{\circ}}{3 \sin \left(60^{\circ}-\beta / 2\right)} .
$$
It follows that $\frac{B F}{B P}=\frac{3 \sin \left(60^{\circ}-\beta / 2\right) \sin \left(60^{\circ}+\beta / 2\right)}{\sin ^{2} 60^{\circ}}=4 \sin \left(60^{\circ}-\beta / 2\right) \sin \left(60^{\circ}+\right.$ $\beta / 2)=2\left(\cos \beta-\cos 120^{\circ}\right)=2 \cos \beta+1=\frac{B F}{B P_{1}}$. Therefore $P \equiv P_{1}$.
6. Let $a, b, c$ be sides of the triangle. Let $A_{1}$ be the intersection of line $A I$ with $B C$. By the known fact, $B A_{1}: A_{1} C=c: b$ and $A I: I A_{1}=A B: B A_{1}$, hence $B A_{1}=\frac{a c}{b+c}$ and $\frac{A I}{I A_{1}}=\frac{A B}{B A_{1}}=\frac{b+c}{a}$. Consequently $\frac{A I}{l_{A}}=\frac{b+c}{a+b+c}$. Put $a=n+p, b=p+m, c=m+n$ : it is obvious that $m, n, p$ are positive. Our inequality becomes
$$
2<\frac{(2 m+n+p)(m+2 n+p)(m+n+2 p)}{(m+n+p)^{3}} \leq \frac{64}{27}
$$
The right side inequality immediately follows from the inequality between arithmetic and geometric means applied on $2 m+n+p, m+2 n+p$ and $m+n+2 p$. For the left side inequality, denote by $T=m+n+p$. Then we can write $(2 m+n+p)(m+2 n+p)(m+n+2 p)=(T+m)(T+n)(T+p)$ and
$(T+m)(T+n)(T+p)=T^{3}+(m+n+p) T^{2}+(m n+n p+p n) T+m n p>2 T^{3}$.
Remark. The inequalities cannot be improved. In fact, $\frac{A I \cdot B I \cdot C I}{l_{A} l_{B} l_{C}}$ is equal to $8 / 27$ for $a=b=c$, while it can be arbitrarily close to $1 / 4$ if $a=b$ and $c$ is sufficiently small.
7. The given equations imply $A B=C D, A C=B D, A D=B C$. Let $L_{1}$, $M_{1}, N_{1}$ be the midpoints of $A D, B D, C D$ respectively. Then the above
equalities yield
$$
\begin{gathered}
L_{1} M_{1}=A B / 2=L M \\
L_{1} M_{1}\|A B\| L M \\
L_{1} M=C D / 2=L M_{1} \\
L_{1} M\|C D\| L M_{1}
\end{gathered}
$$
Thus $L, M, L_{1}, M_{1}$ are coplanar and $L M L_{1} M_{1}$ is a rhombus as well as
$M N M_{1} N_{1}$ and $L N L_{1} N_{1}$. Then the segments $L L_{1}, M M_{1}, N N_{1}$ have the common midpoint $Q$ and $Q L \perp Q M$, $Q L \perp Q N, Q M \perp Q N$. We also infer that the line $N N_{1}$ is perpendicular to the plane $L M L_{1} M_{1}$ and hence to the line $A B$. Thus $Q A=Q B$, and similarly, $Q B=Q C=Q D$, hence $Q$ is just the center $O$, and $\angle L O M=$ $\angle M O N=\angle N O L=90^{\circ}$.
8. Let $P_{1}\left(x_{1}, y_{1}\right), P_{2}\left(x_{2}, y_{2}\right), \ldots, P_{n}\left(x_{n}, y_{n}\right)$ be the $n$ points of $S$ in the coordinate plane. We may assume $x_{1}216$.
We claim that the answer is $n=217$. First notice that the set $A_{2} \cup A_{3} \cup$ $A_{5} \cup A_{7}$ consists of four prime $(2,3,5,7)$ and 212 composite numbers. The set $S \backslash A$ contains exactly 8 composite numbers: namely, $11^{2}, 11 \cdot 13,11$. $17,11 \cdot 19,11 \cdot 23,13^{2}, 13 \cdot 17,13 \cdot 19$. Thus $S$ consists of the unity, 220 composite numbers and 59 primes.
Let $A$ be a 217 -element subset of $S$, and suppose that there are no five pairwise relatively prime numbers in $A$. Then $A$ can contain at most 4 primes (or unity and three primes) and at least 213 composite numbers. Hence the set $S \backslash A$ contains at most 7 composite numbers. Consequently, at least one of the following 8 five-element sets is disjoint with $S \backslash A$, and is thus entirely contained in $A$ :
$$
\begin{array}{ll}
\{2 \cdot 23,3 \cdot 19,5 \cdot 17,7 \cdot 13,11 \cdot 11\}, & \{2 \cdot 29,3 \cdot 23,5 \cdot 19,7 \cdot 17,11 \cdot 13\}, \\
\{2 \cdot 31,3 \cdot 29,5 \cdot 23,7 \cdot 19,11 \cdot 17\}, & \{2 \cdot 37,3 \cdot 31,5 \cdot 29,7 \cdot 23,11 \cdot 19\}, \\
\{2 \cdot 41,3 \cdot 37,5 \cdot 31,7 \cdot 29,11 \cdot 23\}, & \{2 \cdot 43,3 \cdot 41,5 \cdot 37,7 \cdot 31,13 \cdot 17\}, \\
\{2 \cdot 47,3 \cdot 43,5 \cdot 41,7 \cdot 37,13 \cdot 19\}, & \{2 \cdot 2,3 \cdot 3,5 \cdot 5,7 \cdot 7,13 \cdot 13\} .
\end{array}
$$
As each of these sets consists of five numbers relatively prime in pairs, the claim is proved.
13. Call a sequence $e_{1}, \ldots, e_{n}$ good if $e_{1} a_{1}+\cdots+e_{n} a_{n}$ is divisible by $n$. Among the sums $s_{0}=0, s_{1}=a_{1}, s_{2}=a_{1}+a_{2}, \ldots, s_{n}=a_{1}+\cdots+a_{n}$, two give the same remainder modulo $n$, and their difference corresponds to a good sequence. To show that, permuting the $a_{i}$ 's, we can find $n-1$ different sequences, we use the following
Lemma. Let $A$ be a $k \times n(k \leq n-2)$ matrix of zeros and ones, whose every row contains at least one 0 and at least two 1 's. Then it is possible to permute columns of $A$ is such a way that in any row 1 's do not form a block.
Proof. We will use the induction on $k$. The case $k=1$ and arbitrary $n \geq 3$ is trivial. Suppose that $k \geq 2$ and that for $k-1$ and any $n \geq k+1$ the lemma is true. Consider a $k \times n$ matrix $A, n \geq k+2$. We mark an element $a_{i j}$ if either it is the only zero in the $i$-th row, or one of the 1 's in the row if it contains exactly two 1 's. Since $n \geq 4$, every row contains at most two marked elements, which adds up to at most $2 k<2 n$ marked elements in total. It follows that there is a column with at most one marked element. Assume w.l.o.g. that it is the first column and that $a_{1 j}$ isn't marked for $j>1$. The matrix $B$, obtained by omitting the first row and first column from $A$, satisfies the conditions of the lemma. Therefore, we can permute columns of $B$ and get the required form. Considered as a permutation of column of $A$, this permutation may leave a block of 1's only in the first row of $A$. In the case that it is so, if $a_{11}=1$ we put the first column in the last place, otherwise we put it between any two columns having 1's in the first row. The obtained matrix has the required property.
Suppose now that we have got $k$ different nontrivial good sequences $e_{1}^{i}, \ldots, e_{n}^{i}, i=1, \ldots, k$, and that $k \leq n-2$. The matrix $A=\left(e_{j}^{i}\right)$
fulfils the conditions of Lemma, hence there is a permutation $\sigma$ from Lemma. Now among the sums $s_{0}=0, s_{1}=a_{\sigma(1)}, s_{2}=a_{\sigma(1)}+a_{\sigma(2)}$, $\ldots, s_{n}=a_{\sigma(1)}+\cdots+a_{\sigma(n)}$, two give the same remainder modulo $n$. Let $s_{p} \equiv s_{q}(\bmod n), pm>$ 0 , as large as we like, such that $10^{k} \equiv 10^{m}(\bmod T)$, using for example Euler's theorem. It is obvious that $a_{10^{k}-1}=a_{10^{k}}$ and hence, taking $k$ sufficiently large and using the periodicity, we see that
$$
a_{2 \cdot 10^{k}-10^{m}-1}=a_{10^{k}-1}=a_{10^{k}}=a_{2 \cdot 10^{k}-10^{m}}
$$
Since $\left(2 \cdot 10^{k}-10^{m}\right)!=\left(2 \cdot 10^{k}-10^{m}\right)\left(2 \cdot 10^{k}-10^{m}-1\right)!$ and the last nonzero digit of $2 \cdot 10^{k}-10^{m}$ is nine, we must have $a_{2 \cdot 10^{k}-10^{m}-1}=5$ (if $s$ is a digit, the last digit of $9 s$ is $s$ only if $s=5$ ). But this means that 5 divides $n$ ! with a greater power than 2 does, which is impossible. Indeed, if the exponents of these powers are $\alpha_{2}, \alpha_{5}$ respectively, then $\alpha_{5}=$ $[n / 5]+\left[n / 5^{2}\right]+\cdots \leq \alpha_{2}=[n / 2]+\left[n / 2^{2}\right]+\cdots$.
16. Let $p$ be the least prime number that does not divide $n$ : thus $a_{1}=1$ and $a_{2}=p$. Since $a_{2}-a_{1}=a_{3}-a_{2}=\cdots=r$, the $a_{i}$ 's are $1, p, 2 p-1,3 p-2, \ldots$ We have the following cases:
$p=2$. Then $r=1$ and the numbers $1,2,3, \ldots, n-1$ are relatively prime to $n$, hence $n$ is a prime.
$p=3$. Then $r=2$, so every odd number less than $n$ is relatively prime to $n$, from which we deduce that $n$ has no odd divisors. Therefore $n=2^{k}$ for some $k \in \mathbb{N}$.
$p>3$. Then $r=p-1$ and $a_{k+1}=a_{1}+k(p-1)=1+k(p-1)$. Since $n-1$ also must belong to the progression, we have $p-1 \mid n-2$. Let $q$ be any prime divisor of $p-1$. Then also $q \mid n-2$. On the other hand, since $q0)$ modulo 3 , we get that $5^{z} \equiv 1(\bmod 3)$, hence $z$ is even, say $z=2 z_{1}$. The equation then becomes $3^{x}=5^{2 z_{1}}-4^{y}=\left(5^{z_{1}}-2^{y}\right)\left(5^{z_{1}}+2^{y}\right)$. Each factor $5^{z_{1}}-2^{y}$ and $5^{z_{1}}+2^{y}$ is a power of 3 , for which the only possibility is $5^{z_{1}}+2^{y}=3^{x}$ and $5^{z_{1}}-2^{y}=$ 1. Again modulo 3 these equations reduce to $(-1)^{z_{1}}+(-1)^{y}=0$ and $(-1)^{z_{1}}-(-1)^{y}=1$, implying that $z_{1}$ is odd and $y$ is even. Particularly, $y \geq 2$. Reducing the equation $5^{z_{1}}+2^{y}=3^{x}$ modulo 4 we get that $3^{x} \equiv 1$, hence $x$ is even. Now if $y>2$, modulo 8 this equation yields $5 \equiv 5^{z_{1}} \equiv$ $3^{x} \equiv 1$, a contradiction. Hence $y=2, z_{1}=1$. The only solution of the original equation is $x=y=z=2$.
18. For integers $a>0, n>0$ and $\alpha \geq 0$, we shall write $a^{\alpha} \| n$ when $a^{\alpha} \mid n$ and $a^{\alpha+1} \nmid n$.
Lemma. For every odd number $a \geq 3$ and an integer $n \geq 0$ it holds that
$$
a^{n+1} \|(a+1)^{a^{n}}-1 \quad \text { and } \quad a^{n+1} \|(a-1)^{a^{n}}+1
$$
Proof. We shall prove the first relation by induction (the second is analogous). For $n=0$ the statement is obvious. Suppose that it holds for some $n$, i.e. that $(1+a)^{a^{n}}=1+N a^{n+1}, a \nmid N$. Then
$$
(1+a)^{a^{n+1}}=\left(1+N a^{n+1}\right)^{a}=1+a \cdot N a^{n+1}+\binom{a}{2} N^{2} a^{2 n+2}+M a^{3 n+3}
$$
for some integer $M$. Since $\binom{a}{2}$ is divisible by $a$ for $a$ odd, we deduce that the part of the above sum behind $1+a \cdot N a^{n+1}$ is divisible by $a^{n+3}$. Hence $(1+a)^{a^{n+1}}=1+N^{\prime} a^{n+2}$, where $a \nmid N^{\prime}$.
It follows immediately from Lemma that
$$
1991^{1993} \| 1990^{1991^{1992}}+1 \quad \text { and } \quad 1991^{1991} \| 1992^{1991^{1990}}-1
$$
Adding these two relations we obtain immediately that $k=1991$ is the desired value.
19. Set $x=\cos (\pi a)$. The given equation is equivalent to $4 x^{3}+4 x^{2}-3 x-2=0$, which factorizes as $(2 x+1)\left(2 x^{2}+x-2\right)=0$.
The case $2 x+1=0$ yields $\cos (\pi a)=-1 / 2$ and $a=2 / 3$. It remains to show that if $x$ satisfies $2 x^{2}+x-2=0$ then $a$ is not rational. The polynomial equation $2 x^{2}+x-2=0$ has two real roots, $x_{1,2}=\frac{-1 \pm \sqrt{17}}{4}$, and since $|x| \leq 1$ we must have $x=\cos \pi a=\frac{-1+\sqrt{17}}{4}$.
We now prove by induction that, for every integer $n \geq 0, \cos \left(2^{n} \pi a\right)=$ $\frac{a_{n}+b_{n} \sqrt{17}}{4}$ for some odd integers $a_{n}, b_{n}$. The case $n=0$ is trivial. Also, if $\cos \left(2^{n} \pi a\right)=\frac{a_{n}+b_{n} \sqrt{17}}{4}$, then
$$
\begin{aligned}
\cos \left(2^{n+1} \pi a\right) & =2 \cos ^{2}\left(2^{n} \pi a\right)-1 \\
& =\frac{1}{4}\left(\frac{a_{n}^{2}+17 b_{n}^{2}-8}{2}+a_{n} b_{n} \sqrt{17}\right)=\frac{a_{n+1}+b_{n+1} \sqrt{17}}{4} .
\end{aligned}
$$
By the inductive step that $a_{n}, b_{n}$ are odd, it is obvious that $a_{n+1}, b_{n+1}$ are also odd. This proves the claim.
Note also that, since $a_{n+1}=\frac{1}{2}\left(a_{n}^{2}+17 b_{n}^{2}-8\right)>a_{n}$, the sequence $\left\{a_{n}\right\}$ is strictly increasing. Hence the set of values of $\cos \left(2^{n} \pi a\right), n=0,1,2, \ldots$, is infinite (because $\sqrt{17}$ is irrational). However, if $a$ were rational, then the set of values of $\cos m \pi a, m=1,2, \ldots$, would be finite, a contradiction. Therefore the only possible value for $a$ is $2 / 3$.
20. We prove the result with 1991 replaced by any positive integer $k$. For natural numbers $p, q$, let $\epsilon=(\alpha p-[\alpha p])(\alpha q-[\alpha q])$. Then $0<\epsilon<1$ and
$$
\epsilon=\alpha^{2} p q-\alpha(p[\alpha q]+q[\alpha p])+[\alpha p][\alpha q] .
$$
Multiplying this equality by $\alpha-k$ and using $\alpha^{2}=k \alpha+1$, i.e. $\alpha(\alpha-k)=1$, we get
$$
(\alpha-k) \epsilon=\alpha(p q+[\alpha p][\alpha q])-(p[\alpha q]+q[\alpha p]+k[\alpha p][\alpha q])
$$
Since $0<(\alpha-k) \epsilon<1$, we have $[\alpha(p * q)]=p[\alpha q]+q[\alpha p]+k[\alpha p][\alpha q]$. Now
$$
\begin{aligned}
(p * q) * r & =(p * q) r+[\alpha(p * q)][\alpha r]= \\
& =p q r+[\alpha p][\alpha q] r+[\alpha q][\alpha r] p+[\alpha r][\alpha p] q+k[\alpha p][\alpha q][\alpha r] .
\end{aligned}
$$
Since the last expression is symmetric, the same formula is obtained for $p *(q * r)$.
21. The polynomial $g(x)$ factorizes as $g(x)=f(x)^{2}-9=(f(x)-3)(f(x)+3)$. If one of the equations $f(x)+3=0$ and $f(x)-3=0$ has no integer solutions, then the number of integer solutions of $g(x)=0$ clearly does not exceed 1991.
Suppose now that both $f(x)+3=0$ and $f(x)-3=0$ have integer solutions. Let $x_{1}, \ldots, x_{k}$ be distinct integer solutions of the former, and $x_{k+1}, \ldots, x_{k+l}$ be distinct integer solutions of the latter equation. There exist monic polynomials $p(x), q(x)$ with integer coefficients such that $f(x)+3=\left(x-x_{1}\right)\left(x-x_{2}\right) \ldots\left(x-x_{k}\right) p(x)$ and $f(x)-3=$ $\left(x-x_{k+1}\right)\left(x-x_{k+2}\right) \ldots\left(x-x_{k+l}\right) q(x)$. Thus we obtain
$\left(x-x_{1}\right)\left(x-x_{2}\right) \ldots\left(x-x_{k}\right) p(x)-\left(x-x_{k+1}\right)\left(x-x_{k+2}\right) \ldots\left(x-x_{k+l}\right) q(x)=6$.
Putting $x=x_{k+1}$ we get $\left(x_{k+1}-x_{1}\right)\left(x_{k+1}-x_{2}\right) \cdots\left(x_{k+1}-x_{k}\right) \mid 6$, and since the product of more than four distinct integers cannot divide 6 , this implies $k \leq 4$. Similarly $l \leq 4$; hence $g(x)=0$ has at most 8 distinct integer solutions.
Remark. The proposer provided a solution for the upper bound of 1995 roots which was essentially the same as that of (IMO74-6).
22. Suppose w.l.o.g. that the center of the square is at the origin $O(0,0)$. We denote the curve $y=f(x)=x^{3}+a x^{2}+b x+c$ by $\gamma$ and the vertices of the square by $A, B, C, D$ in this order.
At first, the symmetry with respect to the point $O$ maps $\gamma$ into the curve $\bar{\gamma}\left(y=f(-x)=x^{3}-a x^{2}+b x-c\right)$. Obviously $\bar{\gamma}$ also passes through $A, B, C, D$, and thus has four different intersection points with $\gamma$. Then $2 a x^{2}+2 c$ has at least four distinct solution, which implies $a=c=0$. Particularly, $\gamma$ passes through $O$ and intersects all quadrants, and hence $b<0$.
Further, the curve $\gamma^{\prime}$, obtained by rotation of $\gamma$ around $O$ for $90^{\circ}$, has an equation $-x=f(y)$ and also contains the points $A, B, C, D$ and $O$. The intersection points $(x, y)$ of $\gamma \cap \gamma^{\prime}$ are determined by $-x=f(f(x))$, and hence they are roots of a polynomial $p(x)=f(f(x))+x$ of 9-th degree.
But the number of times that one cubic actually crosses the other in each quadrant is in the general case even (draw the picture!), and since $A B C D$ is the only square lying on $\gamma \cap \gamma^{\prime}$, the intersection points $A, B, C, D$ must be double. It follows that
$$
p(x)=x[(x-r)(x+r)(x-s)(x+s)]^{2},
$$
where $r, s$ are the $x$-coordinates of $A$ and $B$. On the other hand, $p(x)$ is defined by $\left(x^{3}+b x\right)^{3}+b\left(x^{3}+b x\right)+x$, and therefore equating of coefficients with (1) yields
$$
\begin{array}{cc}
3 b=-2\left(r^{2}+s^{2}\right), & 3 b^{2}=\left(r^{2}+s^{2}\right)^{2}+2 r^{2} s^{2}, \\
b\left(b^{2}+1\right)=-2 r^{2} s^{2}\left(r^{2}+s^{2}\right), & b^{2}+1=r^{4} s^{4} .
\end{array}
$$
Straightforward solving this system of equations gives $b=-\sqrt{8}$ and $r^{2}+$ $s^{2}=\sqrt{18}$.
The line segment from $O$ to $(r, s)$ is half a diagonal of the square, and thus a side of the square has length $a=\sqrt{2\left(r^{2}+s^{2}\right)}=\sqrt[4]{72}$.
23. From (i), replacing $m$ by $f(f(m))$, we get
$$
f(f(f(m))+f(f(n)))=-f(f(f(f(m))+1))-n
$$
analogously $\quad f(f(f(n))+f(f(m)))=-f(f(f(f(n))+1))-m$.
From these relations we get $f(f(f(f(m))+1))-f(f(f(f(n))+1))=m-n$. Again from (i),
$$
\begin{aligned}
& f(f(f(f(m))+1))=f(-m-f(f(2))) \\
& \text { and } \quad f(f(f(f(n))+1))=f(-n-f(f(2))) .
\end{aligned}
$$
Setting $f(f(2))=k$ we obtain $f(-m-k)-f(-n-k)=m-n$ for all integers $m, n$. This implies $f(m)=f(0)-m$. Then also $f(f(m))=m$, and using this in (i) we finally get
$$
f(n)=-n-1 \quad \text { for all integers } n
$$
Particularly $f(1991)=-1992$.
From (ii) we obtain $g(n)=g(-n-1)$ for all integers $n$. Since $g$ is a polynomial, it must also satisfy $g(x)=g(-x-1)$ for all real $x$. Let us now express $g$ as a polynomial on $x+1 / 2: g(x)=h(x+1 / 2)$. Then $h$ satisfies $h(x+1 / 2)=h(-x-1 / 2)$, i.e. $h(y)=h(-y)$, hence it is a polynomial in $y^{2}$; thus $g$ is a polynomial in $(x+1 / 2)^{2}=x^{2}+x+1 / 4$. Hence $g(n)=p\left(n^{2}+n\right)$ (for some polynomial $p$ ) is the most general form of $g$.
24. Let $y_{k}=a_{k}-a_{k+1}+a_{k+2}-\cdots+a_{k+n-1}$ for $k=1,2, \ldots, n$, where we define $x_{i+n}=x_{i}$ for $1 \leq i \leq n$. We then have $y_{1}+y_{2}=2 a_{1}, y_{2}+y_{3}=$ $2 a_{2}, \ldots, y_{n}+y_{1}=2 a_{n}$.
(i) Let $n=4 k-1$ for some integer $k>0$. Then for each $i=1,2, \ldots, n$ we have that $y_{i}=\left(a_{i}+a_{i+1}+\cdots+a_{i-1}\right)-2\left(a_{i+1}+a_{i+3}+\cdots+a_{i-2}\right)=1+$ $2+\cdots+(4 k-1)-2\left(a_{i+1}+a_{i+3}+\cdots+a_{i-2}\right)$ is even. Suppose now that $a_{1}, \ldots, a_{n}$ is a good permutation. Then each $y_{i}$ is positive and even, so $y_{i} \geq 2$. But for some $t \in\{1, \ldots, n\}$ we must have $a_{t}=1$, and thus $y_{t}+y_{t+1}=2 a_{t}=2$ which is impossible. Hence the numbers $n=4 k-1$ are not good.
(ii) Let $n=4 k+1$ for some integer $k>0$. Then $2,4, \ldots, 4 k, 4 k+1,4 k-$ $1, \ldots, 3,1$ is a permutation with the desired property. Indeed, in this case $y_{1}=y_{4 k+1}=1, y_{2}=y_{4 k}=3, \ldots, y_{2 k}=y_{2 k+2}=4 k-1$, $y_{2 k+1}=4 k+1$.
Therefore all nice numbers are given by $4 k+1, k \in \mathbb{N}$.
25. Since replacing $x_{1}$ by 1 can only reduce the set of indices $i$ for which the desired inequality holds, we may assume $x_{1}=1$. Similarly we may assume $x_{n}=0$. Now we can let $i$ be the largest index such that $x_{i}>1 / 2$. Then $x_{i+1} \leq 1 / 2$, hence
$$
x_{i}\left(1-x_{i+1}\right) \geq \frac{1}{4}=\frac{1}{4} x_{1}\left(1-x_{n}\right) .
$$
26. Without loss of generality we can assume $b_{1} \geq b_{2} \geq \cdots \geq b_{n}$. We denote by $A_{i}$ the product $a_{1} a_{2} \ldots a_{i-1} a_{i+1} \ldots a_{n}$. If for some $i0
$$
because $x_{k-1}+x_{k} \leq \frac{2}{3}\left(x_{1}+x_{k-1}+x_{k-2}\right) \leq \frac{2}{3}$. Repeating this procedure we can reduce the number of nonzero $x_{i}$ 's to two, increasing the value of $F$ in each step. It remains to maximize $F$ over $n$-tuples $\left(x_{1}, x_{2}, 0, \ldots, 0\right)$ with $x_{1}, x_{2} \geq 0, x_{1}+x_{2}=1$ : in this case $F$ equals $x_{1} x_{2}$ and attains its maximum value $\frac{1}{4}$ when $x_{1}=x_{2}=\frac{1}{2}, x_{3}=\ldots, x_{n}=0$.
28. Let $x_{n}=c(n \sqrt{2}-[n \sqrt{2}])$ for some constant $c>0$. For $i>j$, putting $p=[i \sqrt{2}]-[j \sqrt{2}]$, we have
$\left|x_{i}-x_{j}\right|=c|(i-j) \sqrt{2}-p|=\frac{\left|2(i-j)^{2}-p^{2}\right| c}{(i-j) \sqrt{2}+p} \geq \frac{c}{(i-j) \sqrt{2}+p} \geq \frac{c}{4(i-j)}$,
because $p<(i-j) \sqrt{2}+1$. Taking $c=4$, we obtain that for any $i>j$, $(i-j)\left|x_{i}-x_{j}\right| \geq 1$. Of course, this implies $(i-j)^{a}\left|x_{i}-x_{j}\right| \geq 1$ for any $a>1$ 。
Remark. The constant 4 can be replaced with $3 / 2+\sqrt{2}$.
Second solution. Another example of a sequence $\left\{x_{n}\right\}$ is constructed in the following way: $x_{1}=0, x_{2}=1, x_{3}=2$ and $x_{3^{k} i+m}=x_{m}+\frac{i}{3^{k}}$ for $i=1,2$ and $1 \leq m \leq 3^{k}$. It is easily shown that $|i-j| \cdot\left|x_{i}-x_{j}\right| \geq 1 / 3$ for any $i \neq j$.
Third solution. If $n=b_{0}+2 b_{1}+\cdots+2^{k} b_{k}, b_{i} \in\{0,1\}$, then one can set $x_{n}$ to be $=b_{0}+2^{-a} b_{1}+\cdots+2^{-k a} b_{k}$. In this case it holds that $|i-j|^{a}\left|x_{i}-x_{j}\right| \geq$ $\frac{2^{a}-2}{2^{a}-1}$.
29. One easily observes that the following sets are super-invariant: one-point set, its complement, closed and open half-lines or their complements, and the whole real line. To show that these are the only possibilities, we first observe that $S$ is super-invariant if and only if for each $a>0$ there is a $b$ such that $x \in S \Leftrightarrow a x+b \in S$.
(i) Suppose that for some $a$ there are two such $b$ 's: $b_{1}$ and $b_{2}$. Then $x \in$ $S \Leftrightarrow a x+b_{1} \in S$ and $x \in S \Leftrightarrow a x+b_{2} \in S$, which implies that $S$ is periodic: $y \in S \Leftrightarrow y+\frac{b_{1}-b_{2}}{a} \in S$. Since $S$ is identical to a translate of any stretching of $S$, all positive numbers are periods of $S$. Therefore $S \equiv \mathbb{R}$.
(ii) Assume that, for each $a, b=f(a)$ is unique. Then for any $a_{1}$ and $a_{2}$,
$$
\begin{aligned}
x \in S & \Leftrightarrow a_{1} x+f\left(a_{1}\right) \in S \Leftrightarrow a_{1} a_{2} x+a_{2} f\left(a_{1}\right)+f\left(a_{2}\right) \in S \\
& \Leftrightarrow a_{2} x+f\left(a_{2}\right) \in S \Leftrightarrow a_{1} a_{2} x+a_{1} f\left(a_{2}\right)+f\left(a_{1}\right) \in S .
\end{aligned}
$$
As above it follows that $a_{1} f\left(a_{2}\right)+f\left(a_{1}\right)=a_{2} f\left(a_{1}\right)+f\left(a_{2}\right)$, or equivalently $f\left(a_{1}\right)\left(a_{2}-1\right)=f\left(a_{2}\right)\left(a_{1}-1\right)$. Hence (for some $\left.c\right), f(a)=c(a-1)$ for all $a$. Now $x \in S \Leftrightarrow a x+c(a-1) \in S$ actually means that $y-c \in S \Leftrightarrow a y-c \in S$ for all $a$. Then it is easy to conclude that $\{y-c \mid y \in S\}$ is either a half-line or the whole line, and so is $S$.
30. Let $a$ and $b$ be the integers written by $A$ and $B$ respectively, and let $xx-r_{n-1}, B$ would know that $a+b>x$ and hence $a+b=y$, while if $b0$, there exists an $m$ for which $s_{m}-r_{m}<0$, a contradiction.
### 4.33 Solutions to the Shortlisted Problems of IMO 1992
1. Assume that a pair $(x, y)$ with $xy$ is a positive integer). This will imply the desired result, since starting from the pair $(1,1)$ we can obtain arbitrarily many solutions. First, we show that $\operatorname{gcd}\left(x_{1}, y\right)=1$. Suppose to the contrary that $\operatorname{gcd}\left(x_{1}, y\right)$ $=d>1$. Then $d\left|x_{1}\right| y^{2}+m \Rightarrow d \mid m$, which implies $d|y| x^{2}+m \Rightarrow d \mid x$. But this last is impossible, since $\operatorname{gcd}(x, y)=1$. Thus it remains to show that $x_{1} \mid y^{2}+m$ and $y \mid x_{1}^{2}+m$. The former relation is obvious. Since $\operatorname{gcd}(x, y)=1$, the latter is equivalent to $y \mid\left(x x_{1}\right)^{2}+m x^{2}=y^{4}+2 m y^{2}+$ $m^{2}+m x^{2}$, which is true because $y \mid m\left(m+x^{2}\right)$ by the assumption. Hence $\left(y, x_{1}\right)$ indeed satisfies all the required conditions.
Remark. The original problem asked to prove the existence of a pair $(x, y)$ of positive integers satisfying the given conditions such that $x+y \leq m+1$. The problem in this formulation is trivial, since the pair $x=y=1$ satisfies the conditions. Moreover, this is sometimes the only solution with $x+y \leq m+1$. For example, for $m=3$ the least nontrivial solution is $\left(x_{0}, y_{0}\right)=(1,4)$.
2. Let us define $x_{n}$ inductively as $x_{n}=f\left(x_{n-1}\right)$, where $x_{0} \geq 0$ is a fixed real number. It follows from the given equation in $f$ that $x_{n+2}=-a x_{n+1}+$ $b(a+b) x_{n}$. The general solution to this equation is of the form
$$
x_{n}=\lambda_{1} b^{n}+\lambda_{2}(-a-b)^{n},
$$
where $\lambda_{1}, \lambda_{2} \in \mathbb{R}$ satisfy $x_{0}=\lambda_{1}+\lambda_{2}$ and $x_{1}=\lambda_{1} b-\lambda_{2}(a+b)$. In order to have $x_{n} \geq 0$ for all $n$ we must have $\lambda_{2}=0$. Hence $x_{0}=\lambda_{1}$ and $f\left(x_{0}\right)=x_{1}=\lambda_{1} b=b x_{0}$. Since $x_{0}$ was arbitrary, we conclude that $f(x)=b x$ is the only possible solution of the functional equation. It is easily verified that this is indeed a solution.
3. Consider two squares $A B^{\prime} C D^{\prime}$ and $A^{\prime} B C^{\prime} D$. Since $A C \perp B D$, these two squares are homothetic, which implies that the lines $A A^{\prime}, B B^{\prime}, C C^{\prime}, D D^{\prime}$ are concurrent at a certain point $O$. Since the rotation about $A$ by $90^{\circ}$ takes $\triangle A B K$ into $\triangle A F D$, it follows that $B K \perp D F$. Denote by $T$ the intersection of $B K$ and $D F$. The rotation about some point $X$ by $90^{\circ}$ maps $B K$ into $D F$ if and only if $T X$ bisects an angle between $B K$ and $D F$. Therefore $\angle F T A=$
$\angle A T K=45^{\circ}$. Moreover, the quadrilateral $B A^{\prime} D T$ is cyclic, which implies that $\angle B T A^{\prime}=B D A^{\prime}=45^{\circ}$ and consequently that the points $A, T, A^{\prime}$ are collinear. It follows that the
point $O$ lies on a bisector of $\angle B T D$ and therefore the rotation $\mathcal{R}$ about $O$ by $90^{\circ}$ takes $B K$ into $D F$. Analogously, $\mathcal{R}$ maps the lines $C E, D G, A I$ into $A H, B J, C L$. Hence the quadrilateral $P_{1} Q_{1} R_{1} S_{1}$ is the image of the quadrilateral $P_{2} Q_{2} R_{2} S_{2}$, and the result follows.
4. There are 36 possible edges in total. If not more than 3 edges are left undrawn, then we can choose 6 of the given 9 points no two of which are connected by an undrawn edge. These 6 points together with the edges between them form a two-colored complete graph, and thus by a wellknown result there exists at least one monochromatic triangle. It follows that $n \leq 33$.
In order to show that $n=33$, we shall give an example of a graph with 32 edges that does not contain a monochromatic triangle. Let us start with a complete graph $C_{5}$ with 5 vertices. Its edges can be colored in two colors so that there is no monochromatic triangle (Fig. 1). Furthermore, given a graph $\mathcal{H}$ with $k$ vertices without monochromatic triangles, we can add to it a new vertex, join it to all vertices of $\mathcal{H}$ except $A$, and color each edge $B X$ in the same way as $A X$. The obtained graph obviously contains no monochromatic triangles. Applying this construction four times to the graph $C_{5}$ we get an example like that of Fig. 2.
Fig. 1
Fig. 2
Second solution. For simplicity, we call the colors red and blue.
Let $r(k, l)$ be the least positive integer $r$ such that each complete $r$-graph whose edges are colored in red and blue contains either a complete red $k$-graph or a complete blue $l$-graph. Also, let $t(n, k)$ be the greatest possible number of edges in a graph with $n$ vertices that does not contain a complete $k$-graph. These numbers exist by the theorems of Ramsey and Turán.
Let us assume that $r(k, l)0$ for all $x>0$. It also follows that $f(x)<0$ for $x<0$. In other words, $f$ preserves sign.
Now setting $x>0$ and $y=-f(x)$ in the given functional equation we obtain
$$
f(x-f(x))=f\left(\sqrt{x}^{2}+f(-x)\right)=-x+f(\sqrt{x})^{2}=-(x-f(x)) .
$$
But since $f$ preserves sign, this implies that $f(x)=x$ for $x>0$. Moreover, since $f(-x)=-f(x)$, it follows that $f(x)=x$ for all $x$. It is easily verified that this is indeed a solution.
7. Let $G_{1}, G_{2}$ touch the chord $B C$ at $P, Q$ and touch the circle $G$ at $R, S$ respectively. Let $D$ be the midpoint of the complementary $\operatorname{arc} B C$ of $G$. The homothety centered at $R$ mapping $G_{1}$ onto $G$ also maps the line $B C$ onto a tangent of $G$ parallel to $B C$. It follows that this line touches $G$ at point $D$, which is therefore the image of $P$ under the homothety. Hence $R, P$, and $D$ are collinear. Since $\angle D B P=\angle D C B=\angle D R B$, it follows that $\triangle D B P \sim \triangle D R B$ and consequently that $D P \cdot D R=D B^{2}$. Similarly, points $S, Q, D$ are collinear and satisfy $D Q \cdot D S=D B^{2}=D P \cdot D R$.
Hence $D$ lies on the radical axis of the circles $G_{1}$ and $G_{2}$, i.e., on their common tangent $A W$, which also implies that $A W$ bisects the angle $B A D$. Furthermore, since $D B=D C=D W=\sqrt{D P \cdot D R}$, it follows from the lemma of (SL99-14) that $W$ is the incenter of $\triangle A B C$.
Remark. According to the third solution of (SL93-3), both $P W$ and $Q W$ contain the incenter of $\triangle A B C$, and the result is immediate. The problem can also be solved by inversion centered at $W$.
8. For simplicity, we shall write $n$ instead of 1992.
Lemma. There exists a tangent $n$-gon $A_{1} A_{2} \ldots A_{n}$ with sides $A_{1} A_{2}=a_{1}$, $A_{2} A_{3}=a_{2}, \ldots, A_{n} A_{1}=a_{n}$ if and only if the system
$$
x_{1}+x_{2}=a_{1}, x_{2}+x_{3}=a_{2}, \ldots, x_{n}+x_{1}=a_{n}
$$
has a solution $\left(x_{1}, \ldots, x_{n}\right)$ in positive reals.
Proof. Suppose that such an $n$-gon $A_{1} A_{2} \ldots A_{n}$ exists. Let the side $A_{i} A_{i+1}$ touch the inscribed circle at point $P_{i}$ (where $\left.A_{n+1}=A_{1}\right)$. Then $x_{1}=$ $A_{1} P_{n}=A_{1} P_{1}, x_{2}=A_{2} P_{1}=A_{2} P_{2}, \ldots, x_{n}=A_{n} P_{n-1}=A_{n} P_{n}$ is clearly a positive solution of (1).
Now suppose that the system (1) has a positive real solution $\left(x_{1}, \ldots\right.$, $x_{n}$ ). Let us draw a polygonal line $A_{1} A_{2} \ldots A_{n+1}$ touching a circle of radius $r$ at points $P_{1}, P_{2}, \ldots, P_{n}$ respectively such that $A_{1} P_{1}=$ $A_{n+1} P_{n}=x_{1}$ and $A_{i} P_{i}=A_{i} P_{i-1}=x_{i}$ for $i=2, \ldots, n$. Observe that $O A_{1}=O A_{n+1}=\sqrt{x_{1}^{2}+r^{2}}$ and the function $f(r)=\angle A_{1} O A_{2}+$ $\angle A_{2} O A_{3}+\cdots+\angle A_{n} O A_{n+1}=$ $2\left(\arctan \frac{x_{1}}{r}+\cdots+\arctan \frac{x_{n}}{r}\right)$ is continuous. Thus $A_{1} A_{2} \ldots A_{n+1}$ is a closed simple polygonal line if and only if $f(r)=360^{\circ}$. But such an $r$ exists, since $f(r) \rightarrow 0$
when $r \rightarrow \infty$ and $f(r) \rightarrow \infty$ when $r \rightarrow 0$. This proves the second direction of the lemma.
For $n=4 k$, the system (1) is solvable in positive reals if $a_{i}=i$ for $i \equiv 1,2$ $(\bmod 4), a_{i}=i+1$ for $i \equiv 3$ and $a_{i}=i-1$ for $i \equiv 0(\bmod 4)$. Indeed, one solution is given by $x_{i}=1 / 2$ for $i \equiv 1, x_{i}=3 / 2$ for $i \equiv 3$ and $x_{i}=i-3 / 2$ for $i \equiv 0,2(\bmod 4)$.
Remark. For $n=4 k+2$ there is no such $n$-gon. In fact, solvability of the system (1) implies $a_{1}+a_{3}+\cdots=a_{2}+a_{4}+\cdots$, while in the case $n=4 k+2$ the sum $a_{1}+a_{2}+\cdots+a_{n}$ is odd.
9. Since the equation $x^{3}-x-c=0$ has only one real root for every $c>$ $2 /(3 \sqrt{3}), \alpha$ is the unique real root of $x^{3}-x-33^{1992}=0$. Hence $f^{n}(\alpha)=$ $f(\alpha)=\alpha$.
Remark. Consider any irreducible polynomial $g(x)$ in the place of $x^{3}-$ $x-33^{1992}$. The problem amounts to proving that if $\alpha$ and $f(\alpha)$ are roots
of $g$, then any $f^{(n)}(\alpha)$ is also a root of $g$. In fact, since $g(f(x))$ vanishes at $x=\alpha$, it must be divisible by the minimal polynomial of $\alpha$, that is, $g(x)$. It follows by induction that $g\left(f^{(n)}(x)\right)$ is divisible by $g(x)$ for all $n \in \mathbb{N}$, and hence $g\left(f^{(n)}(\alpha)\right)=0$.
10. Let us set $S(x)=\{(y, z) \mid(x, y, z) \in V\}, S_{y}(x)=\left\{z \mid(x, z) \in S_{y}\right\}$ and $S_{z}(x)=\left\{y \mid(x, y) \in S_{z}\right\}$. Clearly $S(x) \subset S_{x}$ and $S(x) \subset S_{y}(x) \times S_{z}(x)$. It follows that
$$
\begin{aligned}
|V| & =\sum_{x}|S(x)| \leq \sum_{x} \sqrt{\left|S_{x}\right|\left|S_{y}(x)\right|\left|S_{z}(x)\right|} \\
& =\sqrt{\left|S_{x}\right|} \sum_{x} \sqrt{\left|S_{y}(x)\right|\left|S_{z}(x)\right|}
\end{aligned}
$$
Using the Cauchy-Schwarz inequality we also get
$$
\sum_{x} \sqrt{\left|S_{y}(x)\right|\left|S_{z}(x)\right|} \leq \sqrt{\sum_{x}\left|S_{y}(x)\right|} \sqrt{\sum_{x}\left|S_{z}(x)\right|}=\sqrt{\left|S_{y}\right|\left|S_{z}\right|}
$$
Now (1) and (2) together yield $|V| \leq \sqrt{\left|S_{x}\right|\left|S_{y}\right|\left|S_{z}\right|}$.
11. Let $I$ be the incenter of $\triangle A B C$. Since $90^{\circ}+\alpha / 2=\angle B I C=\angle D I E=$ $138^{\circ}$, we obtain that $\angle A=96^{\circ}$.
Let $D^{\prime}$ and $E^{\prime}$ be the points symmetric to $D$ and $E$ with respect to $C E$ and $B D$ respectively, and let $S$ be the intersection point of $E D^{\prime}$ and $B D$. Then $\angle B D E^{\prime}=24^{\circ}$ and $\angle D^{\prime} D E^{\prime}=\angle D^{\prime} D E-\angle E^{\prime} D E=24^{\circ}$, which means that $D E^{\prime}$ bisects the angle $S D D^{\prime}$. Moreover, $\angle E^{\prime} S B=\angle E S B=$ $\angle E D S+\angle D E S=60^{\circ}$ and hence $S E^{\prime}$ bisects the angle $D^{\prime} S B$. It follows that $E^{\prime}$ is the excenter of $\triangle D^{\prime} D S$ and consequently $\angle D^{\prime} D C=\angle D D^{\prime} C=$ $\angle S D^{\prime} E^{\prime}=\left(180^{\circ}-72^{\circ}\right) / 2=54^{\circ}$. Finally, $\angle C=180^{\circ}-2 \cdot 54^{\circ}=72^{\circ}$ and $\angle B=12^{\circ}$.
12. Let us set $\operatorname{deg} f=n$ and $\operatorname{deg} g=m$. We shall prove the result by induction on $n$. If $n1$. It is easy to see that $\alpha\left(n_{m}\right)=2^{m}-m$. On the other hand, squaring and simplifying yields $n_{m}^{2}=1+\sum_{i1$ be a constant to be chosen later, and let $N_{i}=2^{m_{i}} n_{i}-1$ where $m_{i}>\alpha\left(n_{i}\right)$ is such that $m_{i} / \alpha\left(n_{i}\right) \rightarrow \theta$ as $i \rightarrow \infty$. Then $\alpha\left(N_{i}\right)=\alpha\left(n_{i}\right)+m_{i}-1$, whereas $N_{i}^{2}=2^{2 m_{i}} n_{i}^{2}-2^{m_{i}+1} n_{i}+1$ and $\alpha\left(N_{i}^{2}\right)=\alpha\left(n_{i}^{2}\right)-\alpha\left(n_{i}\right)+m_{i}$. It follows that
$$
\lim _{i \rightarrow \infty} \frac{\alpha\left(N_{i}^{2}\right)}{\alpha\left(N_{i}\right)}=\lim _{i \rightarrow \infty} \frac{\alpha\left(n_{i}^{2}\right)+(\theta-1) \alpha\left(n_{i}\right)}{(1+\theta) \alpha\left(n_{i}\right)}=\frac{\theta-1}{\theta+1}
$$
which is equal to $\gamma \in[0,1]$ for $\theta=\frac{1+\gamma}{1-\gamma}$ (for $\gamma=1$ we set $m_{i} / \alpha\left(n_{i}\right) \rightarrow$ $\infty)$.
(3) Let be given a sequence $\left(n_{i}\right)_{i=1}^{\infty}$ with $\alpha\left(n_{i}^{2}\right) / \alpha\left(n_{i}\right) \rightarrow \gamma$. Taking $m_{i}>$ $\alpha\left(n_{i}\right)$ and $N_{i}=2^{m_{i}} n_{i}+1$ we easily find that $\alpha\left(N_{i}\right)=\alpha\left(n_{i}\right)+1$ and $\alpha\left(N_{i}^{2}\right)=\alpha\left(n_{i}^{2}\right)+\alpha\left(n_{i}\right)+1$. Hence $\alpha\left(N_{i}^{2}\right) / \alpha\left(N_{i}\right)=\gamma+1$. Continuing this procedure we can construct a sequence $t_{i}$ such that $\alpha\left(t_{i}^{2}\right) / \alpha\left(t_{i}\right)=$ $\gamma+k$ for an arbitrary $k \in \mathbb{N}$.
18. Let us define inductively $f^{1}(x)=f(x)=\frac{1}{x+1}$ and $f^{n}(x)=f\left(f^{n-1}(x)\right)$, and let $g_{n}(x)=x+f(x)+f^{2}(x)+\cdots+f^{n}(x)$. We shall prove first the following statement.
Lemma. The function $g_{n}(x)$ is strictly increasing on $[0,1]$, and $g_{n-1}(1)=$ $F_{1} / F_{2}+F_{2} / F_{3}+\cdots+F_{n} / F_{n+1}$.
Proof. Since $f(x)-f(y)=\frac{y-x}{(1+x)(1+y)}$ is smaller in absolute value than $x-y$, it follows that $x>y$ implies $f^{2 k}(x)>f^{2 k}(y)$ and $f^{2 k+1}(x)<$ $f^{2 k+1}(y)$, and moreover that for every integer $k \geq 0$,
$$
\left[f^{2 k}(x)-f^{2 k}(y)\right]+\left[f^{2 k+1}(x)-f^{2 k+1}(y)\right]>0
$$
Hence if $x>y$, we have $g_{n}(x)-g_{n}(y)=(x-y)+[f(x)-f(y)]+\cdots+$ $\left[f^{n}(x)-f^{n}(y)\right]>0$, which yields the first part of the lemma.
The second part follows by simple induction, since $f^{k}(1)=F_{k+1} / F_{k+2}$. If some $x_{i}=0$ and consequently $x_{j}=0$ for all $j \geq i$, then the problem reduces to the problem with $i-1$ instead of $n$. Thus we may assume that all $x_{1}, \ldots, x_{n}$ are different from 0 . If we write $a_{i}=\left[1 / x_{i}\right]$, then $x_{i}=\frac{1}{a_{i}+x_{i+1}}$. Thus we can regard $x_{i}$ as functions of $x_{n}$ depending on $a_{1}, \ldots, a_{n-1}$.
Suppose that $x_{n}, a_{n-1}, \ldots, a_{3}, a_{2}$ are fixed. Then $x_{2}, x_{3}, \ldots, x_{n}$ are all fixed, and $x_{1}=\frac{1}{a_{1}+x_{2}}$ is maximal when $a_{1}=1$. Hence the sum $S=$ $x_{1}+x_{2}+\cdots+x_{n}$ is maximized for $a_{1}=1$.
We shall show by induction on $i$ that $S$ is maximized for $a_{1}=a_{2}=\cdots=$ $a_{i}=1$. In fact, assuming that the statement holds for $i-1$ and thus $a_{1}=$ $\cdots=a_{i-1}=1$, having $x_{n}, a_{n-1}, \ldots, a_{i+1}$ fixed we have that $x_{n}, \ldots, x_{i+1}$ are also fixed, and that $x_{i-1}=f\left(x_{i}\right), \ldots, x_{1}=f^{i-1}\left(x_{i}\right)$. Hence by the lemma, $S=g_{i-1}\left(x_{i}\right)+x_{i+1}+\cdots+x_{n}$ is maximal when $x_{i}=\frac{1}{a_{i}+x_{i+1}}$ is maximal, that is, for $a_{i}=1$. Thus the induction is complete.
It follows that $x_{1}+\cdots+x_{n}$ is maximal when $a_{1}=\cdots=a_{n-1}=1$, so that $x_{1}+\cdots+x_{n}=g_{n-1}\left(x_{1}\right)$. By the lemma, the latter does not exceed $g_{n-1}(1)$. This completes the proof.
Remark. The upper bound is the best possible, because it is approached by taking $x_{n}$ close to 1 and inductively (in reverse) defining $x_{i-1}=\frac{1}{1+x_{i}}=$ $\frac{1}{a_{i}+x_{i}}$.
19. Observe that $f(x)=\left(x^{4}+2 x^{2}+3\right)^{2}-8\left(x^{2}-1\right)^{2}=\left[x^{4}+2(1-\sqrt{2}) x^{2}+\right.$ $3+2 \sqrt{2}]\left[x^{4}+2(1+\sqrt{2}) x^{2}+3-2 \sqrt{2}\right]$. Now it is easy to find that the roots of $f$ are
$$
x_{1,2,3,4}= \pm i(i \sqrt[4]{2} \pm 1) \quad \text { and } \quad x_{5,6,7,8}= \pm i(\sqrt[4]{2} \pm 1)
$$
In other words, $x_{k}=\alpha_{i}+\beta_{j}$, where $\alpha_{i}^{2}=-1$ and $\beta_{j}^{4}=2$.
We claim that any root of $f$ can be obtained from any other using rational functions. In fact, we have
$$
\begin{aligned}
& x^{3}=-\alpha_{i}-3 \beta_{j}+3 \alpha_{i} \beta_{j}^{2}+\beta_{j}^{3} \\
& x^{5}=11 \alpha_{i}+7 \beta_{j}-10 \alpha_{i} \beta_{j}^{2}-10 \beta_{j}^{3} \\
& x^{7}=-71 \alpha_{i}-49 \beta_{j}+35 \alpha_{i} \beta_{j}^{2}+37 \beta_{j}^{3}
\end{aligned}
$$
from which we easily obtain that
$\alpha_{i}=24^{-1}\left(127 x+5 x^{3}+19 x^{5}+5 x^{7}\right), \quad \beta_{j}=24^{-1}\left(151 x+5 x^{3}+19 x^{5}+5 x^{7}\right)$.
Since all other values of $\alpha$ and $\beta$ can be obtained as rational functions of $\alpha_{i}$ and $\beta_{j}$, it follows that all the roots $x_{l}$ are rational functions of a particular root $x_{k}$.
We now note that if $x_{1}$ is an integer such that $f\left(x_{1}\right)$ is divisible by $p$, then $p>3$ and $x_{1} \in \mathbb{Z}_{p}$ is a root of the polynomial $f$. By the previous consideration, all remaining roots $x_{2}, \ldots, x_{8}$ of $f$ over the field $\mathbb{Z}_{p}$ are rational functions of $x_{1}$, since 24 is invertible in $Z_{p}$. Then $f(x)$ factors as
$$
f(x)=\left(x-x_{1}\right)\left(x-x_{2}\right) \cdots\left(x-x_{8}\right),
$$
and the result follows.
20. Denote by $U$ the point of tangency of the circle $C$ and the line $l$. Let $X$ and $U^{\prime}$ be the points symmetric to $U$ with respect to $S$ and $M$ respectively; these points do not depend on the choice of $P$. Also, let $C^{\prime}$ be the excircle of $\triangle P Q R$ corresponding to $P, S^{\prime}$ the center of $C^{\prime}$, and $W, W^{\prime}$ the points of tangency of $C$ and $C^{\prime}$ with the line $P Q$ respectively. Obviously, $\triangle W S P \sim \triangle W^{\prime} S^{\prime} P$. Since $S X \| S^{\prime} U^{\prime}$ and $S X: S^{\prime} U^{\prime}=$ $S W: S^{\prime} W^{\prime}=S P: S^{\prime} P$, we deduce that $\Delta S X P \sim \Delta S^{\prime} U^{\prime} P$, and consequently that $P$ lies on the line $X U^{\prime}$. On the other hand, it is easy to show that each point $P$ of the ray $U^{\prime} X$ over $X$ satisfies the required condition. Thus the desired locus is
the extension of $U^{\prime} X$ over $X$.
21. (a) Representing $n^{2}$ as a sum of $n^{2}-13$ squares is equivalent to representing 13 as a sum of numbers of the form $x^{2}-1, x \in \mathbb{N}$, such as $0,3,8,15, \ldots$ But it is easy to check that this is impossible, and hence $s(n) \leq n^{2}-14$.
(b) Let us prove that $s(13)=13^{2}-14=155$. Observe that
$$
\begin{aligned}
13^{2} & =8^{2}+8^{2}+4^{2}+4^{2}+3^{2} \\
& =8^{2}+8^{2}+4^{2}+4^{2}+2^{2}+2^{2}+1^{2} \\
& =8^{2}+8^{2}+4^{2}+3^{2}+3^{2}+2^{2}+1^{2}+1^{2}+1^{2}
\end{aligned}
$$
Given any representation of $n^{2}$ as a sum of $m$ squares one of which is even, we can construct a representation as a sum of $m+3$ squares by dividing the odd square into four equal squares. Thus the first equality enables us to construct representations with $5,8,11, \ldots, 155$ squares, the second to construct ones with $7,10,13, \ldots, 154$ squares, and the
third with $9,12, \ldots, 153$ squares. It remains only to represent $13^{2}$ as a sum of $k=2,3,4,6$ squares. This can be done as follows:
$$
\begin{aligned}
13^{2} & =12^{2}+5^{2}=12^{2}+4^{2}+3^{2} \\
& =11^{2}+4^{2}+4^{2}+4^{2}=12^{2}+3^{2}+2^{2}+2^{2}+2^{2}+2^{2}
\end{aligned}
$$
(c) We shall prove that whenever $s(n)=n^{2}-14$ for some $n \geq 13$, it also holds that $s(2 n)=(2 n)^{2}-14$. This will imply that $s(n)=n^{2}-14$ for any $n=2^{t} \cdot 13$.
If $n^{2}=x_{1}^{2}+\cdots+x_{r}^{2}$, then we have $(2 n)^{2}=\left(2 x_{1}\right)^{2}+\cdots+\left(2 x_{r}\right)^{2}$. Replacing $\left(2 x_{i}\right)^{2}$ with $x_{i}^{2}+x_{i}^{2}+x_{i}^{2}+x_{i}^{2}$ as long as it is possible we can obtain representations of $(2 n)^{2}$ consisting of $r, r+3, \ldots, 4 r$ squares. This gives representations of $(2 n)^{2}$ into $k$ squares for any $k \leq 4 n^{2}-62$. Further, we observe that each number $m \geq 14$ can be written as a sum of $k \geq m$ numbers of the form $x^{2}-1, x \in \mathbb{N}$, which is easy to verify. Therefore if $k \leq 4 n^{2}-14$, it follows that $4 n^{2}-k$ is a sum of $k$ numbers of the form $x^{2}-1$ (since $k \geq 4 n^{2}-k \geq 14$ ), and consequently $4 n^{2}$ is a sum of $k$ squares.
Remark. One can find exactly the value of $f(n)$ for each $n$ :
$$
f(n)= \begin{cases}1, & \text { if } n \text { has a prime divisor congruent to } 3 \bmod 4 \\ 2, & \text { if } n \text { is of the form } 5 \cdot 2^{k}, k \text { a positive integer; } \\ n^{2}-14, & \text { otherwise. }\end{cases}
$$
### 4.34 Solutions to the Shortlisted Problems of IMO 1993
1. First we notice that for a rational point $O$ (i.e., with rational coordinates), there exist 1993 rational points in each quadrant of the unit circle centered at $O$. In fact, it suffices to take
$$
X=\left\{\left.O+\left( \pm \frac{t^{2}-1}{t^{2}+1}, \pm \frac{2 t}{t^{2}+1}\right) \right\rvert\, t=1,2, \ldots, 1993\right\}
$$
Now consider the set $A=\{(i / q, j / q) \mid i, j=0,1, \ldots, 2 q\}$, where $q=$ $\prod_{i=1}^{1993}\left(t^{2}+1\right)$. We claim that $A$ gives a solution for the problem. Indeed, for any $P \in A$ there is a quarter of the unit circle centered at $P$ that is contained in the square $[0,2] \times[0,2]$. As explained above, there are 1993 rational points on this quarter circle, and by definition of $q$ they all belong to $A$.
Remark. Substantially the same problem was proposed by Bulgaria for IMO 71: see (SL71-2), where we give another possible construction of a set $A$.
2. It is well known that $r \leq \frac{1}{2} R$. Therefore $\frac{1}{3}(1+r)^{2} \leq \frac{1}{3}\left(1+\frac{1}{2}\right)^{2}=\frac{3}{4}$.
It remains only to show that $p \leq \frac{1}{4}$. We note that $p$ does not exceed one half of the circumradius of $\triangle A^{\prime} B^{\prime} C^{\prime}$. However, by the theorem on the nine-point circle, this circumradius is equal to $\frac{1}{2} R$, and the conclusion follows.
Second solution. By a well-known relation we have $\cos A+\cos B+\cos C=$ $1+\frac{r}{R}(=1+r$ when $R=1)$. Next, recalling that the incenter of $\triangle A^{\prime} B^{\prime} C^{\prime}$ is at the orthocenter of $\triangle A B C$, we easily obtain $p=2 \cos A \cos B \cos C$. Cosines of angles of a triangle satisfy the identity $\cos ^{2} A+\cos ^{2} B+\cos ^{2} C+$ $2 \cos A \cos B \cos C=1$ (the proof is straightforward: see (SL81-11)). Thus
$$
\begin{aligned}
p+\frac{1}{3}(1+r)^{2} & =2 \cos A \cos B \cos C+\frac{1}{3}(\cos A+\cos B+\cos C)^{2} \\
& \leq 2 \cos A \cos B \cos C+\cos ^{2} A+\cos ^{2} B+\cos ^{2} C=1
\end{aligned}
$$
3. Let $O_{1}$ and $\rho$ be the center and radius of $k_{c}$. It is clear that $C, I, O_{1}$ are collinear and $C I / C O_{1}=r / \rho$. By Stewart's theorem applied to $\triangle O C O_{1}$,
$$
O I^{2}=\frac{r}{\rho} O O_{1}^{2}+\left(1-\frac{r}{\rho}\right) O C^{2}-C I \cdot I O_{1} .
$$
Since $O O_{1}=R-\rho, O C=R$ and by Euler's formula $O I^{2}=R^{2}-2 R r$, substituting these values in (1) gives $C I \cdot I O_{1}=r \rho$, or equivalently $C O_{1}$. $I O_{1}=\rho^{2}=D O_{1}^{2}$. Hence the triangles $C O_{1} D$ and $D O_{1} I$ are similar, implying $\angle D I O_{1}=90^{\circ}$. Since $C D=C E$ and the line $C O_{1}$ bisects the segment $D E$, it follows that $I$ is the midpoint of $D E$.
Second solution. Under the inversion with center $C$ and power $a b, k_{c}$ is transformed into the excircle of $\widehat{A} \widehat{B} C$ corresponding to $C$. Thus $C D=$
$\frac{a b}{s}$, where $s$ is the common semiperimeter of $\triangle A B C$ and $\triangle \widehat{A} \widehat{B} C$, and consequently the distance from $D$ to $B C$ is $\frac{a b}{s} \sin C=\frac{2 S_{A B C}}{s}=2 r$. The statement follows immediately.
Third solution. We shall prove a stronger statement: Let $A B C D$ be a convex quadrilateral inscribed in a circle $k$, and $k^{\prime}$ the circle that is tangent to segments $B O, A O$ at $K, L$ respectively (where $O=B D \cap A C$ ), and internally to $k$ at $M$. Then $K L$ contains the incenters $I, J$ of $\triangle A B C$ and $\triangle A B D$.
Let $K^{\prime}, K^{\prime \prime}, L^{\prime}, L^{\prime \prime}, N$ denote the midpoints of arcs $B C, B D, A C, A D, A B$ that don't contain $M ; X^{\prime}, X^{\prime \prime}$ the points on $k$ defined by $X^{\prime} N=N X^{\prime \prime}=$ $K^{\prime} K^{\prime \prime}=L^{\prime} L^{\prime \prime}$ (as oriented arcs); and set $S=A K^{\prime} \cap B L^{\prime \prime}, \bar{M}=N S \cap k$, $\bar{K}=K^{\prime \prime} M \cap B O, \bar{L}=L^{\prime} M \cap A O$.
It is clear that $I=A K^{\prime} \cap B L^{\prime}, J=A K^{\prime \prime} \cap B L^{\prime \prime}$. Furthermore, $X^{\prime} \bar{M}$ contains $I$ (to see this, use the fact that for $A, B, C, D, E, F$ on $k$, lines $A D, B E, C F$ are concurrent if and only if $A B \cdot C D \cdot E F=B C \cdot D E \cdot F A$, and then express $A \bar{M} / \bar{M} B$ by applying this rule to $A M B K^{\prime} N L^{\prime \prime}$ and show that $A K^{\prime}, \bar{M} X^{\prime}, B L^{\prime}$ are concurrent).
Analogously, $X^{\prime \prime} \bar{M}$ contains $J$. Now the points $B, \bar{K}, I, S, \bar{M}$ lie on a circle $(\angle B \overline{K M}=\angle B I \bar{M}=\angle B S \bar{M})$, and points $A, \bar{L}, J, S, \bar{M}$ do so as well. Lines $I \bar{K}, J \bar{L}$ are parallel to $K^{\prime \prime} L^{\prime}$ (because $\angle \overline{M K} I=\angle \bar{M} B I=$ $\left.\angle \bar{M} K^{\prime \prime} L^{\prime}\right)$. On the other hand, the quadrilateral $A B I J$ is cyclic, and simple calculation with angles shows that $I J$ is also parallel to $K^{\prime \prime} L^{\prime}$. Hence $\bar{K}, I, J, \bar{L}$ are collinear.
Finally, $\bar{K} \equiv K, \bar{L} \equiv L$, and $\bar{M} \equiv M$ because the homothety centered at $M$ that maps $k^{\prime}$ to $k$ sends $K$ to $K^{\prime \prime}$ and $L$ to $L^{\prime}$ (thus $M, K, K^{\prime \prime}$, as well as $M, L, L^{\prime}$, must be collinear). As is seen now, the deciphered picture yields many other interesting properties. Thus, for example, $N, S, M$ are collinear, i.e., $\angle A M S=\angle B M S$.
Fourth solution. We give an alternative proof of the more general statement in the third solution. Let $W$ be the foot of the perpendicular from $B$ to $A C$. We define $q=C W, h=B W, t=O L=O K, x=A L$, $\theta=\measuredangle W B O(\theta$ is negative if $\mathcal{B}(O, W, A), \theta=0$ if $W=O)$, and as usual, $a=B C, b=A C, c=A B$. Let $\alpha=\measuredangle K L C$ and $\beta=\measuredangle I L C$ (both angles must be acute). Our goal is to prove $\alpha=\beta$. We note that $90^{\circ}-\theta=2 \alpha$. One easily gets
$$
\tan \alpha=\frac{\cos \theta}{1+\sin \theta}, \quad \tan \beta=\frac{\frac{2 S_{A B C}}{a+b+c}}{\frac{b+c-a}{2}-x}
$$
Applying Casey's theorem to $A, B, C, k^{\prime}$, we get $A C \cdot B K+A L \cdot B C=$ $A B \cdot C L$, i.e., $b\left(\frac{h}{\cos \theta}-t\right)+x a=c(b-x)$. Using that $t=b-x-q-h \tan \theta$ we get
$$
x=\frac{b(b+c-q)-b h\left(\frac{1}{\cos \theta}+\tan \theta\right)}{a+b+c} .
$$
Plugging (2) into the second equation of (1) and using $b h=2 S_{A B C}$ and $c^{2}=b^{2}+a^{2}-2 b q$, we obtain $\tan \alpha=\tan \beta$, i.e., $\alpha=\beta$, which completes our proof.
4. Let $h$ be the altitude from $A$ and $\varphi=\angle B A D$. We have $B M=\frac{1}{2}(B D+$ $A B-A D)$ and $M D=\frac{1}{2}(B D-A B+A D)$, so
$$
\begin{aligned}
\frac{1}{M B}+\frac{1}{M D} & =\frac{B D}{M B \cdot M D}=\frac{4 B D}{B D^{2}-A B^{2}-A D^{2}+2 A B \cdot A D} \\
& =\frac{4 B D}{2 A B \cdot A D(1-\cos \varphi)}=\frac{2 B D \sin \varphi}{2 S_{A B D}(1-\cos \varphi)} \\
& =\frac{2 B D \sin \varphi}{B D \cdot h(1-\cos \varphi)}=\frac{2}{h \tan \frac{\varphi}{2}} .
\end{aligned}
$$
It follows that $\frac{1}{M B}+\frac{1}{M D}$ depends only on $h$ and $\varphi$. Specially, $\frac{1}{N C}+\frac{1}{N E}=$ $\frac{2}{h \tan (\varphi / 2)}$ as well.
5. For $n=1$ the game is trivially over. If $n=2$, it can end, for example, in the following way:
Fig. 1
The sequence of moves shown in Fig. 2 enables us to remove three pieces placed in a $1 \times 3$ rectangle, using one more piece and one more free cell. In that way, for any $n \geq 4$ we can reduce an $(n+3) \times(n+3)$ square to an $n \times n$ square (Fig. 3). Therefore the game can end for every $n$ that is not divisible by 3 .
Fig. 2
Fig. 3
Suppose now that one can play the game on a $3 k \times 3 k$ square so that at the end only one piece remains. Denote the cells by $(i, j), i, j \in\{1, \ldots, 3 k\}$, and let $S_{0}, S_{1}, S_{2}$ denote the numbers of pieces on those squares $(i, j)$ for
which $i+j$ gives remainder $0,1,2$ respectively upon division by 3 . Initially $S_{0}=S_{1}=S_{2}=3 k^{2}$. After each move, two of $S_{0}, S_{1}, S_{2}$ diminish and one increases by one. Thus each move reverses the parity of the $S_{i}$ 's, so that $S_{0}, S_{1}, S_{2}$ are always of the same parity. But in the final position one of the $S_{i}$ 's must be equal to 1 and the other two must be 0 , which is impossible.
6. Notice that for $\alpha=\frac{1+\sqrt{5}}{2}, \alpha^{2} n=\alpha n+n$ for all $n \in \mathbb{N}$. We shall show that $f(n)=\left[\alpha n+\frac{1}{2}\right]$ (the closest integer to $\alpha n$ ) satisfies the requirements. Observe that $f$ is strictly increasing and $f(1)=2$. By the definition of $f$, $|f(n)-\alpha n| \leq \frac{1}{2}$ and $f(f(n))-f(n)-n$ is an integer. On the other hand,
$$
\begin{aligned}
|f(f(n))-f(n)-n| & =\left|f(f(n))-f(n)-\alpha^{2} n+\alpha n\right| \\
& =\left|f(f(n))-\alpha f(n)+\alpha f(n)-\alpha^{2} n-f(n)+\alpha n\right| \\
& =|(\alpha-1)(f(n)-\alpha n)+(f(f(n))-\alpha f(n))| \\
& \leq(\alpha-1)|f(n)-\alpha n|+|f(f(n))-\alpha f(n)| \\
& \leq \frac{1}{2}(\alpha-1)+\frac{1}{2}=\frac{1}{2} \alpha<1,
\end{aligned}
$$
which implies that $f(f(n))-f(n)-n=0$.
7. Multiplying by $a$ and $c$ the equation
$$
a x^{2}+2 b x y+c y^{2}=P^{k} n
$$
gives $(a x+b y)^{2}+P y^{2}=a P^{k} n$ and $(b x+c y)^{2}+P x^{2}=c P^{k} n$.
It follows immediately that $M(n)$ is finite; moreover, $(a x+b y)^{2}$ and $(b x+$ $c y)^{2}$ are divisible by $P$, and consequently $a x+b y, b x+c y$ are divisible by $P$ because $P$ is not divisible by a square greater than 1 . Thus there exist integers $X, Y$ such that $b x+c y=P X, a x+b y=-P Y$. Then $x=-b X-c Y$ and $y=a X+b Y$. Introducing these values into (1) and simplifying the expression obtained we get
$$
a X^{2}+2 b X Y+c Y^{2}=P^{k-1} n
$$
Hence $(x, y) \mapsto(X, Y)$ is a bijective correspondence between integral solutions of (1) and (2), so that $M\left(P^{k} n\right)=M\left(P^{k-1} n\right)=\cdots=M(n)$.
8. Suppose that $f(n)=1$ for some $n>0$. Then $f(n+1)=n+2, f(n+$ $2)=2 n+4, f(n+3)=n+1, f(n+4)=2 n+5, f(n+5)=n$, and so by induction $f(n+2 k)=2 n+3+k, f(n+2 k-1)=n+3-k$ for $k=1,2, \ldots, n+2$. Particularly, $n^{\prime}=3 n+3$ is the smallest value greater than $n$ for which $f\left(n^{\prime}\right)=1$. It follows that all numbers $n$ with $f(n)=1$ are given by $n=b_{i}$, where $b_{0}=1, b_{n}=3 b_{n-1}+3$. Furthermore, $b_{n}=3+3 b_{n-1}=3+3^{2}+3^{2} b_{n-2}=\cdots=3+3^{2}+\cdots+3^{n}+3^{n}=$ $=\frac{1}{2}\left(5 \cdot 3^{n}-3\right)$.
It is seen from above that if $n \leq b_{i}$, then $f(n) \leq f\left(b_{i}-1\right)=b_{i}+1$. Hence if $f(n)=1993$, then $n \geq b_{i} \geq 1992$ for some $i$. The smallest such $b_{i}$ is
$b_{7}=5466$, and $f\left(b_{i}+2 k-1\right)=b_{i}+3-k=1993$ implies $k=3476$. Thus the least integer in $S$ is $n_{1}=5466+2 \cdot 3476-1=12417$.
All the elements of $S$ are given by $n_{i}=b_{i+6}+2 k-1$, where $b_{i+6}+3-k=$ 1993, i.e., $k=b_{i+6}-1990$. Therefore $n_{i}=3 b_{i+6}-3981=\frac{1}{2}\left(5 \cdot 3^{i+7}-7971\right)$. Clearly $S$ is infinite and $\lim _{i \rightarrow \infty} \frac{n_{i+1}}{n_{i}}=3$.
9. We shall first complete the "multiplication table" for the sets $A, B, C$. It is clear that this multiplication is commutative and associative, so that we have the following relations:
$$
\begin{aligned}
& A C=(A B) B=B B=C \\
& A^{2}=A A=(A B) C=B C=A \\
& C^{2}=C C=B(B C)=B A=B
\end{aligned}
$$
(a) Now put 1 in $A$ and distribute the primes arbitrarily in $A, B, C$. This distribution uniquely determines the partition of $\mathbb{Q}^{+}$with the stated property. Indeed, if an arbitrary rational number
$$
x=p_{1}^{\alpha_{1}} \cdots p_{k}^{\alpha_{k}} q_{1}^{\beta_{1}} \cdots q_{l}^{\beta_{l}} r_{1}^{\gamma_{1}} \cdots r_{m}^{\gamma_{m}}
$$
is given, where $p_{i} \in A, q_{i} \in B, r_{i} \in C$ are primes, it is easy to see that $x$ belongs to $A, B$, or $C$ according as $\beta_{1}+\cdots+\beta_{l}+2 \gamma_{1}+\cdots+2 \gamma_{m}$ is congruent to 0,1 , or $2(\bmod 3)$.
(b) In every such partition, cubes all belong to $A$. In fact, $A^{3}=A^{2} A=$ $A A=A, B^{3}=B^{2} B=C B=A, C^{3}=C^{2} C=B C=A$.
(c) By (b) we have $1,8,27 \in A$. Then $2 \notin A$, and since the problem is symmetric with respect to $B, C$, we can assume $2 \in B$ and consequently $4 \in C$. Also $7 \notin A$, and also $7 \notin B$ (otherwise, $28=4 \cdot 7 \in A$ and $27 \in A$ ), so $7 \in C, 14 \in A, 28 \in B$. Further, we see that $3 \notin A$ (since otherwise $9 \in A$ and $8 \in A$ ). Put 3 in $C$. Then $5 \notin B$ (otherwise $15 \in A$ and $14 \in A$ ), so let $5 \in C$ too. Consequently $6,10 \in A$. Also $13 \notin A$, and $13 \notin C$ because $26 \notin A$, so $13 \in B$. Now it is easy to distribute the remaining primes $11,17,19,23,29,31$ : one possibility is
$$
\begin{aligned}
& A=\{1,6,8,10,14,19,23,27,29,31,33, \ldots\}, \\
& C=\{3,4,5,7,18,22,24,26,30,32,34, \ldots\} \\
& B=\{2,9,11,12,13,15,16,17,20,21,25,28,35, \ldots\}
\end{aligned}
$$
Remark. It can be proved that $\min \{n \in \mathbb{N} \mid n \in A, n+1 \in A\} \leq 77$.
10. (a) Let $n=p$ be a prime and let $p \mid a^{p}-1$. By Fermat's theorem $p \mid$ $a^{p-1}-1$, so that $p \mid a^{\operatorname{gcd}(p, p-1)}-1=a-1$, i.e., $a \equiv 1(\bmod p)$. Since then $a^{i} \equiv 1(\bmod p)$, we obtain $p \mid a^{p-1}+\cdots+a+1$ and hence $p^{2} \mid a^{p}-1=(a-1)\left(a^{p-1}+\cdots+a+1\right)$.
(b) Let $n=p_{1} \cdots p_{k}$ be a product of distinct primes and let $n \mid a^{n}-1$. Then from $p_{i} \mid a^{n}-1=\left(a^{\left(n / p_{i}\right)}\right)^{p_{i}}-1$ and part (a) we conclude that $p_{i}^{2} \mid a^{n}-1$. Since this is true for all indices $i$, we also have $n^{2} \mid a^{n}-1$; hence $n$ has the property $P$.
11. Due to the extended Eisenstein criterion, $f$ must have an irreducible factor of degree not less than $n-1$. Since $f$ has no integral zeros, it must be irreducible.
Second solution. The proposer's solution was as follows. Suppose that $f(x)=g(x) h(x)$, where $g, h$ are nonconstant polynomials with integer coefficients. Since $|f(0)|=3$, either $|g(0)|=1$ or $|h(0)|=1$. We may assume $|g(0)|=1$ and that $g(x)=\left(x-\alpha_{1}\right) \cdots\left(x-\alpha_{k}\right)$. Then $\left|\alpha_{1} \cdots \alpha_{k}\right|=$ 1. Since $\alpha_{i}^{n-1}\left(\alpha_{i}+5\right)=-3$, taking the product over $i=1,2, \ldots, k$ yields $\left|\left(\alpha_{1}+5\right) \cdots\left(\alpha_{k}+5\right)\right|=|g(-5)|=3^{k}$. But $f(-5)=g(-5) h(-5)=3$, so the only possibility is $\operatorname{deg} g=k=1$. This is impossible, because $f$ has no integral zeros.
Remark. Generalizing this solution, it can be shown that if $a, m, n$ are positive integers and $p1$, let $S_{N}=\{(m, n) \in S \mid m+n=N\}$. If $f(m, n)=\left(m_{1}, n_{1}\right)$, then $m_{1}+n_{1}=m+n$ with $m_{1}$ odd and $m_{1} \leq \frac{n}{2}<$ $\frac{N}{2}n_{2}>\cdots>n_{s}$.
First, suppose that $n_{i} \equiv n_{j}(\bmod n), i \neq j$. Consider the number
$$
B=A-a_{i} b^{n_{i}}-a_{j} b^{n_{j}}+\left(a_{i}+a_{j}\right) b^{n_{j}+k n}
$$
with $k$ chosen large enough so that $n_{j}+k n>n_{1}$ : this number is divisible by $b^{n}-1$ as well. But if $a_{i}+a_{j}-\frac{c_{i}}{n}$ implies
$$
\sum_{i=1}^{n} q_{i}>-\sum_{i=1}^{n} \frac{c_{i}}{n} \geq-1
$$
which leads to $\sum_{i=1}^{n} q_{i} \geq 0$. The proof is complete.
21. Assume that $S$ is a circle with center $O$ that cuts $S_{i}$ diametrically in points $P_{i}, Q_{i}, i \in\{A, B, C\}$, and denote by $r_{i}, r$ the radii of $S_{i}$ and $S$ respectively. Since $O A$ is perpendicular to $P_{A} Q_{A}$, it follows by Pythagoras's theorem that $O A^{2}+A P_{A}^{2}=O P_{A}^{2}$, i.e., $r_{A}^{2}+O A^{2}=r^{2}$. Analogously $r_{B}^{2}+O B^{2}=r^{2}$ and $r_{C}^{2}+O C^{2}=r^{2}$. Thus if $O_{A}, O_{B}, O_{C}$ are the feet of perpendiculars from $O$ to $B C, C A, A B$ respectively, then $O_{C} A^{2}-O_{C} B^{2}=r_{B}^{2}-r_{A}^{2}$. Since the left-hand side is a monotonic function of $O_{C} \in A B$, the point $O_{C}$ is uniquely determined by the imposed conditions. The same holds for $O_{A}$ and $O_{B}$.
If $A, B, C$ are not collinear, then the positions of $O_{A}, O_{B}, O_{C}$ uniquely determine the point $O$, and therefore the circle $S$ also. On the other hand, if $A, B, C$ are collinear, all one can deduce is that $O$ lies on the lines $l_{A}, l_{B}, l_{C}$ through $O_{A}, O_{B}, O_{C}$, perpendicular to $B C, C A, A B$ respectively. By this, $l_{A}, l_{B}, l_{C}$ are parallel, so $O$ can be either anywhere on the line if these lines coincide, or
nowhere if they don't coincide. So if there exists more than one circle $S$, $A, B, C$ lie on a line and the foot $O^{\prime}$ of the perpendicular from $O$ to the line $A B C$ is fixed. If $X, Y$ are the intersection points of $S$ and the line $A B C$, then $r^{2}=O X^{2}=O A^{2}+r_{A}^{2}$ and consequently $O^{\prime} X^{2}=O^{\prime} A^{2}+r_{A}^{2}$, which implies that $X, Y$ are fixed.
22. Let $M$ be the point inside $\angle A D B$ that satisfies $D M=D B$ and $D M \perp$
$D B$. Then $\angle A D M=\angle A C B$ and $A D / D M=A C / C B$. It follows that the triangles $A D M, A C B$ are similar; hence $\angle C A D=\angle B A M$ (because $\angle C A B=\angle D A M$ ) and $A B / A M=A C / A D$. Consequently the triangles $C A D, B A M$ are similar and therefore $\frac{A C}{A B}=\frac{C D}{B M}=$ $\frac{C D}{\sqrt{2} B D}$. Hence $\frac{A B \cdot C D}{A C \cdot B D}=\sqrt{2}$.
Let $C T, C U$ be the tangents at $C$ to the circles $A C D, B C D$ respectively. Then (in oriented angles) $\angle T C U=\angle T C D+\angle D C U=\angle C A D+\angle C B D=$ $90^{\circ}$, as required.
Second solution to the first part. Denote by $E, F, G$ the feet of the perpendiculars from $D$ to $B C, C A, A B$. Consider the pedal triangle $E F G$. Since $F G=A D \sin \angle A$, from the sine theorem we have $F G: G E: E F=$ $(C D \cdot A B):(B D \cdot A C):(A D \cdot B C)$. Thus $E G=F G$. On the other hand, $\angle E G F=\angle E G D+\angle D G F=\angle C B D+\angle C A D=90^{\circ}$ implies that $E F: E G=\sqrt{2}: 1$; hence the required ratio is $\sqrt{2}$.
Third solution to the first part. Under inversion centered at $C$ and with power $r^{2}=C A \cdot C B$, the triangle $D A B$ maps into a right-angled isosceles triangle $D^{*} A^{*} B^{*}$, where
$$
D^{*} A^{*}=\frac{A D \cdot B C}{C D}, D^{*} B^{*}=\frac{A C \cdot B D}{C D}, A^{*} B^{*}=\frac{A B \cdot C D}{C D} .
$$
Thus $D^{*} B^{*}: A^{*} B^{*}=\sqrt{2}$, and this is the required ratio.
23. Let the given numbers be $a_{1}, \ldots, a_{n}$. Put $s=a_{1}+\cdots+a_{n}$ and $m=$ $\operatorname{lcm}\left(a_{1}, \ldots, a_{n}\right)$ and write $m=2^{k} r$ with $k \geq 0$ and $r$ odd. Let the binary expansion of $r$ be $r=2^{k_{0}}+2^{k_{1}}+\cdots+2^{k_{t}}$, with $0=k_{0}<\cdots1$ (since the case $a<-1$ is reduced to $a>1$ by taking $a^{\prime}=-a, x_{i}^{\prime}=-x_{i}$ ). Since the left sides of the equations are nonnegative, we have $x_{i} \geq-\frac{1}{a}>-1, i=1, \ldots, 1000$. Suppose w.l.o.g. that $x_{1}=\max \left\{x_{i}\right\}$. In particular, $x_{1} \geq x_{2}, x_{3}$. If $x_{1} \geq 0$, then we deduce that $x_{1000}^{2} \geq 1 \Rightarrow x_{1000} \geq 1$; further, from this we deduce that $x_{999}>1$ etc., so either $x_{i}>1$ for all $i$ or $x_{i}<0$ for all $i$.
(i) $x_{i}>1$ for every $i$. Then $x_{1} \geq x_{2}$ implies $x_{1}^{2} \geq x_{2}^{2}$, so $x_{2} \geq x_{3}$. Thus $x_{1} \geq x_{2} \geq \cdots \geq x_{1000} \geq x_{1}$, and consequently $x_{1}=\cdots=x_{1000}$. In this case the only solution is $x_{i}=\frac{1}{2}\left(a+\sqrt{a^{2}+4}\right)$ for all $i$.
(ii) $x_{i}<0$ for every $i$. Then $x_{1} \geq x_{3}$ implies $x_{1}^{2} \leq x_{3}^{2} \Rightarrow x_{2} \leq x_{4}$. Similarly, this leads to $x_{3} \geq x_{5}$, etc. Hence $x_{1} \geq x_{3} \geq x_{5} \geq \cdots \geq x_{999} \geq x_{1}$ and $x_{2} \leq x_{4} \leq \cdots \leq x_{2}$, so we deduce that $x_{1}=x_{3}=\cdots$ and $x_{2}=x_{4}=$ $\cdots$. Therefore the system is reduced to $x_{1}^{2}=a x_{2}+1, x_{2}^{2}=a x_{1}+1$. Subtracting these equations, one obtains $\left(x_{1}-x_{2}\right)\left(x_{1}+x_{2}+a\right)=0$. There are two possibilities:
(1) If $x_{1}=x_{2}$, then $x_{1}=x_{2}=\cdots=\frac{1}{2}\left(a-\sqrt{a^{2}+4}\right)$.
(2) $x_{1}+x_{2}+a=0$ is equivalent to $x_{1}^{2}+a x_{1}+\left(a^{2}-1\right)=0$. The discriminant of the last equation is $4-3 a^{2}$. Therefore if $a>\frac{2}{\sqrt{3}}$, this case yields no solutions, while if $a \leq \frac{2}{\sqrt{3}}$, we obtain $x_{1}=$ $\frac{1}{2}\left(-a-\sqrt{4-3 a^{2}}\right), x_{2}=\frac{1}{2}\left(-a+\sqrt{4-3 a^{2}}\right)$, or vice versa.
26. Set
$$
\begin{aligned}
f(a, b, c, d) & =a b c+b c d+c d a+d a b-\frac{176}{27} a b c d \\
& =a b(c+d)+c d\left(a+b-\frac{176}{27} a b\right) .
\end{aligned}
$$
If $a+b-\frac{176}{a} b \leq 0$, by the arithmetic-geometric inequality we have $f(a, b, c, d) \leq a b(c+d) \leq \frac{1}{27}$.
On the other hand, if $a+b-\frac{176}{a} b>0$, the value of $f$ increases if $c, d$ are replaced by $\frac{c+d}{2}, \frac{c+d}{2}$. Consider now the following fourtuplets:
$$
\begin{gathered}
P_{0}(a, b, c, d), P_{1}\left(a, b, \frac{c+d}{2}, \frac{c+d}{2}\right), P_{2}\left(\frac{a+b}{2}, \frac{a+b}{2}, \frac{c+d}{2}, \frac{c+d}{2}\right), \\
P_{3}\left(\frac{1}{4}, \frac{a+b}{2}, \frac{c+d}{2}, \frac{1}{4}\right), P_{4}\left(\frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}\right)
\end{gathered}
$$
From the above considerations we deduce that for $i=0,1,2,3$ either $f\left(P_{i}\right) \leq f\left(P_{i+1}\right)$, or directly $f\left(P_{i}\right) \leq 1 / 27$. Since $f\left(P_{4}\right)=1 / 27$, in every case we are led to
$$
f(a, b, c, d)=f\left(P_{0}\right) \leq \frac{1}{27}
$$
Equality occurs only in the cases $(0,1 / 3,1 / 3,1 / 3)$ (with permutations) and ( $1 / 4,1 / 4,1 / 4,1 / 4)$.
Remark. Lagrange multipliers also work. On the boundary of the set one of the numbers $a, b, c, d$ is 0 , and the inequality immediately follows, while for an extremum point in the interior, among $a, b, c, d$ there are at most two distinct values, in which case one easily verifies the inequality.
### 4.35 Solutions to the Shortlisted Problems of IMO 1994
1. Obviously $a_{0}>a_{1}>a_{2}>\cdots$. Since $a_{k}-a_{k+1}=1-\frac{1}{a_{k}+1}$, we have $a_{n}=a_{0}+\left(a_{1}-a_{0}\right)+\cdots+\left(a_{n}-a_{n-1}\right)=1994-n+\frac{1}{a_{0}+1}+\cdots+\frac{1}{a_{n-1}+1}>$ $1994-n$. Also, for $1 \leq n \leq 998$,
$$
\frac{1}{a_{0}+1}+\cdots+\frac{1}{a_{n-1}+1}<\frac{n}{a_{n-1}+1}<\frac{998}{a_{997}+1}<1
$$
because as above, $a_{997}>997$. Hence $\left\lfloor a_{n}\right\rfloor=1994-n$.
2. We may assume that $a_{1}>a_{2}>\cdots>a_{m}$. We claim that for $i=1, \ldots, m$, $a_{i}+a_{m+1-i} \geq n+1$. Indeed, otherwise $a_{i}+a_{m+1-i}, \ldots, a_{i}+a_{m-1}, a_{i}+a_{m}$ are $i$ different elements of $A$ greater than $a_{i}$, which is impossible. Now by adding for $i=1, \ldots, m$ we obtain $2\left(a_{1}+\cdots+a_{m}\right) \geq m(n+1)$, and the result follows.
3. The last condition implies that $f(x)=x$ has at most one solution in $(-1,0)$ and at most one solution in $(0, \infty)$. Suppose that for $u \in(-1,0)$, $f(u)=u$. Then putting $x=y=u$ in the given functional equation yields $f\left(u^{2}+2 u\right)=u^{2}+2 u$. Since $u \in(-1,0) \Rightarrow u^{2}+2 u \in(-1,0)$, we deduce that $u^{2}+2 u=u$, i.e., $u=-1$ or $u=0$, which is impossible. Similarly, if $f(v)=v$ for $v \in(0, \infty)$, we are led to the same contradiction.
However, for all $x \in S, f(x+(1+x) f(x))=x+(1+x) f(x)$, so we must have $x+(1+x) f(x)=0$. Therefore $f(x)=-\frac{x}{1+x}$ for all $x \in S$. It is directly verified that this function satisfies all the conditions.
4. Suppose that $\alpha=\beta$. The given functional equation for $x=y$ yields $f(x / 2)=x^{-\alpha} f(x)^{2} / 2$; hence the functional equation can be written as
$$
f(x) f(y)=\frac{1}{2} x^{\alpha} y^{-\alpha} f(y)^{2}+\frac{1}{2} y^{\alpha} x^{-\alpha} f(x)^{2}
$$
i.e.,
$$
\left((x / y)^{\alpha / 2} f(y)-(y / x)^{\alpha / 2} f(x)\right)^{2}=0
$$
Hence $f(x) / x^{\alpha}=f(y) / y^{\alpha}$ for all $x, y \in \mathbb{R}^{+}$, so $f(x)=\lambda x^{\alpha}$ for some $\lambda$. Substituting into the functional equation we obtain that $\lambda=2^{1-\alpha}$ or $\lambda=0$. Thus either $f(x) \equiv 2^{1-\alpha} x^{\alpha}$ or $f(x) \equiv 0$.
Now let $\alpha \neq \beta$. Interchanging $x$ with $y$ in the given equation and subtracting these equalities from each other, we get $\left(x^{\alpha}-x^{\beta}\right) f(y / 2)=\left(y^{\alpha}-\right.$ $\left.y^{\beta}\right) f(x / 2)$, so for some constant $\lambda \geq 0$ and all $x \neq 1, f(x / 2)=\lambda\left(x^{\alpha}-x^{\beta}\right)$. Substituting this into the given equation, we obtain that only $\lambda=0$ is possible, i.e., $f(x) \equiv 0$.
5. If $f^{(n)}(x)=\frac{p_{n}(x)}{q_{n}(x)}$ for some positive integer $n$ and polynomials $p_{n}, q_{n}$, then
$$
f^{(n+1)}(x)=f\left(\frac{p_{n}(x)}{q_{n}(x)}\right)=\frac{p_{n}(x)^{2}+q_{n}(x)^{2}}{2 p_{n}(x) q_{n}(x)}
$$
Note that $f^{(0)}(x)=x / 1$. Thus $f^{(n)}(x)=\frac{p_{n}(x)}{q_{n}(x)}$, where the sequence of polynomials $p_{n}, q_{n}$ is defined recursively by
$$
\begin{gathered}
p_{0}(x)=x, \quad q_{0}(x)=1, \text { and } \\
p_{n+1}(x)=p_{n}(x)^{2}+q_{n}(x)^{2}, q_{n+1}(x)=2 p_{n}(x) q_{n}(x) .
\end{gathered}
$$
Furthermore, $p_{0}(x) \pm q_{0}(x)=x \pm 1$ and $p_{n+1}(x) \pm q_{n+1}(x)=p_{n}(x)^{2}+$ $q_{n}(x)^{2} \pm 2 p_{n}(x) q_{n}(x)=\left(p_{n}(x) \pm q_{n}(x)\right)^{2}$, so $p_{n}(x) \pm q_{n}(x)=(x \pm 1)^{2^{n}}$ for all $n$. Hence
$$
p_{n}(x)=\frac{(x+1)^{2^{n}}+(x-1)^{2^{n}}}{2} \quad \text { and } \quad q_{n}(x)=\frac{(x+1)^{2^{n}}-(x-1)^{2^{n}}}{2} .
$$
Finally,
$$
\begin{aligned}
\frac{f^{(n)}(x)}{f^{(n+1)}(x)} & =\frac{p_{n}(x) q_{n+1}(x)}{q_{n}(x) p_{n+1}(x)}=\frac{2 p_{n}(x)^{2}}{p_{n+1}(x)}=\frac{\left((x+1)^{2^{n}}+(x-1)^{2^{n}}\right)^{2}}{(x+1)^{2^{n+1}}+(x-1)^{2^{n+1}}} \\
& =1+\frac{2\left(\frac{x+1}{x-1}\right)^{2^{n}}}{1+\left(\frac{x+1}{x-1}\right)^{2^{n+1}}}=1+\frac{1}{f\left(\left(\frac{x+1}{x-1}\right)^{2^{n}}\right)}
\end{aligned}
$$
6. Call the first and second player $M$ and $N$ respectively. $N$ can keep $A \leq 6$. Indeed, let 10 dominoes be placed as shown in the picture, and whenever $M$ marks a 1 in a cell of some domino, let $N$ mark 0 in the other cell of that domino if it is still empty. Since any $3 \times 3$ square contains at least three complete domi-
noes, there are at least three 0 's inside. Hence $A \leq 6$.
We now show that $M$ can make $A=6$. Let him start by marking 1 in $c 3$. By symmetry, we may assume that $N$ 's response is made in row 4 or 5 . Then $M$ marks 1 in $c 2$. If $N$ puts 0 in $c 1$, then $M$ can always mark two 1's in $b \times\{1,2,3\}$ as well as three 1's in $\{a, d\} \times\{1,2,3\}$. Thus either $\{a, b, c\} \times\{1,2,3\}$ or $\{b, c, d\} \times\{1,2,3\}$ will contain six 1's. However, if $N$ does not play his second move in $c 1$, then $M$ plays there, and thus he can easily achieve to have six 1's either in $\{a, b, c\} \times\{1,2,3\}$ or $\{c, d, e\} \times\{1,2,3\}$.
7. Let $a_{1}, a_{2}, \ldots, a_{m}$ be the ages of the male citizens $(m \geq 1)$. We claim that the age of each female citizen can be expressed in the form $c_{1} a_{1}+\cdots+c_{m} a_{m}$ for some constants $c_{i} \geq 0$, and we will prove this by induction on the number $n$ of female citizens.
The claim is clear if $n=1$. Suppose it holds for $n$ and consider the case of $n+1$ female citizens. Choose any of them, say $A$ of age $x$ who knows $k$
citizens (at least one male). By the induction hypothesis, the age of each of the other $n$ females is expressible as $c_{1} a_{1}+\cdots+c_{m} a_{m}+c_{0} x$, where $c_{i} \geq 0$ and $c_{0}+c_{1}+\cdots+c_{m}=1$. Consequently, the sum of ages of the $k$ citizens who know $A$ is $k x=b_{1} a_{1}+\cdots+b_{m} a_{m}+b_{0} x$ for some constants $b_{i} \geq 0$ with sum $k$. But $A$ knows at least one male citizen (who does not contribute to the coefficient of $x)$, so $b_{0} \leq k-1$. Hence $x=\frac{b_{1} a_{1}+\cdots+b_{m} a_{m}}{k-b_{0}}$, and the claim follows.
8. (a) Let $a, b, c, a \leq b \leq c$ be the amounts of money in dollars in Peter's first, second, and third account, respectively. If $a=0$, then we are done, so suppose that $a>0$. Let Peter make transfers of money into the first account as follows. Write $b=a q+r$ with $0 \leq r1994$ is trivial. Suppose that $n=1994$. Label the girls $G_{1}$ to $G_{1994}$, and let $G_{1}$ initially hold all the cards. At any moment give to each card the value $i, i=1, \ldots, 1994$, if $G_{i}$ holds it. Define the characteristic $C$ of a position as the sum of all these values. Initially $C=1994$. In each move, if $G_{i}$ passes cards to $G_{i-1}$ and $G_{i+1}$ (where $G_{0}=G_{1994}$ and $G_{1995}=G_{1}$ ), $C$ changes for $\pm 1994$ or does not change, so that it remains divisible by 1994. But if the game ends, the characteristic of the final position will be $C=1+2+\cdots+1994=$ $997 \cdot 1995$, which is not divisible by 1994.
(b) Whenever a card is passed from one girl to another for the first time, let the girls sign their names on it. Thereafter, if one of them passes a card to her neighbor, we shall assume that the passed card is exactly the one signed by both of them. Thus each signed card is stuck between two neighboring girls, so if $n<1994$, there are two neighbors who never exchange cards. Consequently, there is a girl $G$ who played only a finite number of times. If her neighbor plays infinitely often, then after her last move, $G$ will continue to accumulate cards indefinitely, which is impossible. Hence every girl plays finitely many times.
11. Tile the table with dominoes and numbers as shown in the picture. The second player will not lose if whenever the first player plays in a cell of a domino, he plays in the other cell of the domino, and whenever the first player plays on a number, he plays on the same number that is diagonally adjacent.
12. Define $S_{n}$ recursively as follows: Let $S_{2}=\{(0,0),(1,1)\}$ and $S_{n+1}=$ $S_{n} \cup T_{n}$, where $T_{n}=\left\{\left(x+2^{n-1}, y+M_{n}\right) \mid(x, y) \in S_{n}\right\}$, with $M_{n}$ chosen large enough so that the entire set $T_{n}$ lies above every line passing through two points of $S_{n}$. By definition, $S_{n}$ has exactly $2^{n-1}$ points and contains no three collinear points. We claim that no $2 n$ points of this set are the vertices of a convex $2 n$-gon.
Consider an arbitrary convex polygon $\mathcal{P}$ with vertices in $S_{n}$. Join by a diagonal $d$ the two vertices of $\mathcal{P}$ having the smallest and greatest $x$ coordinates. This diagonal divides $\mathcal{P}$ into two convex polygons $\mathcal{P}_{1}, \mathcal{P}_{2}$, the former lying above $d$. We shall show by induction that both $\mathcal{P}_{1}, \mathcal{P}_{2}$ have at most $n$ vertices. Assume to the contrary that $\mathcal{P}_{1}$ has at least $n+1$ vertices $A_{1}\left(x_{1}, y_{1}\right), \ldots, A_{n+1}\left(x_{n+1}, y_{n+1}\right)$ in $S_{n}$, with $x_{1}<\cdots\cdots>\frac{y_{n+1}-y_{n}}{x_{n+1}-x_{n}}$. By the induction hypothesis, not more than $n-1$ of these vertices belong to $S_{n-1}$ or $T_{n-1}$, so let $A_{k-1}, A_{k} \in S_{n-1}$, $A_{k+1} \in T_{n-1}$. But by the construction of $T_{n-1}, \frac{y_{k+1}-y_{k}}{x_{k+1}-x_{k}}>\frac{y_{k}-y_{k-1}}{x_{k}-x_{k-1}}$, which
gives a contradiction. Similarly, $\mathcal{P}_{2}$ has no more than $n$ vertices, and therefore $\mathcal{P}$ itself has at most $2 n-2$ vertices.
13. Extend $A D$ and $B C$ to meet at $P$, and let $Q$ be the foot of the perpendicular from $P$ to $A B$. Denote by $O$ the center of $\Gamma$. Since $\triangle P A Q \sim \triangle O A D$ and $\triangle P B Q \sim \triangle O B C$, we obtain $\frac{A Q}{A D}=\frac{P Q}{O D}=\frac{P Q}{O C}=\frac{B Q}{B C}$. Therefore $\frac{A Q}{Q B} \cdot \frac{B C}{C P} \cdot \frac{P D}{D A}=1$, so by the converse Ceva theorem, $A C, B D$, and $P Q$ are concurrent. It follows that $Q \equiv F$. Finally, since the points $O, C, P, D, F$ are concyclic, we have $\angle D F P=\angle D O P=\angle P O C=\angle P F C$.
14. Although it does not seem to have been noticed at the jury, the statement of the problem is false. For $A(0,0), B(0,4), C(1,4), D(7,0)$, we have $M(4,2), P(2,1), Q(2,3)$ and $N(9 / 2,1 / 2) \notin \triangle A B M$.
The official solution, if it can be called so, actually shows that $N$ lies inside $A B C D$ and goes as follows: The case $A D=B C$ is trivial, so let $A D>B C$. Let $L$ be the midpoint of $A B$. Complete the parallelograms $A D M X$ and $B C M Y$. Now $N=D X \cap C Y$, so let $C Y$ and $D X$ intersect $A B$ at $K$ and $H$ respectively. From $L X=L Y$ and
$$
\frac{H L}{L X}=\frac{H A}{A D}<\frac{L A}{A D}<\frac{K B}{A D}<\frac{K B}{B C}=\frac{K L}{L Y}
$$
we get $H Ln \geq 2$. Since $m^{3}+1 \equiv 1$ and
$m n-1 \equiv-1(\bmod n)$, we deduce $\frac{n^{3}+1}{m n-1}=k n-1$ for some integer $k>0$. On the other hand, $k n-1<\frac{n^{3}+1}{n^{2}-1}=n+\frac{1}{n-1} \leq 2 n-1$ gives that $k=1$, and therefore $n^{3}+1=(m n-1)(n-1)$. This yields $m=\frac{n^{2}+1}{n-1}=n+1+\frac{2}{n-1} \in \mathbb{N}$, so $n \in\{2,3\}$ and $m=5$. The solutions with $mz_{n}$ when $x_{n-1}$ is even and $2 y_{n}>z_{n}>y_{n}$ when $x_{n-1}$ is odd. In fact, $n=1$ is the trivial case, while if it holds for $n \geq 1$, then $y_{n+1}=2 y_{n}>z_{n}=z_{n+1}$ if $x_{n}$ is even, and $2 y_{n+1}=2 y_{n}>y_{n}+z_{n}=z_{n+1}$ if $x_{n}$ is odd (since then $x_{n-1}$ is even).
If $x_{1}=0$, then $x_{0}=3$ is good. Suppose $x_{n}=0$ for some $n \geq 2$. Then $x_{n-1}$ is odd and $x_{n-2}$ is even, so that $y_{n-1}>z_{n-1}$. We claim that a pair $\left(y_{n-1}, z_{n-1}\right)$, where $2^{k}=y_{n-1}>z_{n-1}>0$ and $z_{n-1} \equiv 1$ $(\bmod 4)$, uniquely determines $x_{0}=f\left(y_{n-1}, z_{n-1}\right)$. We see that $x_{n-1}=$ $\frac{1}{2} y_{n-1}+z_{n-1}$, and define $\left(x_{k}, y_{k}, z_{k}\right)$ backwards as follows, until we get $\left(y_{k}, z_{k}\right)=(4,1)$. If $y_{k}>z_{k}$, then $x_{k-1}$ must have been even, so we define $\left(x_{k-1}, y_{k-1}, z_{k-1}\right)=\left(2 x_{k}, y_{k} / 2, z_{k}\right)$; otherwise $x_{k-1}$ must have been odd, so we put $\left(x_{k-1}, y_{k-1}, z_{k-1}\right)=\left(x_{k}-y_{k} / 2+z_{k}, y_{k}, z_{k}-y_{k}\right)$. We eventually arrive at $\left(y_{0}, z_{0}\right)=(4,1)$ and a good integer $x_{0}=f\left(y_{n-1}, z_{n-1}\right)$, as claimed. Thus for example $\left(y_{n-1}, z_{n-1}\right)=(64,61)$ implies $x_{n-1}=93$, $\left(x_{n-2}, y_{n-2}, z_{n-2}\right)=(186,32,61)$ etc., and $x_{0}=1953$, while in the case of $\left(y_{n-1}, z_{n-1}\right)=(128,1)$ we get $x_{0}=2080$.
Note that $y^{\prime}>y \Rightarrow f\left(y^{\prime}, z^{\prime}\right)>f(y, z)$ and $z^{\prime}>z \Rightarrow f\left(y, z^{\prime}\right)>f(y, z)$. Therefore there are no $y, z$ for which $19531$, and let $x_{j} \equiv u_{j}(\bmod p)$ where $0 \leq u_{j} \leq p-1$ (particularly $u_{i} \equiv 0$ ). Then $u_{j+1} \equiv u_{j} u_{j-1}+$ $1(\bmod p)$. The number of possible pairs $\left(u_{j}, u_{j+1}\right)$ is finite, so $u_{j}$ is eventually periodic. We claim that for some $d_{p}>0, u_{i+d_{p}}=0$. Indeed, suppose the contrary and let $\left(u_{m}, u_{m+1}, \ldots, u_{m+d-1}\right)$ be the first period for $m \geq i$. Then $m \neq i$. By the assumption $u_{m-1} \not \equiv$ $u_{m+d-1}$, but $u_{m-1} u_{m} \equiv u_{m+1}-1 \equiv u_{m+d+1}-1 \equiv u_{m+d-1} u_{m+d} \equiv$ $u_{m+d-1} u_{m}(\bmod p)$, which is impossible if $p \nmid u_{m}$. Hence there is a $d_{p}$ with $u_{i}=u_{i+d_{p}}=0$ and moreover $u_{i+1}=u_{i+d_{p}+1}=1$, so the sequence $u_{j}$ is periodic with period $d_{p}$ starting from $u_{i}$. Let $m$ be the least common multiple of all $d_{p}$ 's, where $p$ goes through all prime divisors of $x_{i}$. Then the same primes divide every $x_{i+k m}, k=1,2, \ldots$, so for large enough $k$ and $j=i+k m, x_{i}^{i} \mid x_{j}^{j}$.
(b) If $i=1$, we cannot deduce that $x_{i+1} \equiv 1(\bmod p)$. The following example shows that the statement from (a) need not be true in this case. Take $x_{1}=22$ and $x_{2}=9$. Then $x_{n}$ is even if and only if $n \equiv 1(\bmod$ 3 ), but modulo 11 the sequence $\left\{x_{n}\right\}$ is $0,9,1,10,0,1,1,2,3,7,0, \ldots$, so $11 \mid x_{n}(n>1)$ if and only if $n \equiv 5(\bmod 6)$. Thus for no $n>1$ can we have $22 \mid x_{n}$.
24. A multiple of 10 does not divide any wobbly number. Also, if $25 \mid n$, then every multiple of $n$ ends with $25,50,75$, or 00 ; hence it is not wobbly. We now show that every other number $n$ divides some wobbly number.
(i) Let $n$ be odd and not divisible by 5 . For any $k \geq 1$ there exists $l$ such that $\left(10^{k}-1\right) n$ divides $10^{l}-1$, and thus also divides $10^{k l}-1$. Consequently, $v_{k}=\frac{10^{k l}-1}{10^{k}-1}$ is divisible by $n$, and it is wobbly when $k=2$ (indeed, $v_{2}=101 \ldots 01$ ).
If $n$ is divisible by 5 , one can simply take $5 v_{2}$ instead.
(ii) Let $n$ be a power of 2 . We prove by induction on $m$ that $2^{2 m+1}$ has a wobbly multiple $w_{m}$ with exactly $m$ nonzero digits. For $m=1$, take $w_{1}=8$. Suppose that for some $m \geq 1$ there is a wobbly $w_{m}=$ $2^{2 m+1} d_{m}$. Then the numbers $a \cdot 10^{2 m}+w_{m}$ are wobbly and divisible by $2^{2 m+1}$ when $a \in\{2,4,6,8\}$. Moreover, one of these numbers is divisible by $2^{2 m+3}$. Indeed, it suffices to choose $a$ such that $\frac{a}{2}+d_{m}$ is divisible by 4 . This proves the induction step.
(iii) Let $n=2^{m} r$, where $m \geq 1$ and $r$ is odd, $5 \nmid r$. Then $v_{2 m} w_{m}$ is wobbly and divisible by both $2^{m}$ and $r$ (using notation from (i), $r \mid v_{2 m}$ ).
### 4.36 Solutions to the Shortlisted Problems of IMO 1995
1. Let $x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}$. Then $x y z=1$ and
$$
S=\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)}=\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y}
$$
We must prove that $S \geq \frac{3}{2}$. From the Cauchy-Schwarz inequality,
$$
[(y+z)+(z+x)+(x+y)] \cdot S \geq(x+y+z)^{2} \Rightarrow S \geq \frac{x+y+z}{2}
$$
It follows from the A-G mean inequality that $\frac{x+y+z}{2} \geq \frac{3}{2} \sqrt[3]{x y z}=\frac{3}{2}$; hence the proof is complete. Equality holds if and only if $x=y=z=1$, i.e., $a=b=c=1$.
Remark. After reducing the problem to $\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \geq \frac{3}{2}$, we can solve the problem using Jensen's inequality applied to the function $g(u, v)=$ $u^{2} / v$. The problem can also be solved using Muirhead's inequality.
2. We may assume $c \geq 0$ (otherwise, we may simply put $-y_{i}$ in the place of $\left.y_{i}\right)$. Also, we may assume $a \geq b$. If $b \geq c$, it is enough to take $n=a+b-c$, $x_{1}=\cdots=x_{a}=1, y_{1}=\cdots=y_{c}=y_{a+1}=\cdots=y_{a+b-c}=1$, and the other $x_{i}$ 's and $y_{i}$ 's equal to 0 , so we need only consider the case $a>c>b$. We proceed to prove the statement of the problem by induction on $a+b$. The case $a+b=1$ is trivial. Assume that the statement is true when $a+b \leq$ $N$, and let $a+b=N+1$. The triple $(a+b-2 c, b, c-b)$ satisfies the condition (since $(a+b-2 c) b-(c-b)^{2}=a b-c^{2}$ ), so by the induction hypothesis there are $n$-tuples $\left(x_{i}\right)_{i=1}^{n}$ and $\left(y_{i}\right)_{i=1}^{n}$ with the wanted property. It is easy to verify that $\left(x_{i}+y_{i}\right)_{i=1}^{n}$ and $\left(y_{i}\right)_{i=1}^{n}$ give a solution for $(a, b, c)$.
3. Write $A_{i}=\frac{a_{i}^{2}+a_{i+1}^{2}-a_{i+2}^{2}}{a_{i}+a_{i+1}-a_{i+2}}=a_{i}+a_{i+1}+a_{i+2}-\frac{2 a_{i} a_{i+1}}{a_{i}+a_{i+1}-a_{i+2}}$. Since $2 a_{i} a_{i+1} \geq$ $4\left(a_{i}+a_{i+1}-2\right)$ (which is equivalent to $\left.\left(a_{i}-2\right)\left(a_{i+1}-2\right) \geq 0\right)$, it follows that $A_{i} \leq a_{i}+a_{i+1}+a_{i+2}-4\left(1+\frac{a_{i+2}-2}{a_{i}+a_{i+1}-a_{i+2}}\right) \leq a_{i}+a_{i+1}+a_{i+2}-$ $4\left(1+\frac{a_{i+2}-2}{4}\right)$, because $1 \leq a_{i}+a_{i+1}-a_{i+2} \leq 4$. Therefore $A_{i} \leq a_{i}+$ $a_{i+1}-2$, so $\sum_{i=1}^{n} A_{i} \leq 2 s-2 n$ as required.
4. The second equation is equivalent to $\frac{a^{2}}{y z}+\frac{b^{2}}{z x}+\frac{c^{2}}{x y}+\frac{a b c}{x y z}=4$. Let $x_{1}=$ $\frac{a}{\sqrt{y z}}, y_{1}=\frac{b}{\sqrt{z x}}, z_{1}=\frac{c}{\sqrt{x y}}$. Then $x_{1}^{2}+y_{1}^{2}+z_{1}^{2}+x_{1} y_{1} z_{1}=4$, where $00$ is clearly impossible. On the other hand, if $u v w \leq 0$, then two of $u, v, w$ are nonnegative, say $u, v \geq 0$. Taking into account $w=-u-v$, the above equality reduces to $2\left[(a+c-2 v) u^{2}+(b+c-2 u) v^{2}+2 c u v\right]=0$, so $u=v=0$.
Third solution. The fact that we are given two equations and three variables suggests that this is essentially a problem on inequalities. Setting $f(x, y, z)=4 x y z-a^{2} x-b^{2} y-c^{2} z$, we should show that $\max f(x, y, z)=$ $a b c$, for $0\frac{n-1}{4}$. Then $f(1 / x)=f\left(x+1 / x^{2}\right)-f(x)<1 / 4$, so $f(1 / x)>-1 / 2$. On the other hand, this implies $\left(\frac{n-1}{4}\right)^{2}0$. Since $\sum_{i=1}^{n-1} y_{i}=Y$ and $\sum_{i=1}^{n-1}(n-i)=\frac{n(n-1)}{2}$, we have $\sum z_{i}=0$. Note that $Y<\sum_{j=2}^{n}(j-1) x_{n}=\frac{n(n-1)}{2} x_{n}$, and consequently $z_{n-1}=$ $\frac{n(n-1)}{2} x_{n}-Y>0$. Furthermore, we have
$$
\frac{z_{i+1}}{n-i-1}-\frac{z_{i}}{n-i}=\frac{n(n-1)}{2}\left(\frac{y_{i+1}}{n-i-1}-\frac{y_{i}}{n-i}\right)>0
$$
which means that $\frac{z_{1}}{n-1}<\frac{z_{2}}{n-2}<\cdots<\frac{z_{n-1}}{1}$. Therefore there is a $k$ for which $z_{1}, \ldots, z_{k} \leq 0$ and $z_{k+1}, \ldots, z_{n-1}>0$. But then $z_{i}\left(x_{i}-x_{k}\right) \geq 0$, i.e., $x_{i} z_{i} \geq x_{k} z_{i}$ for all $i$, so $\sum_{i=1}^{n-1} x_{i} z_{i}>\sum_{i=1}^{n-1} x_{k} z_{i}=0$ as required.
Second solution. Set $X=\sum_{j=1}^{n-1}(n-j) x_{j}$ and $Y=\sum_{j=2}^{n}(j-1) x_{j}$. Since $4 X Y=(X+Y)^{2}-(X-Y)^{2}$, the RHS of the inequality becomes
$$
X Y=\frac{1}{4}\left[(n-1)^{2}\left(\sum_{i=1}^{n} x_{i}\right)^{2}-\left(\sum_{i=1}^{n}(2 i-1-n) x_{i}\right)^{2}\right]
$$
The LHS equals $\frac{1}{4}\left((n-1)^{2}\left(\sum_{i=1}^{n} x_{i}\right)^{2}-(n-1) \sum_{i(n-1) \sum_{i0$ (so, $x_{j}-x_{i}=d_{i}+d_{i+1}+\cdots+d_{j-1}$ ) and expanding the obtained expressions, we reduce this inequality to $\sum_{k} k^{2}(n-k)^{2} d_{k}^{2}+2 \sum_{k\sum_{k}(n-1) k(n-k) d_{k}^{2}+$ $2 \sum_{k3$. Then either $a_{k}^{2} \equiv-7\left(\bmod 2^{k+1}\right)$ or $a_{k}^{2} \equiv 2^{k}-7$ $\left(\bmod 2^{k+1}\right)$. In the former case, take $a_{k+1}=a_{k}$. In the latter case, set $a_{k+1}=a_{k}+2^{k-1}$. Then $a_{k+1}^{2}=a_{k}^{2}+2^{k} a_{k}+2^{2 k-2} \equiv a_{k}^{2}+2^{k} \equiv-7(\bmod$ $2^{k+1}$ ) because $a_{k}$ is odd.
16. If $A$ is odd, then every number in $M_{1}$ is of the form $x(x+A)+B \equiv B$ $(\bmod 2)$, while numbers in $M_{2}$ are congruent to $C$ modulo 2 . Thus it is enough to take $C \equiv B+1(\bmod 2)$.
If $A$ is even, then all numbers in $M_{1}$ have the form $\left(X+\frac{A}{2}\right)^{2}+B-\frac{A^{2}}{4}$ and are congruent to $B-\frac{A^{2}}{4}$ or $B-\frac{A^{2}}{4}+1$ modulo 4 , while numbers in $M_{2}$ are congruent to $C$ modulo 4 . So one can choose any $C \equiv B-\frac{A^{2}}{4}+2$ $(\bmod 4)$.
17. For $n=4$, the vertices of a unit square $A_{1} A_{2} A_{3} A_{4}$ and $p_{1}=p_{2}=p_{3}=$ $p_{4}=\frac{1}{6}$ satisfy the conditions. We claim that there are no solutions for $n=5$ (and thus for any $n \geq 5$ ).
Suppose to the contrary that points $A_{i}$ and $p_{i}, i=1, \ldots, 5$, satisfy the conditions. Denote the area of $\triangle A_{i} A_{j} A_{k}$ by $S_{i j k}=p_{i}+p_{j}+p_{k}, 1 \leq i<$ $j2$. Therefore $c=\frac{b^{2}+z}{z^{2} b-1}<1$, a contradiction.
The only solutions for $(x, y)$ are $(4,2),(4,6),(5,2),(5,3)$.
19. For each two people let $n$ be the number of people exchanging greetings with both of them. To determine $n$ in terms of $k$, we shall count in two ways the number of triples $(A, B, C)$ of people such that $A$ exchanged greetings with both $B$ and $C$, but $B$ and $C$ mutually did not.
There are $12 k$ possibilities for $A$, and for each $A$ there are $(3 k+6)$ possibilities for $B$. Since there are $n$ people who exchanged greetings with both
$A$ and $B$, there are $3 k+5-n$ who did so with $A$ but not with $B$. Thus the number of triples $(A, B, C)$ is $12 k(3 k+6)(3 k+5-n)$. On the other hand, there are $12 k$ possible choices of $B$, and $12 k-1-(3 k+6)=9 k-7$ possible choices of $C$; for every $B, C, A$ can be chosen in $n$ ways, so the number of considered triples equals $12 k n(9 k-7)$.
Hence $(3 k+6)(3 k+5-n)=n(9 k-7)$, i.e., $n=\frac{3(k+2)(3 k+5)}{12 k-1}$. This gives us that $\frac{4 n}{3}=\frac{12 k^{2}+44 k+40}{12 k-1}=k+4-\frac{3 k-44}{12 k-1}$ is an integer too. It is directly verified that only $k=3$ gives an integer value for $n$, namely $n=6$.
Remark. The solution is complete under the assumption that such a $k$ exists. We give an example of such a party with 36 persons, $k=3$. Let the people sit in a $6 \times 6$ array $\left[P_{i j}\right]_{i, j=1}^{6}$, and suppose that two persons $P_{i j}, P_{k l}$ exchanged greetings if and only if $i=k$ or $j=l$ or $i-j \equiv k-l$ (mod 6). Thus each person exchanged greetings with exactly 15 others, and it is easily verified that this party satisfies the conditions.
20. We shall consider the set $M=\{0,1, \ldots, 2 p-1\}$ instead. Let $M_{1}=$ $\{0,1, \ldots, p-1\}$ and $M_{2}=\{p, p+1, \ldots, 2 p-1\}$. We shall denote by $|A|$ and $\sigma(A)$ the number of elements and the sum of elements of the set $A$; also, let $C_{p}$ be the family of all $p$-element subsets of $M$. Define the mapping $T: C_{p} \rightarrow C_{p}$ as $T(A)=\left\{x+1 \mid x \in A \cap M_{1}\right\} \cup\left\{A \cap M_{2}\right\}$, the addition being modulo $p$. There are exactly two fixed points of $T$ : these are $M_{1}$ and $M_{2}$. Now if $A$ is any subset from $C_{p}$ distinct from $M_{1}, M_{2}$, and $k=\left|A \cap M_{1}\right|$ with $1 \leq k \leq p-1$, then for $i=0,1, \ldots, p-1$, $\sigma\left(T^{i}(A)\right)=\sigma(A)+i k(\bmod p)$. Hence subsets $A, T(A), \ldots, T^{p-1}(A)$ are distinct, and exactly one of them has sum of elements divisible by $p$. Since $\sigma\left(M_{1}\right), \sigma\left(M_{2}\right)$ are divisible by $p$ and $C_{p} \backslash\left\{M_{1}, M_{2}\right\}$ decomposes into families of the form $\left\{A, T(A), \ldots, T^{p-1}(A)\right\}$, we conclude that the required number is $\frac{1}{p}\left(\left|C_{p}\right|-2\right)+2=\frac{1}{p}\left(\binom{2 p}{p}-2\right)+2$.
Second solution. Let $C_{k}$ be the family of all $k$-element subsets of $\{1,2, \ldots, 2 p\}$. Denote by $M_{k}(k=1,2, \ldots, p)$ the family of $p$-element multisets with $k$ distinct elements from $\{1,2, \ldots, 2 p\}$, exactly one of which appears more than once, that have sum of elements divisible by $p$. It is clear that every subset from $C_{k}, k0$, so $f(x)=$ $x p(x) / q(x)$ is a positive integer too. Let $\left\{p_{0}, p_{1}, p_{2}, \ldots\right\}$ be all prime numbers in increasing order. Since it easily follows by induction that all $x_{n}$ 's are square-free, we can assign to each of them a unique code according to which primes divide it: if $p_{m}$ is the largest prime dividing $x_{n}$, the code corresponding to $x_{n}$ will be $\ldots 0 s_{m} s_{m-1} \ldots s_{0}$, with $s_{i}=1$ if $p_{i} \mid x_{n}$ and $s_{i}=0$ otherwise. Let us investigate how $f$ acts on these codes. If the code of $x_{n}$ ends with 0 , then $x_{n}$ is odd, so the code of $f\left(x_{n}\right)=x_{n+1}$ is obtained from that of $x_{n}$ by replacing $s_{0}=0$ by $s_{0}=1$. Furthermore, if the code of
$x_{n}$ ends with $011 \ldots 1$, then the code of $x_{n+1}$ ends with $100 \ldots 0$ instead. Thus if we consider the codes as binary numbers, $f$ acts on them as an addition of 1 . Hence the code of $x_{n}$ is the binary representation of $n$ and thus $x_{n}$ uniquely determines $n$.
Specifically, if $x_{n}=1995=3 \cdot 5 \cdot 7 \cdot 19$, then its code is 10001110 and corresponds to $n=142$.
26. For $n=1$ the result is trivial, since $x_{1}=1$. Suppose now that $n \geq 2$ and let $f_{n}(x)=x^{n}-\sum_{i=0}^{n-1} x^{i}$. Note that $x_{n}$ is the unique positive real root of $f_{n}$, because $\frac{f_{n}(x)}{x^{n-1}}=x-1-\frac{1}{x}-\cdots-\frac{1}{x^{n-1}}$ is strictly increasing on $\mathbb{R}^{+}$. Consider $g_{n}(x)=(x-1) f_{n}(x)=(x-2) x^{n}+1$. Obviously $g_{n}(x)$ has no positive roots other than 1 and $x_{n}>1$. Observe that $\left(1-\frac{1}{2^{n}}\right)^{n}>$ $1-\frac{n}{2^{n}} \geq \frac{1}{2}$ for $n \geq 2$ (by Bernoulli's inequality). Since then
$$
g_{n}\left(2-\frac{1}{2^{n}}\right)=-\frac{1}{2^{n}}\left(2-\frac{1}{2^{n}}\right)^{n}+1=1-\left(1-\frac{1}{2^{n+1}}\right)^{n}>0
$$
and
$$
g_{n}\left(2-\frac{1}{2^{n-1}}\right)=-\frac{1}{2^{n-1}}\left(2-\frac{1}{2^{n-1}}\right)^{n}+1=1-2\left(1-\frac{1}{2^{n}}\right)^{n}<0
$$
we conclude that $x_{n}$ is between $2-\frac{1}{2^{n-1}}$ and $2-\frac{1}{2^{n}}$, as required.
Remark. Moreover, $\lim _{n \rightarrow \infty} 2^{n}\left(2-x_{n}\right)=1$.
27. Computing the first few values of $f(n)$, we observe the following pattern:
$$
\begin{aligned}
f(4 k) & =k, k \geq 3, & f(8) & =3 ; \\
f(4 k+1) & =1, k \geq 4, & f(5) & =f(13)=2 ; \\
f(4 k+2) & =k-3, k \geq 7, & f(2) & =1, f(6)=f(10)=2, \\
& & f(14) & =f(18)=3, f(26)=4 ; \\
f(4 k+3) & =2 . & &
\end{aligned}
$$
We shall prove these statements simultaneously by induction on $n$, having verified them for $k \leq 7$.
(i) Let $n=4 k$. Since $f(3)=f(7)=\cdots=f(4 k-1)=2$, we have $f(4 k) \geq k$. But $f(n) \leq \max _{m95$ for all $n$. If to the contrary $f(n) \leq 95$, we have $f(m)=n+f(m+95-f(n))$, so by induction $f(m)=k n+f(m+k(95-f(n))) \geq k n$ for all $k$, which is impossible. Now for $m>95$ we have $f(m+f(n)-95)=n+f(m)$, and again by induction $f(m+k(f(n)-95))=k n+f(m)$ for all $m, n, k$. It follows that with $n$ fixed,
$$
(\forall m) \lim _{k \rightarrow \infty} \frac{f(m+k(f(n)-95))}{m+k(f(n)-95)}=\frac{n}{f(n)-95}
$$
hence
$$
\lim _{s \rightarrow \infty} \frac{f(s)}{s}=\frac{n}{f(n)-95}
$$
Hence $\frac{n}{f(n)-95}$ does not depend on $n$, i.e., $f(n) \equiv c n+95$ for some constant $c$. It is easily checked that only $c=1$ is possible.
### 4.37 Solutions to the Shortlisted Problems of IMO 1996
1. We have $a^{5}+b^{5}-a^{2} b^{2}(a+b)=\left(a^{3}-b^{3}\right)\left(a^{2}-b^{2}\right) \geq 0$, i.e. $a^{5}+b^{5} \geq$ $a^{2} b^{2}(a+b)$. Hence
$$
\frac{a b}{a^{5}+b^{5}+a b} \leq \frac{a b}{a^{2} b^{2}(a+b)+a b}=\frac{a b c^{2}}{a^{2} b^{2} c^{2}(a+b)+a b c^{2}}=\frac{c}{a+b+c} .
$$
Now, the left side of the inequality to be proved does not exceed $\frac{c}{a+b+c}+$ $\frac{a}{a+b+c}+\frac{b}{a+b+c}=1$. Equality holds if and only if $a=b=c$.
2. Clearly $a_{1}>0$, and if $p \neq a_{1}$, we must have $a_{n}<0,\left|a_{n}\right|>\left|a_{1}\right|$, and $p=-a_{n}$. But then for sufficiently large odd $k,-a_{n}^{k}=\left|a_{n}\right|^{k}>(n-1)\left|a_{1}\right|^{k}$, so that $a_{1}^{k}+\cdots+a_{n}^{k} \leq(n-1)\left|a_{1}\right|^{k}-\left|a_{n}\right|^{k}<0$, a contradiction. Hence $p=a_{1}$.
Now let $x>a_{1}$. From $a_{1}+\cdots+a_{n} \geq 0$ we deduce $\sum_{j=2}^{n}\left(x-a_{j}\right) \leq$ $(n-1)\left(x+\frac{a_{1}}{n-1}\right)$, so by the AM-GM inequality,
$$
\left(x-a_{2}\right) \cdots\left(x-a_{n}\right) \leq\left(x+\frac{a_{1}}{n-1}\right)^{n-1} \leq x^{n-1}+x^{n-2} a_{1}+\cdots+a_{1}^{n-1}
$$
The last inequality holds because $\binom{n-1}{r} \leq(n-1)^{r}$ for all $r \geq 0$. Multiplying (1) by $\left(x-a_{1}\right)$ yields the desired inequality.
3. Since $a_{1}>2$, it can be written as $a_{1}=b+b^{-1}$ for some $b>0$. Furthermore, $a_{1}^{2}-2=b^{2}+b^{-2}$ and hence $a_{2}=\left(b^{2}+b^{-2}\right)\left(b+b^{-1}\right)$. We prove that
$$
a_{n}=\left(b+b^{-1}\right)\left(b^{2}+b^{-2}\right)\left(b^{4}+b^{-4}\right) \cdots\left(b^{2^{n-1}}+b^{-2^{n-1}}\right)
$$
by induction. Indeed, $\frac{a_{n+1}}{a_{n}}=\left(\frac{a_{n}}{a_{n-1}}\right)^{2}-2=\left(b^{2^{n-1}}+b^{-2^{n-1}}\right)^{2}-2=$ $b^{2^{n}}+b^{-2^{n}}$.
Now we have
$$
\begin{aligned}
\sum_{i=1}^{n} \frac{1}{a_{i}}= & 1+\frac{b}{b^{2}+1}+\frac{b^{3}}{\left(b^{2}+1\right)\left(b^{4}+1\right)}+\cdots \\
& \cdots+\frac{b^{2^{n}-1}}{\left(b^{2}+1\right)\left(b^{4}+1\right) \ldots\left(b^{2}+1\right)}
\end{aligned}
$$
Note that $\frac{1}{2}\left(a+2-\sqrt{a^{2}-4}\right)=1+\frac{1}{b}$; hence we must prove that the right side in (1) is less than $\frac{1}{b}$. This follows from the fact that
$$
\begin{aligned}
& \frac{b^{2^{k}}}{\left(b^{2}+1\right)\left(b^{4}+1\right) \cdots\left(b^{2^{k}}+1\right)} \\
& \quad=\frac{1}{\left(b^{2}+1\right)\left(b^{4}+1\right) \cdots\left(b^{2^{k-1}}+1\right)}-\frac{1}{\left(b^{2}+1\right)\left(b^{4}+1\right) \cdots\left(b^{2^{k}}+1\right)}
\end{aligned}
$$
hence the right side in (1) equals $\frac{1}{b}\left(1-\frac{1}{\left(b^{2}+1\right)\left(b^{4}+1\right) \ldots\left(b^{2^{n}}+1\right)}\right)$, and this is clearly less than $1 / b$.
4. Consider the function
$$
f(x)=\frac{a_{1}}{x}+\frac{a_{2}}{x^{2}}+\cdots+\frac{a_{n}}{x^{n}} .
$$
Since $f$ is strictly decreasing from $+\infty$ to 0 on the interval $(0,+\infty)$, there exists exactly one $R>0$ for which $f(R)=1$. This $R$ is also the only positive real root of the given polynomial.
Since $\ln x$ is a concave function on $(0,+\infty)$, Jensen's inequality gives us
$$
\sum_{j=1}^{n} \frac{a_{j}}{A}\left(\ln \frac{A}{R^{j}}\right) \leq \ln \left(\sum_{j=1}^{n} \frac{a_{j}}{A} \cdot \frac{A}{R^{j}}\right)=\ln f(R)=0 .
$$
Therefore $\sum_{j=1}^{n} a_{j}(\ln A-j \ln R) \leq 0$, which is equivalent to $A \ln A \leq$ $B \ln R$, i.e., $A^{A} \leq R^{B}$.
5. Considering the polynomials $\pm P( \pm x)$ we may assume w.l.o.g. that $a, b \geq$ 0 . We have four cases:
(1) $c \geq 0, d \geq 0$. Then $|a|+|b|+|c|+|d|=a+b+c+d=P(1) \leq 1$.
(2) $c \geq 0, d<0$. Then $|a|+|b|+|c|+|d|=a+b+c-d=P(1)-2 P(0) \leq 3$.
(3) $c<0, d \geq 0$. Then
$$
\begin{aligned}
|a|+|b|+|c|+|d| & =a+b-c+d \\
& =\frac{4}{3} P(1)-\frac{1}{3} P(-1)-\frac{8}{3} P(1 / 2)+\frac{8}{3} P(-1 / 2) \leq 7 .
\end{aligned}
$$
(4) $c<0, d<0$. Then
$$
\begin{aligned}
|a|+|b|+|c|+|d| & =a+b-c-d \\
& =\frac{5}{3} P(1)-4 P(1 / 2)+\frac{4}{3} P(-1 / 2) \leq 7
\end{aligned}
$$
Remark. It can be shown that the maximum of 7 is attained only for $P(x)= \pm\left(4 x^{3}-3 x\right)$.
6. Let $f(x), g(x)$ be polynomials with integer coefficients such that
$$
f(x)(x+1)^{n}+g(x)\left(x^{n}+1\right)=k_{0}
$$
Write $n=2^{r} m$ for $m$ odd and note that $x^{n}+1=\left(x^{2^{r}}+1\right) B(x)$, where $B(x)=x^{2^{r}(m-1)}-x^{2^{r}(m-2)}+\cdots-x^{2^{r}}+1$. Moreover, $B(-1)=1$; hence $B(x)-1=(x+1) c(x)$ and thus
$$
R(x) B(x)+1=(B(x)-1)^{n}=(x+1)^{n} c(x)^{n}
$$
for some polynomials $c(x)$ and $R(x)$.
The zeros of the polynomial $x^{2^{r}}+1$ are $\omega_{j}$, with $\omega_{1}=\cos \frac{\pi}{2^{r}}+i \sin \frac{\pi}{2^{r}}$, and $\omega_{j}=\omega^{2 j-1}$ for $1 \leq j \leq 2^{r}$. We have
$$
\left(\omega_{1}+1\right)\left(\omega_{2}+1\right) \cdots\left(\omega_{2^{r+1}}+1\right)=2
$$
From $(*)$ we also get $f\left(\omega_{j}\right)\left(\omega_{j}+1\right)^{n}=k_{0}$ for $j=1,2, \ldots, 2^{r}$. Since $A=f\left(\omega_{1}\right) f\left(\omega_{2}\right) \cdots f\left(\omega_{2^{r}}\right)$ is a symmetric polynomial in $\omega_{1}, \ldots, \omega_{2^{r}}$ with integer coefficients, $A$ is an integer. Consequently, taking the product over $j=1,2, \ldots, 2^{r}$ and using (2) we deduce that $2^{n} A=k_{0}^{2^{r}}$ is divisible by $2^{n}=2^{2^{r} m}$. Hence $2^{m} \mid k_{0}$.
Furthermore, since $\omega_{j}+1=\left(\omega_{1}+1\right) p_{j}\left(\omega_{1}\right)$ for some polynomial $p_{j}$ with integer coefficients, (2) gives $\left(\omega_{1}+1\right)^{2^{r}} p\left(\omega_{1}\right)=2$, where $p(x)=$ $p_{2}(x) \cdots p_{2^{r}}(x)$ has integer coefficients. But then the polynomial $(x+$ $1)^{2^{2}} p(x)-2$ has a zero $x=\omega_{1}$, so it is divisible by its minimal polynomial $x^{2^{r}}+1$. Therefore
$$
(x+1)^{2^{r}} p(x)=2+\left(x^{2^{r}}+1\right) q(x)
$$
for some polynomial $q(x)$. Raising (3) to the $m$ th power we get $(x+$ $1)^{n} p(x)^{n}=2^{m}+\left(x^{2^{r}}+1\right) Q(x)$ for some polynomial $Q(x)$ with integer coefficients. Now using (1) we obtain
$$
\begin{aligned}
(x+1)^{n} c(x)^{n}\left(x^{2^{r}}+1\right) Q(x) & =\left(x^{2^{r}}+1\right) Q(x)+\left(x^{2^{r}}+1\right) Q(x) B(x) R(x) \\
& =(x+1)^{n} p(x)^{n}-2^{m}+\left(x^{n}+1\right) Q(X) R(x) .
\end{aligned}
$$
Therefore $(x+1)^{n} f(x)+\left(x^{n}+1\right) g(x)=2^{m}$ for some polynomials $f(x), g(x)$ with integer coefficients, and $k_{0}=2^{m}$.
7. We are given that $f(x+a+b)-f(x+a)=f(x+b)-f(x)$, where $a=1 / 6$ and $b=1 / 7$. Summing up these equations for $x, x+b, \ldots, x+6 b$ we obtain $f(x+a+1)-f(x+a)=f(x+1)-f(x)$. Summing up the new equations for $x, x+a, \ldots, x+5 a$ we obtain that
$$
f(x+2)-f(x+1)=f(x+1)-f(x) .
$$
It follows by induction that $f(x+n)-f(x)=n[f(x+1)-f(x)]$. If $f(x+1) \neq f(x)$, then $f(x+n)-f(x)$ will exceed in absolute value an arbitrarily large number for a sufficiently large $n$, contradicting the assumption that $f$ is bounded. Hence $f(x+1)=f(x)$ for all $x$.
8. Putting $m=n=0$ we obtain $f(0)=0$ and consequently $f(f(n))=f(n)$ for all $n$. Thus the given functional equation is equivalent to
$$
f(m+f(n))=f(m)+f(n), \quad f(0)=0
$$
Clearly one solution is $(\forall x) f(x)=0$. Suppose $f$ is not the zero function. We observe that $f$ has nonzero fixed points (for example, any $f(n)$ is a fixed point). Let $a$ be the smallest nonzero fixed point of $f$. By induction, each $k a(k \in \mathbb{N})$ is a fixed point too. We claim that all fixed points of $f$ are of this form. Indeed, suppose that $b=k a+i$ is a fixed point, where $i45^{\circ}$, then $\angle F H P=\angle A$. If $\angle A=45^{\circ}$, the point $P$ escapes to infinity. If $\angle A<45^{\circ}$, the point $P$ appears on the extension of $A C$ over $C$, and $\angle F H P=180^{\circ}-\angle A$.
13. By the law of cosines applied to $\triangle C A_{1} B_{1}$, we obtain
$$
A_{1} B_{1}^{2}=A_{1} C^{2}+B_{1} C^{2}-A_{1} C \cdot B_{1} C \geq A_{1} C \cdot B_{1} C
$$
Analogously, $B_{1} C_{1}^{2} \geq B_{1} A \cdot C_{1} A$ and $C_{1} A_{1}^{2} \geq C_{1} B \cdot A_{1} B$, so that multiplying these inequalities yields
$$
A_{1} B_{1}^{2} \cdot B_{1} C_{1}^{2} \cdot C_{1} A_{1}^{2} \geq A_{1} B \cdot A_{1} C \cdot B_{1} A \cdot B_{1} C \cdot C_{1} A \cdot C_{1} B
$$
Now, the lines $A A_{1}, B B_{1}, C C_{1}$ concur, so by Ceva's theorem, $A_{1} B \cdot B_{1} C$. $C_{1} A=A B_{1} \cdot B C_{1} \cdot C A_{1}$, which together with (1) gives the desired inequality. Equality holds if and only if $C A_{1}=C B_{1}$, etc.
14. Let $a, b, c, d, e$, and $f$ denote the lengths of the sides $A B, B C, C D, D E$, $E F$, and $F A$ respectively.
Note that $\angle A=\angle D, \angle B=\angle E$, and $\angle C=\angle F$. Draw the lines $P Q$ and $R S$ through $A$ and $D$ perpendicular to $B C$ and $E F$ respectively $(P, R \in B C, Q, S \in E F)$. Then $B F \geq P Q=R S$. Therefore $2 B F \geq$ $P Q+R S$, or
$$
\begin{array}{ll}
& 2 B F \geq(a \sin B+f \sin C)+(c \sin C+d \sin B), \\
\text { and similarly, } & 2 B D \geq(c \sin A+b \sin B)+(e \sin B+f \sin A), \\
& 2 D F \geq(e \sin C+d \sin A)+(a \sin A+b \sin C)
\end{array}
$$
Next, we have the following formulas for the considered circumradii:
$$
R_{A}=\frac{B F}{2 \sin A}, \quad R_{C}=\frac{B D}{2 \sin C}, \quad R_{E}=\frac{D F}{2 \sin E}
$$
It follows from (1) that
$$
\begin{aligned}
R_{A}+R_{C}+R_{E} & \geq \frac{1}{4} a\left(\frac{\sin B}{\sin A}+\frac{\sin A}{\sin B}\right)+\frac{1}{4} b\left(\frac{\sin C}{\sin B}+\frac{\sin B}{\sin C}\right)+\cdots \\
& \geq \frac{1}{2}(a+b+\cdots)=\frac{P}{2}
\end{aligned}
$$
with equality if and only if $\angle A=\angle B=\angle C=120^{\circ}$ and $F B \perp B C$ etc., i.e., if and only if the hexagon is regular.
Second solution. Let us construct points $A^{\prime \prime}, C^{\prime \prime}, E^{\prime \prime}$ such that $A B A^{\prime \prime} F$, $C D C^{\prime \prime} B$, and $E F E^{\prime \prime} D$ are parallelograms. It follows that $A^{\prime \prime}, C^{\prime \prime}, B$ are collinear and also $C^{\prime \prime}, E^{\prime \prime}, B$ and $E^{\prime \prime}, A^{\prime \prime}, F$. Furthermore, let $A^{\prime}$ be the intersection of the perpendiculars through $F$ and $B$ to $F A^{\prime \prime}$ and $B A^{\prime \prime}$, respectively, and let $C^{\prime}$ and $E^{\prime}$ be analogously defined. Since $A^{\prime} F A^{\prime \prime} B$ is cyclic with the diameter being $A^{\prime} A^{\prime \prime}$ and since $\triangle F A^{\prime \prime} B \cong$ $\triangle B A F$, it follows that $2 R_{A}=$
$A^{\prime} A^{\prime \prime}=x$.
Similarly, $2 R_{C}=C^{\prime} C^{\prime \prime}=y$ and $2 R_{E}=E^{\prime} E^{\prime \prime}=z$. We also have $A B=$ $F A^{\prime \prime}=y_{a}, A F=A^{\prime \prime} B=z_{a}, C D=C^{\prime \prime} B=z_{c}, C B=C^{\prime \prime} D=x_{c}$, $E F=E^{\prime \prime} D=x_{e}$, and $E D=E^{\prime \prime} F=y_{e}$. The original inequality we must prove now becomes
$$
x+y+z \geq y_{a}+z_{a}+z_{c}+x_{c}+x_{e}+y_{e} .
$$
We now follow and generalize the standard proof of the Erdős-Mordell inequality (for the triangle $A^{\prime} C^{\prime} E^{\prime}$ ), which is what (1) is equivalent to when $A^{\prime \prime}=C^{\prime \prime}=E^{\prime \prime}$.
We set $C^{\prime} E^{\prime}=a, A^{\prime} E^{\prime}=c$ and $A^{\prime} C^{\prime}=e$. Let $A_{1}$ be the point symmetric to $A^{\prime \prime}$ with respect to the bisector of $\angle E^{\prime} A^{\prime} C^{\prime}$. Let $F_{1}$ and $B_{1}$ be the feet of the perpendiculars from $A_{1}$ to $A^{\prime} C^{\prime}$ and $A^{\prime} E^{\prime}$, respectively. In that case, $A_{1} F_{1}=A^{\prime \prime} F=y_{a}$ and $A_{1} B_{1}=A^{\prime \prime} B=z_{a}$. We have
$$
\begin{aligned}
a x=A^{\prime} A_{1} \cdot E^{\prime} C^{\prime} \geq 2 S_{A^{\prime} E^{\prime} A_{1} C^{\prime}} & =2 S_{A^{\prime} E^{\prime} A_{1}}+2 S_{A^{\prime} C^{\prime} A_{1}} \\
& =c z_{a}+e y_{a} .
\end{aligned}
$$
Similarly, $c y \geq e x_{c}+a z_{c}$ and $e z \geq a y_{e}+c x_{e}$. Thus
$$
\begin{aligned}
x+y+z & \geq \frac{c}{a} z_{a}+\frac{a}{c} z_{c}+\frac{e}{c} x_{c}+\frac{c}{e} x_{e}+\frac{a}{e} y_{e}+\frac{e}{a} y_{a} \\
& =\left(\frac{c}{a}+\frac{a}{c}\right)\left(\frac{z_{a}+z_{c}}{2}\right)+\left(\frac{c}{a}-\frac{a}{c}\right)\left(\frac{z_{a}-z_{c}}{2}\right)+\cdots .
\end{aligned}
$$
Let us set $a_{1}=\frac{x_{c}-x_{e}}{2}, c_{1}=\frac{y_{e}-y_{a}}{2}, e_{1}=\frac{z_{a}-z_{c}}{2}$. We note that $\triangle A^{\prime \prime} C^{\prime \prime} E^{\prime \prime} \sim$ $\triangle A^{\prime} C^{\prime} E^{\prime}$ and hence $a_{1} / a=c_{1} / c=e_{1} / e=k$. Thus $\left(\frac{c}{a}-\frac{a}{c}\right) e_{1}+$ $\left(\frac{e}{c}-\frac{c}{e}\right) a_{1}+\left(\frac{a}{e}-\frac{e}{a}\right) c_{1}=k\left(\frac{c e}{a}-\frac{a e}{c}+\frac{e a}{c}-\frac{c a}{e}+\frac{a c}{e}-\frac{e c}{a}\right)=0$. Equation (2) reduces to
$$
\begin{aligned}
x+y+z \geq & \left(\frac{c}{a}+\frac{a}{c}\right)\left(\frac{z_{a}+z_{c}}{2}\right)+\left(\frac{e}{c}+\frac{c}{e}\right)\left(\frac{x_{e}+x_{c}}{2}\right) \\
& +\left(\frac{a}{e}+\frac{e}{a}\right)\left(\frac{y_{a}+y_{e}}{2}\right) .
\end{aligned}
$$
Using $c / a+a / c, e / c+c / e, a / e+e / a \geq 2$ we finally get $x+y+z \geq$ $y_{a}+z_{a}+z_{c}+x_{c}+x_{e}+y_{e}$.
Equality holds if and only if $a=c=e$ and $A^{\prime \prime}=C^{\prime \prime}=E^{\prime \prime}=$ center of $\triangle A^{\prime} C^{\prime} E^{\prime}$, i.e., if and only if $A B C D E F$ is regular.
Remark. From the second proof it is evident that the Erdős-Mordell inequality is a special case of the problem. if $P_{a}, P_{b}, P_{c}$ are the feet of the perpendiculars from a point $P$ inside $\triangle A B C$ to the sides $B C, C A, A B$, and $P_{a} P P_{b} P_{c}^{\prime}, P_{b} P P_{c} P_{a}^{\prime}, P_{c} P P_{a} P_{b}^{\prime}$ parallelograms, we can apply the problem to the hexagon $P_{a} P_{c}^{\prime} P_{b} P_{a}^{\prime} P_{c} P_{b}^{\prime}$ to prove the Erdős-Mordell inequality for $\triangle A B C$ and point $P$.
15. Denote by $A B C D$ and $E F G H$ the two rectangles, where $A B=a, B C=$ $b, E F=c$, and $F G=d$. Obviously, the first rectangle can be placed within the second one with the angle $\alpha$ between $A B$ and $E F$ if and only if
$$
a \cos \alpha+b \sin \alpha \leq c, \quad a \sin \alpha+b \cos \alpha \leq d
$$
Hence $A B C D$ can be placed within $E F G H$ if and only if there is an $\alpha \in[0, \pi / 2]$ for which (1) holds.
The lines $l_{1}(a x+b y=c)$ and $l_{2}(b x+a y=d)$ and the axes $x$ and $y$ bound a region $\mathcal{R}$. By (1), the desired placement of the rectangles is possible if and only if $\mathcal{R}$ contains some point $(\cos \alpha, \sin \alpha)$ of the unit circle centered at the origin $(0,0)$. This in turn holds if and only if the intersection point $L$ of $l_{1}$ and $l_{2}$ lies outside the unit circle. It is easily computed that $L$ has coordinates $\left(\frac{b d-a c}{b^{2}-a^{2}}, \frac{b c-a d}{b^{2}-a^{2}}\right)$. Now $L$ being outside the unit circle is exactly equivalent to the inequality we want to prove.
Remark. If equality holds, there is exactly one way of placing. This happens, for example, when $(a, b)=(5,20)$ and $(c, d)=(16,19)$.
Second remark. This problem is essentially very similar to (SL89-2).
16. Let $A_{1}$ be the point of intersection of $O A^{\prime}$ and $B C$; similarly define $B_{1}$ and $C_{1}$. From the similarity of triangles $O B A_{1}$ and $O A^{\prime} B$ we obtain $O A_{1}$. $O A^{\prime}=R^{2}$. Now it is enough to show that $8 O A_{1} \cdot O B^{\prime} \cdot O C^{\prime} \leq R^{3}$. Thus we must prove that
$$
\lambda \mu \nu \leq \frac{1}{8}, \quad \text { where } \quad \frac{O A_{1}}{O A}=\lambda, \quad \frac{O B_{1}}{O B}=\mu, \quad \frac{O C_{1}}{O C}=\nu
$$
On the other hand, we have
$$
\frac{\lambda}{1+\lambda}+\frac{\mu}{1+\mu}+\frac{\nu}{1+\nu}=\frac{S_{O B C}}{S_{A B C}}+\frac{S_{A O C}}{S_{A B C}}+\frac{S_{A B O}}{S_{A B C}}=1 .
$$
Simplifying this relation, we get
$$
1=\lambda \mu+\mu \nu+\nu \lambda+2 \lambda \mu \nu \geq 3(\lambda \mu \nu)^{2 / 3}+2 \lambda \mu \nu
$$
which cannot hold if $\lambda \mu \nu>\frac{1}{8}$. Hence $\lambda \mu \nu \leq \frac{1}{8}$, with equality if and only if $\lambda=\mu=\nu=\frac{1}{2}$. This implies that $O$ is the centroid of $A B C$, and consequently, that the triangle is equilateral.
Second solution. In the official solution, the inequality to be proved is transformed into
$$
\cos (A-B) \cos (B-C) \cos (C-A) \geq 8 \cos A \cos B \cos C
$$
Since $\frac{\cos (B-C)}{\cos A}=-\frac{\cos (B-C)}{\cos (B+C)}=\frac{\tan B \tan C+1}{\tan B \tan C-1}$, the last inequality becomes $(x y+1)(y z+1)(z x+1) \geq 8(x y-1)(y z-1)(z x-1)$, where we write $x, y, z$ for $\tan A, \tan B, \tan C$. Using the relation $x+y+z=x y z$, we can reduce this inequality to
$$
(2 x+y+z)(x+2 y+z)(x+y+2 z) \geq 8(x+y)(y+z)(z+x)
$$
This follows from the AM-GM inequality: $2 x+y+z=(x+y)+(x+z) \geq$ $2 \sqrt{(x+y)(x+z)}$, etc.
17. Let the diagonals $A C$ and $B D$ meet in $X$. Either $\angle A X B$ or $\angle A X D$ is geater than or equal to $90^{\circ}$, so we assume w.l.o.g. that $\angle A X B \geq 90^{\circ}$. Let $\alpha, \beta, \alpha^{\prime}, \beta^{\prime}$ denote $\angle C A B, \angle A B D, \angle B D C, \angle D C A$. These angles are all acute and satisfy $\alpha+\beta=\alpha^{\prime}+\beta^{\prime}$. Furthermore,
$$
R_{A}=\frac{A D}{2 \sin \beta}, \quad R_{B}=\frac{B C}{2 \sin \alpha}, \quad R_{C}=\frac{B C}{2 \sin \alpha^{\prime}}, \quad R_{D}=\frac{A D}{2 \sin \beta^{\prime}}
$$
Let $\angle B+\angle D=180^{\circ}$. Then $A, B, C, D$ are concyclic and trivially $R_{A}+$ $R_{C}=R_{B}+R_{D}$.
Let $\angle B+\angle D>180^{\circ}$. Then $D$ lies within the circumcircle of $A B C$, which implies that $\beta>\beta^{\prime}$. Similarly $\alpha<\alpha^{\prime}$, so we obtain $R_{A}R_{D}$ and $R_{C}>R_{B}$, so $R_{A}+R_{C}>R_{B}+R_{D}$.
18. We first prove the result in the simplest case. Given a 2 -gon $A B A$ and a point $O$, let $a, b, c, h$ denote $O A, O B, A B$, and the distance of $O$ from $A B$. Then $D=a+b, P=2 c$, and $H=2 h$, so we should show that
$$
(a+b)^{2} \geq 4 h^{2}+c^{2}
$$
Indeed, let $l$ be the line through $O$ parallel to $A B$, and $D$ the point symmetric to $B$ with respect to $l$. Then $(a+b)^{2}=(O A+O B)^{2}=(O A+$ $O D)^{2} \geq A D^{2}=c^{2}+4 h^{2}$.
Now we pass to the general case. Let $A_{1} A_{2} \ldots A_{n}$ be the polygon $\mathcal{F}$ and denote by $d_{i}, p_{i}$, and $h_{i}$ respectively $O A_{i}, A_{i} A_{i+1}$, and the distance of $O$ from $A_{i} A_{i+1}$ (where $A_{n+1}=A_{1}$ ). By the case proved above, we have for each $i, d_{i}+d_{i+1} \geq \sqrt{4 h_{i}^{2}+p_{i}^{2}}$. Summing these inequalities for $i=1, \ldots, n$ and squaring, we obtain
$$
4 D^{2} \geq\left(\sum_{i=1}^{n} \sqrt{4 h_{i}^{2}+p_{i}^{2}}\right)^{2}
$$
It remains only to prove that $\sum_{i=1}^{n} \sqrt{4 h_{i}^{2}+p_{i}^{2}} \geq \sqrt{\sum_{i=1}^{n}\left(4 h_{i}^{2}+p_{i}^{2}\right)}=$ $\sqrt{4 H^{2}+D^{2}}$. But this follows immediately from the Minkowski inequality. Equality holds if and only if it holds in (1) and in the Minkowski inequality, i.e., if and only if $d_{1}=\cdots=d_{n}$ and $h_{1} / p_{1}=\cdots=h_{n} / p_{n}$. This means that $\mathcal{F}$ is inscribed in a circle with center at $O$ and $p_{1}=\cdots=p_{n}$, so $\mathcal{F}$ is a regular polygon and $O$ its center.
19. It is easy to check that after 4 steps we will have all $a, b, c, d$ even. Thus $|a b-c d|,|a c-b d|,|a d-b c|$ remain divisible by 4 , and clearly are not prime. The answer is no.
Second solution. After one step we have $a+b+c+d=0$. Then $a c-b d=$ $a c+b(a+b+c)=(a+b)(b+c)$ etc., so
$$
|a b-c d| \cdot|a c-b d| \cdot|a d-b c|=(a+b)^{2}(a+c)^{2}(b+c)^{2}
$$
However, the product of three primes cannot be a square, hence the answer is $n o$.
20. Let $15 a+16 b=x^{2}$ and $16 a-15 b=y^{2}$, where $x, y \in \mathbb{N}$. Then we obtain $x^{4}+y^{4}=(15 a+16 b)^{2}+(16 a-15 b)^{2}=\left(15^{2}+16^{2}\right)\left(a^{2}+b^{2}\right)=481\left(a^{2}+b^{2}\right)$. In particular, $481=13 \cdot 37 \mid x^{4}+y^{4}$. We have the following lemma.
Lemma. Suppose that $p \mid x^{4}+y^{4}$, where $x, y \in \mathbb{Z}$ and $p$ is an odd prime, where $p \not \equiv 1(\bmod 8)$. Then $p \mid x$ and $p \mid y$.
Proof. Since $p \mid x^{8}-y^{8}$ and by Fermat's theorem $p \mid x^{p-1}-y^{p-1}$, we deduce that $p \mid x^{d}-y^{d}$, where $d=(p-1,8)$. But $d \neq 8$, so $d \mid 4$. Thus $p \mid x^{4}-y^{4}$, which implies that $p \mid 2 y^{4}$, i.e., $p \mid y$ and $p \mid x$.
In particular, we can conclude that $13 \mid x, y$ and $37 \mid x, y$. Hence $x$ and $y$ are divisible by 481 . Thus each of them is at least 481.
On the other hand, $x=y=481$ is possible. It is sufficient to take $a=$ $31 \cdot 481$ and $b=481$.
Second solution. Note that $15 x^{2}+16 y^{2}=481 a^{2}$. It can be directly verified that the divisibility of $15 x^{2}+16 y^{2}$ by 13 and by 37 implies that both $x$ and $y$ are divisible by both primes. Thus $481 \mid x, y$.
21. (a) It clearly suffices to show that for every integer $c$ there exists a quadratic sequence with $a_{0}=0$ and $a_{n}=c$, i.e., that $c$ can be expressed as $\pm 1^{2} \pm 2^{2} \pm \cdots \pm n^{2}$. Since
$$
(n+1)^{2}-(n+2)^{2}-(n+3)^{2}+(n+4)^{2}=4
$$
we observe that if our claim is true for $c$, then it is also true for $c \pm 4$. Thus it remains only to prove the claim for $c=0,1,2,3$. But one immediately finds $1=1^{2}, 2=-1^{2}-2^{2}-3^{2}+4^{2}$, and $3=-1^{2}+2^{2}$, while the case $c=0$ is trivial.
(b) We have $a_{0}=0$ and $a_{n}=1996$. Since $a_{n} \leq 1^{2}+2^{2}+\cdots+n^{2}=$ $\frac{1}{6} n(n+1)(2 n+1)$, we get $a_{17} \leq 1785$, so $n \geq 18$. On the other hand, $a_{18}$ is of the same parity as $1^{2}+2^{2}+\cdots+18^{2}=2109$, so it cannot be equal to 1996. Therefore we must have $n \geq 19$. To construct a required sequence with $n=19$, we note that $1^{2}+2^{2}+\cdots+19^{2}=$ $2470=1996+2 \cdot 237$; hence it is enough to write 237 as a sum of distinct squares. Since $237=14^{2}+5^{2}+4^{2}$, we finally obtain
$$
1996=1^{2}+2^{2}+3^{2}-4^{2}-5^{2}+6^{2}+\cdots+13^{2}-14^{2}+15^{2}+\cdots+19^{2}
$$
22. Let $a, b \in \mathbb{N}$ satisfy the given equation. It is not possible that $a=b$ (since it leads to $a^{2}+2=2 a$ ), so we assume w.l.o.g. that $a>b$. Next, for $a>b=1$ the equation becomes $a^{2}=2 a$, and one obtains a solution $(a, b)=(2,1)$.
Let $b>1$. If $\left[\frac{a^{2}}{b}\right]=\alpha$ and $\left[\frac{b^{2}}{a}\right]=\beta$, then we trivially have $a b \geq$ $\alpha \beta$. Since also $\frac{a^{2}+b^{2}}{a b} \geq 2$, we obtain $\alpha+\beta \geq \alpha \beta+2$, or equivalently
$(\alpha-1)(\beta-1) \leq-1$. But $\alpha \geq 1$, and therefore $\beta=0$. It follows that $a>b^{2}$, i.e., $a=b^{2}+c$ for some $c>0$. Now the given equation becomes $b^{3}+2 b c+\left[\frac{c^{2}}{b}\right]=\left[\frac{b^{4}+2 b^{2} c+b^{2}+c^{2}}{b^{3}+b c}\right]+b^{3}+b c$, which reduces to
$$
(c-1) b+\left[\frac{c^{2}}{b}\right]=\left[\frac{b^{2}(c+1)+c^{2}}{b^{3}+b c}\right]
$$
If $c=1$, then (1) always holds, since both sides are 0 . We obtain a family of solutions $(a, b)=\left(n, n^{2}+1\right)$ or $(a, b)=\left(n^{2}+1, n\right)$. Note that the solution $(1,2)$ found earlier is obtained for $n=1$.
If $c>1$, then $(1)$ implies that $\frac{b^{2}(c+1)+c^{2}}{b^{3}+b c} \geq(c-1) b$. This simplifies to
$$
c^{2}\left(b^{2}-1\right)+b^{2}\left(c\left(b^{2}-2\right)-\left(b^{2}+1\right)\right) \leq 0
$$
Since $c \geq 2$ and $b^{2}-2 \geq 0$, the only possibility is $b=2$. But then (2) becomes $3 c^{2}+8 c-20 \leq 0$, which does not hold for $c \geq 2$.
Hence the only solutions are $\left(n, n^{2}+1\right)$ and $\left(n^{2}+1, n\right), n \in \mathbb{N}$.
23. We first observe that the given functional equation is equivalent to
$$
4 f\left(\frac{(3 m+1)(3 n+1)-1}{3}\right)+1=(4 f(m)+1)(4 f(n)+1) .
$$
This gives us the idea of introducing a function $g: 3 \mathbb{N}_{0}+1 \rightarrow 4 \mathbb{N}_{0}+1$ defined as $g(x)=4 f\left(\frac{x-1}{3}\right)+1$. By the above equality, $g$ will be multiplicative, i.e.,
$$
g(x y)=g(x) g(y) \quad \text { for all } x, y \in 3 \mathbb{N}_{0}+1
$$
Conversely, any multiplicative bijection $g$ from $3 \mathbb{N}_{0}+1$ onto $4 \mathbb{N}_{0}+1$ gives us a function $f$ with the required property: $f(x)=\frac{g(3 x+1)-1}{4}$.
It remains to give an example of such a function $g$. Let $P_{1}, P_{2}, Q_{1}, Q_{2}$ be the sets of primes of the forms $3 k+1,3 k+2,4 k+1$, and $4 k+3$, respectively. It is well known that these sets are infinite. Take any bijection $h$ from $P_{1} \cup P_{2}$ onto $Q_{1} \cup Q_{2}$ that maps $P_{1}$ bijectively onto $Q_{1}$ and $P_{2}$ bijectively onto $Q_{2}$. Now define $g$ as follows: $g(1)=1$, and for $n=p_{1} p_{2} \cdots p_{m}\left(p_{i}\right.$ 's need not be different) define $g(n)=h\left(p_{1}\right) h\left(p_{2}\right) \cdots h\left(p_{m}\right)$. Note that $g$ is well-defined. Indeed, among the $p_{i}$ 's an even number are of the form $3 k+2$, and consequently an even number of $h\left(p_{i}\right) \mathrm{s}$ are of the form $4 k+3$. Hence the product of the $h\left(p_{i}\right)$ 's is of the form $4 k+1$. Also, it is obvious that $g$ is multiplicative. Thus, the defined $g$ satisfies all the required properties.
24. We shall work on the array of lattice points defined by $\mathcal{A}=\left\{(x, y) \in \mathbb{Z}^{2} \mid\right.$ $0 \leq x \leq 19,0 \leq y \leq 11\}$. Our task is to move from $(0,0)$ to $(19,0)$ via the points of $\mathcal{A}$ so that each move has the form $(x, y) \rightarrow(x+a, y+b)$, where $a, b \in \mathbb{Z}$ and $a^{2}+b^{2}=r$.
(a) If $r$ is even, then $a+b$ is even whenever $a^{2}+b^{2}=r(a, b \in \mathbb{Z})$. Thus the parity of $x+y$ does not change after each move, so we cannot reach $(19,0)$ from $(0,0)$.
If $3 \mid r$, then both $a$ and $b$ are divisible by 3 , so if a point $(x, y)$ can be reached from $(0,0)$, we must have $3 \mid x$. Since $3 \nmid 19$, we cannot get to $(19,0)$.
(b) We have $r=73=8^{2}+3^{2}$, so each move is either $(x, y) \rightarrow(x \pm 8, y \pm 3)$ or $(x, y) \rightarrow(x \pm 3, y \pm 8)$. One possible solution is shown in Fig. 1.
(c) We have $97=9^{2}+4^{2}$. Let us partition $\mathcal{A}$ as $\mathcal{B} \cup \mathcal{C}$, where $\mathcal{B}=$ $\{(x, y) \in \mathcal{A} \mid 4 \leq y \leq 7\}$. It is easily seen that moves of the type $(x, y) \rightarrow(x \pm 9, y \pm 4)$ always take us from the set $\mathcal{B}$ to $\mathcal{C}$ and vice versa, while the moves $(x, y) \rightarrow(x \pm 4, y \pm 9)$ always take us from $\mathcal{C}$ to $\mathcal{C}$. Furthermore, each move of the type $(x, y) \rightarrow(x \pm 9, y \pm 4)$ changes the parity of $x$, so to get from $(0,0)$ to $(19,0)$ we must have an odd number of such moves. On the other hand, with an odd number of such moves, starting from $\mathcal{C}$ we can end up only in $\mathcal{B}$, although the point $(19,0)$ is not in $\mathcal{B}$. Hence, the answer is no.
Remark. Part (c) can also be solved by examining all cells that can be reached from $(0,0)$. All these cells are marked in Fig. 2.
Fig. 1
Fig. 2
25. Let the vertices in the bottom row be assigned an arbitrary coloring, and suppose that some two adjacent vertices receive the same color. The number of such colorings equals $2^{n}-2$. It is easy to see that then the colors of the remaining vertices get fixed uniquely in order to satisfy the requirement. So in this case there are $2^{n}-2$ possible colorings.
Next, suppose that the vertices in the bottom row are colored alternately red and blue. There are two such colorings. In this case, the same must hold for every row, and thus we get $2^{n}$ possible colorings.
It follows that the total number of considered colorings is $\left(2^{n}-2\right)+2^{n}=$ $2^{n+1}-2$.
26. Denote the required maximum size by $M_{k}(m, n)$. If $m<\frac{n(n+1)}{2}$, then trivially $M=k$, so from now on we assume that $m \geq \frac{n(n+1)}{2}$.
First we give a lower bound for $M$. Let $r=r_{k}(m, n)$ be the largest integer such that $r+(r+1)+\cdots+(r+n-1) \leq m$. This is equivalent to $n r \leq m-\frac{n(n-1)}{2} \leq n(r+1)$, so $r=\left[\frac{m}{n}-\frac{n-1}{2}\right]$. Clearly no $n$ elements from $\{r+1, r+2, \ldots, k\}$ add up to $m$, so
$$
M \geq k-r_{k}(m, n)=k-\left[\frac{m}{n}-\frac{n-1}{2}\right] .
$$
We claim that $M$ is actually equal to $k-r_{k}(m, n)$. To show this, we shall prove by induction on $n$ that if no $n$ elements of a set $S \subseteq\{1,2, \ldots, k\}$ add up to $m$, then $|S| \leq k-r_{k}(m, n)$.
For $n=2$ the claim is true, because then for each $i=1, \ldots, r_{k}(m, 2)=$ $\left[\frac{m-1}{2}\right]$ at least one of $i$ and $m-i$ must be excluded from $S$. Now let us assume that $n>2$ and that the result holds for $n-1$. Suppose that $S \subseteq\{1,2, \ldots, k\}$ does not contain $n$ distinct elements with the sum $m$, and let $x$ be the smallest element of $S$. We may assume that $x \leq r_{k}(m, n)$, because otherwise the statement is clear. Consider the set $S^{\prime}=\{y-x \mid$ $y \in S, y \neq x\}$. Then $S^{\prime}$ is a subset of $\{1,2, \ldots, k-x\}$ no $n-1$ elements of which have the sum $m-n x$. Also, it is easily checked that $n-1 \leq$ $m-n x-1 \leq k-x$, so we may apply the induction hypothesis, which yields that
$$
|S| \leq 1+k-x-r_{k}(m-n x, n-1)=k-\left[\frac{m-x}{n-1}-\frac{n}{2}\right] .
$$
On the other hand, $\left(\frac{m-x}{n-1}-\frac{n}{2}\right)-r_{k}(m, n)=\frac{m-n x-\frac{n(n-1)}{2}}{n(n-1)} \geq 0$ because $x \leq r_{k}(m, n)$; hence (2) implies $|S| \leq k-r_{k}(m, n)$ as claimed.
27. Suppose that such sets of points $\mathcal{A}, \mathcal{B}$ exist.
First, we observe that there exist five points $A, B, C, D, E$ in $\mathcal{A}$ such that their convex hull does not contain any other point of $\mathcal{A}$. Indeed, take any point $A \in \mathcal{A}$. Since any two points of $\mathcal{A}$ are at distance at least 1 , the number of points $X \in \mathcal{A}$ with $X A \leq r$ is finite for every $r>0$. Thus it is enough to choose four points $B, C, D, E$ of $\mathcal{A}$ that are closest to $A$. Now consider the convex hull $\mathcal{C}$ of $A, B, C, D, E$.
Suppose that $\mathcal{C}$ is a pentagon, say $A B C D E$. Then each of the disjoint triangles $A B C, A C D, A D E$ contains a point of $\mathcal{B}$. Denote these points by $P, Q, R$. Then $\triangle P Q R$ contains some point $F \in \mathcal{A}$, so $F$ is inside $A B C D E$, a contradiction.
Suppose that $\mathcal{C}$ is a quadrilateral, say $A B C D$, with $E$ lying within $A B C D$. Then the triangles $A B E, B C E, C D E, D A E$ contain some points $P, Q, R, S$ of $\mathcal{B}$ that form two disjoint triangles. It follows that there are two points of $\mathcal{A}$ inside $A B C D$, which is a contradiction.
Finally, suppose that $\mathcal{C}$ is a triangle with two points of $\mathcal{A}$ inside. Then $\mathcal{C}$ is the union of five disjoint triangles with vertices in $\mathcal{A}$, so there are at least five points of $\mathcal{B}$ inside $\mathcal{C}$. These five points make at least three disjoint triangles containing three points of $\mathcal{A}$. This is again a contradiction. It follows that no such sets $\mathcal{A}, \mathcal{B}$ exist.
28. Note that w.l.o.g., we can assume that $p$ and $q$ are coprime. Indeed, otherwise it suffices to consider the problem in which all $x_{i}$ 's and $p, q$ are divided by $\operatorname{gcd}(p, q)$.
Let $k, l$ be the number of indices $i$ with $x_{i+1}-x_{i}=p$ and the number of those $i$ with $x_{i+1}-x_{i}=-q(0 \leq i1, k=q t, l=p t$, and $n=(p+q) t$.
Consider the sequence $y_{i}=x_{i+p+q}-x_{i}, i=0, \ldots, n-p-q$. We claim that at least one of the $y_{i}$ 's equals zero. We begin by noting that each $y_{i}$ is of the form $u p-v q$, where $u+v=p+q$; therefore $y_{i}=(u+v) p-$ $v(p+q)=(p-v)(p+q)$ is always divisible by $p+q$. Moreover, $y_{i+1}-y_{i}=$ $\left(x_{i+p+q+1}-x_{i+p+q}\right)-\left(x_{i+1}-x_{i}\right)$ is 0 or $\pm(p+q)$. We conclude that if no $y_{i}$ is 0 then all $y_{i}$ 's are of the same sign. But this is in contradiction with the relation $y_{0}+y_{p+q}+\cdots+y_{n-p-q}=x_{n}-x_{0}=0$. Consequently some $y_{i}$ is zero, as claimed.
Second solution. As before we assume $(p, q)=1$. Let us define a sequence of points $A_{i}\left(y_{i}, z_{i}\right)(i=0,1, \ldots, n)$ in $\mathbb{N}_{0}^{2}$ inductively as follows. Set $A_{0}=$ $(0,0)$ and define $\left(y_{i+1}, z_{i+1}\right)$ as $\left(y_{i}, z_{i}+1\right)$ if $x_{i+1}=x_{i}+p$ and $\left(y_{i}+1, z_{i}\right)$ otherwise. The points $A_{i}$ form a trajectory $L$ in $\mathbb{N}_{0}^{2}$ continuously moving upwards and rightwards by steps of length 1 . Clearly, $x_{i}=p z_{i}-q y_{i}$ for all $i$. Since $x_{n}=0$, it follows that $\left(z_{n}, y_{n}\right)=(k q, k p), k \in \mathbb{N}$. Since $y_{n}+z_{n}=n>p+q$, it follows that $k>1$. We observe that $x_{i}=x_{j}$ if and only if $A_{i} A_{j} \| A_{0} A_{n}$. We shall show that such $i, j$ with $i1$. Such an index $j$ exists, since otherwise the game is over. Then one must make at least one move in the $j$ th cell, which implies that $x_{j}, x_{j}^{\prime} \geq 1$. However, then the sequences $\left\{x_{i}\right\}$ and $\left\{x_{i}^{\prime}\right\}$ with $x_{j}$ and $x_{j}^{\prime}$ decreased by 1 also satisfy (1) for a sequence $\left\{b_{i}\right\}$ where $b_{j-1}, b_{j}, b_{j+1}$ is replaced with $b_{j-1}+1, b_{j}-2, b_{j+1}+1$. This contradicts the assumption of minimal $\min \left\{\sum_{i \in \mathbb{Z}} x_{i}, \sum_{i \in \mathbb{Z}} x_{i}^{\prime}\right\}$ for the initial $\left\{b_{i}\right\}$.
30. For convenience, we shall write $f^{2}, f g, \ldots$ for the functions $f \circ f, f \circ g, \ldots$ We need two lemmas.
Lemma 1. If $f(x) \in S$ and $g(x) \in T$, then $x \in S \cap T$.
Proof. The given condition means that $f^{3}(x)=g^{2} f(x)$ and $g f g(x)=$ $f g^{2}(x)$. Since $x \in S \cup T=U$, we have two cases:
$x \in S$. Then $f^{2}(x)=g^{2}(x)$, which also implies $f^{3}(x)=f g^{2}(x)$. Therefore $g f g(x)=f g^{2}(x)=f^{3}(x)=g^{2} f(x)$, and since $g$ is a bijection, we obtain $f g(x)=g f(x)$, i.e., $x \in T$.
$x \in T$. Then $f g(x)=g f(x)$, so $g^{2} f(x)=g f g(x)$. It follows that $f^{3}(x)=g^{2} f(x)=g f g(x)=f g^{2}(x)$, and since $f$ is a bijection, we obtain $x \in S$.
Hence $x \in S \cap T$ in both cases. Similarly, $f(x) \in T$ and $g(x) \in S$ again imply $x \in S \cap T$.
Lemma 2. $f(S \cap T)=g(S \cap T)=S \cap T$.
Proof. By symmetry, it is enough to prove $f(S \cap T)=S \cap T$, or in other words that $f^{-1}(S \cap T)=S \cap T$. Since $S \cap T$ is finite, this is equivalent to $f(S \cap T) \subseteq S \cap T$.
Let $f(x) \in S \cap T$. Then if $g(x) \in S$ (since $f(x) \in T$ ), Lemma 1 gives $x \in S \cap T$; similarly, if $g(x) \in T$, then by Lemma $1, x \in S \cap T$.
Now we return to the problem. Assume that $f(x) \in S$. If $g(x) \notin S$, then $g(x) \in T$, so from Lemma 1 we deduce that $x \in S \cap T$. Then Lemma 2 claims that $g(x) \in S \cap T$ too, a contradiction. Analogously, from $g(x) \in S$ we are led to $f(x) \in S$. This finishes the proof.
### 4.38 Solutions to the Shortlisted Problems of IMO 1997
1. Let $A B C$ be the given triangle, with $\angle B=90^{\circ}$ and $A B=m, B C=n$. For an arbitrary polygon $\mathcal{P}$ we denote by $w(\mathcal{P})$ and $b(\mathcal{P})$ respectively the total areas of the white and black parts of $\mathcal{P}$.
(a) Let $D$ be the fourth vertex of the rectangle $A B C D$. When $m$ and $n$ are of the same parity, the coloring of the rectangle $A B C D$ is centrally symmetric with respect to the midpoint of $A C$. It follows that $w(A B C)=\frac{1}{2} w(A B C D)$ and $b(A B C)=\frac{1}{2} b(A B C D)$; thus $f(m, n)=\frac{1}{2}|w(A B C D)-b(A B C D)|$. Hence $f(m, n)$ equals $\frac{1}{2}$ if $m$ and $n$ are both odd, and 0 otherwise.
(b) The result when $m, n$ are of the same parity follows from (a). Suppose that $m>n$, where $m$ and $n$ are of different parity. Choose a point $E$ on $A B$ such that $A E=1$. Since by (a) $|w(E B C)-b(E B C)|=$ $f(m-1, n) \leq \frac{1}{2}$, we have $f(m, n) \leq \frac{1}{2}+|w(E A C)-b(E A C)| \leq$ $\frac{1}{2}+S(E A C)=\frac{1}{2}+\frac{n-1}{2}=\frac{n}{2}$. Therefore $f(m, n) \leq \frac{1}{2} \min (m, n)$.
(c) Let us calculate $f(m, n)$ for $m=2 k+1, n=2 k, k \in \mathbb{N}$. With $E$ defined as in (b), we have $B E=B C=2 k$. If the square at $B$ is w.l.o.g. white, $C E$ passes only through black squares. The white part of $\triangle E A C$ then consists of $2 k$ similar triangles with areas $\frac{1}{2} \frac{i}{2 k} \frac{i}{2 k+1}=\frac{i^{2}}{4 k(2 k+1)}$, where $i=1,2, \ldots, 2 k$. The total white area of $E A C$ is
$$
\frac{1}{4 k(2 k+1)}\left(1^{2}+2^{2}+\cdots+(2 k)^{2}\right)=\frac{4 k+1}{12} .
$$
Therefore the black area is $(8 k-1) / 12$, and $f(2 k+1,2 k)=(2 k-1) / 6$, which is not bounded.
2. For any sequence $X=\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ let us define
$$
\bar{X}=\left(1,2, \ldots, x_{1}, 1,2, \ldots, x_{2}, \ldots, 1,2, \ldots, x_{n}\right)
$$
Also, for any two sequences $A, B$ we denote their concatenation by $A B$. It clearly holds that $\overline{A B}=\bar{A} \bar{B}$. The sequences $R_{1}, R_{2}, \ldots$ are given by $R_{1}=(1)$ and $R_{n}=\overline{R_{n-1}}(n)$ for $n>1$.
We consider the family of sequences $Q_{n i}$ for $n, i \in \mathbb{N}, i \leq n$, defined as follows:
$Q_{n 1}=(1), \quad Q_{n n}=(n), \quad$ and $\quad Q_{n i}=Q_{n-1, i-1} Q_{n-1, i} \quad$ if $11$. Let $x \in\{1,2, \ldots, 2 n-1\}$ be a fixed number that does not appear on the fixed diagonal of $A$. Such an element must exist, since the diagonal can contain at most $n$ different numbers. Let us call the union of the $i$ th row and the $i$ th column the $i$ th cross. There are $n$ crosses, and each of them contains exactly one $x$. On the other hand, each entry $x$ of $A$ is contained in exactly two crosses. Hence $n$ must be even. However, 1997 is an odd number; hence no coveralls matrix exists for $n=1997$.
(b) For $n=2, A_{2}=\left[\begin{array}{ll}1 & 2 \\ 3 & 1\end{array}\right]$ is a coveralls matrix. For $n=4$, one such matrix is, for example,
$$
A_{4}=\left[\begin{array}{llll}
1 & 2 & 5 & 6 \\
3 & 1 & 7 & 5 \\
4 & 6 & 1 & 2 \\
7 & 4 & 3 & 1
\end{array}\right]
$$
This construction can be generalized. Suppose that we are given an $n \times n$ coveralls matrix $A_{n}$. Let $B_{n}$ be the matrix obtained from $A_{n}$ by adding $2 n$ to each entry, and $C_{n}$ the matrix obtained from $B_{n}$ by replacing each diagonal entry (equal to $2 n+1$ by induction) with $2 n$. Then the matrix
$$
A_{2 n}=\left[\begin{array}{ll}
A_{n} & B_{n} \\
C_{n} & A_{n}
\end{array}\right]
$$
is coveralls. To show this, suppose that $i \leq n$ (the case $i>n$ is similar). The $i$ th cross is composed of the $i$ th cross of $A_{n}$, the $i$ th row of $B_{n}$, and the $i$ th column of $C_{n}$. The $i$ th cross of $A_{i}$ covers $1,2, \ldots, 2 n-1$. The $i$ th row of $B_{n}$ covers all numbers of the form $2 n+j$, where $j$ is covered by the $i$ th row of $A_{n}$ (including $j=1$ ). Similarly, the $i$ th column of $C_{n}$ covers $2 n$ and all numbers of the form $2 n+k$, where $k>1$ is covered by the $i$ th column of $A_{n}$. Thus we see that all numbers are accounted for in the $i$ th cross of $A_{2 n}$, and hence $A_{2 n}$ is a desired coveralls matrix. It follows that we can find a coveralls matrix whenever $n$ is a power of 2 .
Second solution for part $b$. We construct a coveralls matrix explicitly for $n=2^{k}$. We consider the coordinates/cells of the matrix elements modulo $n$ throughout the solution. We define the $i$-diagonal $(0 \leq i<$ $n$ ) to be the set of cells of the form $(j, j+i)$, for all $j$. We note that each cross contains exactly one cell from the 0 -diagonal (the main diagonal) and two cells from each $i$-diagonal. For two cells within an $i$ diagonal, $x$ and $y$, we define $x$ and $y$ to be related if there exists a cross containing both $x$ and $y$. Evidently, for every cell $x$ not on the 0 -diagonal there are exactly two other cells related to it. The relation thus breaks up each $i$-diagonal $(i>0)$ into cycles of length larger than 1 . Due to the diagonal translational symmetry (modulo $n$ ), all the cycles within a given $i$-diagonal must be of equal length and thus of an even length, since $n=2^{k}$.
The construction of a coveralls matrix is now obvious. We select a number, say 1, to place on all the cells of the 0-diagonal. We pair up the remaining numbers and assign each pair to an $i$-diagonal, say $(2 i, 2 i+1)$. Going along each cycle within the $i$-diagonal we alternately assign values of $2 i$ and $2 i+1$. Since the cycle has an even length, a cell will be related only to a cell of a different number, and hence each cross will contain both $2 i$ and $2 i+1$.
5. We shall prove first the 2-dimensional analogue:
Lemma. Given an equilateral triangle $A B C$ and two points $M, N$ on the sides $A B$ and $A C$ respectively, there exists a triangle with sides $C M, B N, M N$.
Proof. Consider a regular tetrahedron $A B C D$. Since $C M=D M$ and $B N=D N$, one such triangle is $D M N$.
Now, to solve the problem for a regular tetrahedron $A B C D$, we consider a 4-dimensional polytope $A B C D E$ whose faces $A B C D, A B C E, A B D E$, $A C D E, B C D E$ are regular tetrahedra. We don't know what it looks like, but it yields a desired triangle: for $M \in A B C$ and $N \in A D C$, we have $D M=E M$ and $B N=E N$; hence the desired triangle is $E M N$.
Remark. A solution that avoids embedding in $\mathbb{R}^{4}$ is possible, but no longer so short.
6. (a) One solution is
$$
x=2^{n^{2}} 3^{n+1}, \quad y=2^{n^{2}-n} 3^{n}, \quad z=2^{n^{2}-2 n+2} 3^{n-1} .
$$
(b) Suppose w.l.o.g. that $\operatorname{gcd}(c, a)=1$. We look for a solution of the form
$$
x=p^{m}, \quad y=p^{n}, \quad z=q p^{r}, \quad p, q, m, n, r \in \mathbb{N} .
$$
Then $x^{a}+y^{b}=p^{m a}+p^{n b}$ and $z^{c}=q^{c} p^{r c}$, and we see that it is enough to assume $m a-1=n b=r c$ (there are infinitely many such triples $(m, n, r))$ and $q^{c}=p+1$.
7. Let us set $A C=a, C E=b, E A=c$. Applying Ptolemy's inequality for the quadrilateral $A C E F$ we get
$$
A C \cdot E F+C E \cdot A F \geq A E \cdot C F
$$
Since $E F=A F$, this implies $\frac{F A}{F C} \geq \frac{c}{a+b}$. Similarly $\frac{B C}{B E} \geq \frac{a}{b+c}$ and $\frac{D E}{D A} \geq$ $\frac{b}{c+a}$. Now,
$$
\frac{B C}{B E}+\frac{D E}{D A}+\frac{F A}{F C} \geq \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}
$$
Hence it is enough to prove that
$$
\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geq \frac{3}{2}
$$
If we now substitute $x=b+c, y=c+a, z=a+b$ and $S=a+b+c$ the inequality (1) becomes equivalent to $S(1 / x+1 / y+1 / y)-3 \geq 3 / 2$ which follows immediately form $1 / x+1 / y+1 / z \geq 9 /(x+y+z)=9 /(2 S)$. Equality occurs if it holds in Ptolemy's inequalities and also $a=b=c$. The former happens if and only if the hexagon is cyclic. Hence the only case of equality is when $A B C D E F$ is regular.
8. (a) Denote by $b$ and $c$ the perpendicular bisectors of $A B$ and $A C$ respectively. If w.l.o.g. $b$ and $A D$ do not intersect (are parallel), then $\angle B C D=\angle B A D=90^{\circ}$, a contradiction. Hence $V, W$ are well-defined. Now, $\angle D W B=2 \angle D A B$ and $\angle D V C=2 \angle D A C$ as oriented angles, and therefore $\angle(W B, V C)=2(\angle D V C-\angle D W B)=2 \angle B A C=$ $2 \angle B C D$ is not equal to 0 . Consequently $C V$ and $B W$ meet at some $T$ with $\angle B T C=2 \angle B A C$.
(b) Let $B^{\prime}$ be the second point of intersection of $B W$ with $\Gamma$. Clearly $A D=B B^{\prime}$. But we also have $\angle B T C=2 \angle B A C=2 \angle B B^{\prime} C$, which implies that $C T=T B^{\prime}$. It follows that $A D=B B^{\prime}=\left|B T \pm T B^{\prime}\right|=$ $|B T \pm C T|$.
Remark. This problem is also solved easily using trigonometry.
9. For $i=1,2,3$ (all indices in this problem will be modulo 3 ) we denote by $O_{i}$ the center of $C_{i}$ and by $M_{i}$ the midpoint of the $\operatorname{arc} A_{i+1} A_{i+2}$ that does not contain $A_{i}$. First we have that $O_{i+1} O_{i+2}$ is the perpendicular bisector of $I B_{i}$, and thus it contains the circumcenter $R_{i}$ of $A_{i} B_{i} I$. Additionally, it is easy to show that $T_{i+1} A_{i}=T_{i+1} I$ and $T_{i+2} A_{i}=$ $T_{i+2} I$, which implies that $R_{i}$ lies on the line $T_{i+1} T_{i+2}$. Therefore $R_{i}=$ $O_{i+1} O_{i+2} \cap T_{i+1} T_{i+2}$.
Now, the lines $T_{1} O_{1}, T_{2} O_{2}, T_{3} O_{3}$ are concurrent at $I$. By Desargues's theorem, the points of intersection of $O_{i+1} O_{i+2}$ and $T_{i+1} T_{i+2}$, i.e., the $R_{i}$ 's, lie on a line for $i=1,2,3$.
Second solution. The centers of three circles passing through the same point $I$ and not touching each other are collinear if and only if they have another common point. Hence it is enough to show that the circles $A_{i} B_{i} I$ have a common point other than $I$.
Now apply inversion at center $I$ and with an arbitrary power. We shall denote by $X^{\prime}$ the image of $X$ under this inversion. In our case, the image of the circle $C_{i}$ is the line $B_{i+1}^{\prime} B_{i+2}^{\prime}$ while the image of the line $A_{i+1} A_{i+2}$ is the circle $I A_{i+1}^{\prime} A_{i+2}^{\prime}$ that is tangent to $B_{i}^{\prime} B_{i+2}^{\prime}$, and $B_{i}^{\prime} B_{i+2}^{\prime}$. These three circles have equal radii, so their centers $P_{1}, P_{2}, P_{3}$ form a triangle also homothetic to $\triangle B_{1}^{\prime} B_{2}^{\prime} B_{3}^{\prime}$. Consequently, points $A_{1}^{\prime}, A_{2}^{\prime}, A_{3}^{\prime}$, that are the reflections of $I$ across the sides of $P_{1} P_{2} P_{3}$, are vertices of a triangle also homothetic to $B_{1}^{\prime} B_{2}^{\prime} B_{3}^{\prime}$. It follows that $A_{1}^{\prime} B_{1}^{\prime}, A_{2}^{\prime} B_{2}^{\prime}, A_{3}^{\prime} B_{3}^{\prime}$ are concurrent at some point $J^{\prime}$, i.e., that the circles $A_{i} B_{i} I$ all pass through $J$.
10. Suppose that $k \geq 4$. Consider any polynomial $F(x)$ with integer coefficients such that $0 \leq F(x) \leq k$ for $x=0,1, \ldots, k+1$. Since $F(k+1)-F(0)$ is divisible by $k+1$, we must have $F(k+1)=F(0)$. Hence
$$
F(x)-F(0)=x(x-k-1) Q(x)
$$
for some polynomial $Q(x)$ with integer coefficients. In particular, $F(x)-$ $F(0)$ is divisible by $x(k+1-x)>k+1$ for every $x=2,3, \ldots, k-1$, so $F(x)=F(0)$ must hold for any $x=2,3, \ldots, k-1$. It follows that
$$
F(x)-F(0)=x(x-2)(x-3) \cdots(x-k+1)(x-k-1) R(x)
$$
for some polynomial $R(x)$ with integer coefficients. Thus $k \geq \mid F(1)-$ $F(0)|=k(k-2)!| R(1) \mid$, although $k(k-2)!>k$ for $k \geq 4$. In this case we have $F(1)=F(0)$ and similarly $F(k)=F(0)$. Hence, the statement is true for $k \geq 4$.
It is easy to find counterexamples for $k \leq 3$. These are, for example,
$$
F(x)= \begin{cases}x(2-x) & \text { for } k=1 \\ x(3-x) & \text { for } k=2 \\ x(2-x)^{2}(4-x) & \text { for } k=3\end{cases}
$$
11. All real roots of $P(x)$ (if any) are negative: say $-a_{1},-a_{2}, \ldots,-a_{k}$. Then $P(x)$ can be factored as
$$
P(x)=C\left(x+a_{1}\right) \cdots\left(x+a_{k}\right)\left(x^{2}-b_{1} x+c_{1}\right) \cdots\left(x^{2}-b_{m} x+c_{m}\right)
$$
where $x^{2}-b_{i} x+c_{i}$ are quadratic polynomials without real roots.
Since the product of polynomials with positive coefficients is again a polynomial with positive coefficients, it will be sufficient to prove the result for each of the factors in (1). The case of $x+a_{j}$ is trivial. It remains only to prove the claim for every polynomial $x^{2}-b x+c$ with $b^{2}<4 c$.
From the binomial formula, we have for any $n \in \mathbb{N}$,
$$
(1+x)^{n}\left(x^{2}-b x+c\right)=\sum_{i=0}^{n+2}\left[\binom{n}{i-2}-b\binom{n}{i-1}+c\binom{n}{i}\right] x^{i}=\sum_{i=0}^{n+2} C_{i} x^{i}
$$
where
$$
C_{i}=\frac{n!\left((b+c+1) i^{2}-((b+2 c) n+(2 b+3 c+1)) i+c\left(n^{2}+3 n+2\right)\right) x^{i}}{i!(n-i+2)!}
$$
The coefficients $C_{i}$ of $x^{i}$ appear in the form of a quadratic polynomial in $i$ depending on $n$. We claim that for large enough $n$ this polynomial has negative discriminant, and is thus positive for every $i$. Indeed, this discriminant equals $D=((b+2 c) n+(2 b+3 c+1))^{2}-4(b+c+1) c\left(n^{2}+\right.$ $3 n+2)=\left(b^{2}-4 c\right) n^{2}-2 U n+V$, where $U=2 b^{2}+b c+b-4 c$ and $V=(2 b+c+1)^{2}-4 c$, and since $b^{2}-4 c<0$, for large $n$ it clearly holds that $D<0$.
12. Lemma. For any polynomial $P$ of degree at most $n$, the following equality holds:
$$
\sum_{i=0}^{n+1}(-1)^{i}\binom{n+1}{i} P(i)=0
$$
Proof. See (SL81-13).
Suppose to the contrary that the degree of $f$ is at most $p-2$. Then it follows from the lemma that
$$
0=\sum_{i=0}^{p-1}(-1)^{i}\binom{p-1}{i} f(i) \equiv \sum_{i=0}^{p-1} f(i)(\bmod p)
$$
since $\binom{p-1}{i}=\frac{(p-1)(p-2) \cdots(p-i)}{i!} \equiv(-1)^{i}(\bmod p)$. But this is clearly impossible if $f(i)$ equals 0 or 1 modulo $p$ and $f(0)=0, f(1)=1$.
Remark. In proving the essential relation $\sum_{i=0}^{p-1} f(i) \equiv 0(\bmod p)$, it is clearly enough to show that $S_{k}=1^{k}+2^{k}+\cdots+(p-1)^{k}$ is divisible by $p$ for every $k \leq p-2$. This can be shown in two other ways.
(1) By induction. Assume that $S_{0} \equiv \cdots \equiv S_{k-1}(\bmod p)$. By the binomial formula we have
$$
0 \equiv \sum_{n=0}^{p-1}\left[(n+1)^{k+1}-n^{k+1}\right] \equiv(k+1) S_{k}+\sum_{i=0}^{k-1}\binom{k+1}{i} S_{i}(\bmod p),
$$
and the inductive step follows.
(2) Using the primitive root $g$ modulo $p$. Then
$$
S_{k} \equiv 1+g^{k}+\cdots+g^{k(p-2)}=\frac{g^{k(p-1)}-1}{g^{k}-1} \equiv 0(\bmod p) .
$$
13. Denote $A(r)$ and $B(r)$ by $A(n, r)$ and $B(n, r)$ respectively. The numbers $A(n, r)$ can be found directly: one can choose $r$ girls and $r$ boys in $\binom{n}{r}^{2}$ ways, and pair them in $r$ ! ways. Hence
$$
A(n, r)=\binom{n}{r}^{2} \cdot r!=\frac{n!^{2}}{(n-r)!^{2} r!}
$$
Now we establish a recurrence relation between the $B(n, r)$ 's. Let $n \geq 2$ and $2 \leq r \leq n$. There are two cases for a desired selection of $r$ pairs of girls and boys:
(i) One of the girls dancing is $g_{n}$. Then the other $r-1$ girls can choose their partners in $B(n-1, r-1)$ ways and $g_{n}$ can choose any of the remaining $2 n-r$ boys. Thus, the total number of choices in this case is $(2 n-r) B(n-1, r-1)$.
(ii) $g_{n}$ is not dancing. Then there are exactly $B(n-1, r)$ possible choices. Therefore, for every $n \geq 2$ it holds that
$$
B(n, r)=(2 n-r) B(n-1, r-1)+B(n-1, r) \quad \text { for } r=2, \ldots, n
$$
Here we assume that $B(n, r)=0$ for $r>n$, while $B(n, 1)=1+3+\cdots+$ $(2 n-1)=n^{2}$ 。
It is directly verified that the numbers $A(n, r)$ satisfy the same initial conditions and recurrence relations, from which it follows that $A(n, r)=$ $B(n, r)$ for all $n$ and $r \leq n$.
14. We use the following nonstandard notation: ( $1^{\circ}$ ) for $x, y \in \mathbb{N}, x \sim y$ means that $x$ and $y$ have the same prime divisors; $\left(2^{\circ}\right)$ for a prime $p$ and integers $r \geq 0$ and $x>0, p^{r} \| x$ means that $x$ is divisible by $p^{r}$, but not by $p^{r+1}$. First, $b^{m}-1 \sim b^{n}-1$ is obviously equivalent to $b^{m}-1 \sim \operatorname{gcd}\left(b^{m}-1, b^{n}-\right.$ $1)=b^{d}-1$, where $d=\operatorname{gcd}(m, n)$. Setting $b^{d}=a$ and $m=k d$, we reduce
the condition of the problem to $a^{k}-1 \sim a-1$. We are going to show that this implies that $a+1$ is a power of 2 . This will imply that $d$ is odd (for even $d$, $a+1=b^{d}+1$ cannot be divisible by 4 , and consequently $b+1$, as a divisor of $a+1$, is also a power of 2 . But before that, we need the following important lemma (Theorem 2.126).
Lemma. Let $a, k$ be positive integers and $p$ an odd prime. If $\alpha \geq 1$ and $\beta \geq 0$ are such that $p^{\alpha} \| a-1$ and $p^{\beta} \| k$, then $p^{\alpha+\beta} \| a^{k}-1$.
Proof. We use induction on $\beta$. If $\beta=0$, then $\frac{a^{k}-1}{a-1}=a^{k-1}+\cdots+a+1 \equiv k$ $(\bmod p)($ because $a \equiv 1$ ), and it is not divisible by $p$.
Suppose that the lemma is true for some $\beta \geq 0$, and let $k=p^{\beta+1} t$ where $p \nmid t$. By the induction hypothesis, $a^{k / p}=a^{p^{\beta} t}=m p^{\alpha+\beta}+1$ for some $m$ not divisible by $p$. Furthermore,
$$
a^{k}-1=\left(m p^{\alpha+\beta}+1\right)^{p}-1=\left(m p^{\alpha+\beta}\right)^{p}+\cdots+\binom{p}{2}\left(m p^{\alpha+\beta}\right)^{2}+m p^{\alpha+\beta+1}
$$
Since $p \left\lvert\,\binom{ p}{2}=\frac{p(p-1)}{2}\right.$, all summands except for the last one are divisible by $p^{\alpha+\beta+2}$. Hence $p^{\alpha+\beta+1} \| a^{k}-1$, completing the induction. Now let $a^{k}-1 \sim a-1$ for some $a, k>1$. Suppose that $p$ is an odd prime divisor of $k$, with $p^{\beta} \| k$. Then putting $X=a^{p^{\beta}-1}+\cdots+a+1$ we also have $(a-1) X=a^{p^{\beta}}-1 \sim a-1$; hence each prime divisor $q$ of $X$ must also divide $a-1$. But then $a^{i} \equiv 1(\bmod q)$ for each $i \in \mathbb{N}_{0}$, which gives us $X \equiv p^{\beta}(\bmod q)$. Therefore $q \mid p^{\beta}$, i.e., $q=p$; hence $X$ is a power of $p$. On the other hand, since $p \mid a-1$, we put $p^{\alpha} \| a-1$. From the lemma we obtain $p^{\alpha+\beta} \| a^{p^{\beta}}-1$, and deduce that $p^{\beta} \| X$. But $X$ has no prime divisors other than $p$, so we must have $X=p^{\beta}$. This is clearly impossible, because $X>p^{\beta}$ for $a>1$. Thus our assumption that $k$ has an odd prime divisor leads to a contradiction: in other words, $k$ must be a power of 2 . Now $a^{k}-1 \sim a-1$ implies $a-1 \sim a^{2}-1=(a-1)(a+1)$, and thus every prime divisor $q$ of $a+1$ must also divide $a-1$. Consequently $q=2$, so it follows that $a+1$ is a power of 2 . As we explained above, this gives that $b+1$ is also a power of 2 .
Remark. In fact, one can continue and show that $k$ must be equal to 2 . It is not possible for $a^{4}-1 \sim a^{2}-1$ to hold. Similarly, we must have $d=1$. Therefore all possible triples $(b, m, n)$ with $m>n$ are $\left(2^{s}-1,2,1\right)$.
15. Let $a+b t, t=0,1,2, \ldots$, be a given arithmetic progression that contains a square and a cube $(a, b>0)$. We use induction on the progression step $b$ to prove that the progression contains a sixth power.
(i) $b=1$ : this case is trivial.
(ii) $b=p^{m}$ for some prime $p$ and $m>0$. The case $p^{m} \mid a$ trivially reduces to the previous case, so let us have $p^{m} \nmid a$.
Suppose that $\operatorname{gcd}(a, p)=1$. If $x, y$ are integers such that $x^{2} \equiv y^{3} \equiv a$ (here all the congruences will be $\bmod p^{m}$ ), then $x^{6} \equiv a^{3}$ and $y^{6} \equiv a^{2}$. Consider an integer $y_{1}$ such that $y y_{1} \equiv 1$. It satisfies $a^{2}\left(x y_{1}\right)^{6} \equiv$
$x^{6} y^{6} y_{1}^{6} \equiv x^{6} \equiv a^{3}$, and consequently $\left(x y_{1}\right)^{6} \equiv a$. Hence a sixth power exists in the progression.
If $\operatorname{gcd}(a, p)>1$, we can write $a=p^{k} c$, where $k1$ and $\operatorname{gcd}\left(b_{1}, b_{2}\right)=1$. It is given that progressions $a+b_{1} t$ and $a+b_{2} t$ both contain a square and a cube, and therefore by the inductive hypothesis they both contain sixth powers: say $z_{1}^{6}$ and $z_{2}^{6}$, respectively. By the Chinese remainder theorem, there exists $z \in \mathbb{N}$ such that $z \equiv z_{1}\left(\bmod b_{1}\right)$ and $z \equiv z_{2}\left(\bmod b_{2}\right)$. But then $z^{6}$ belongs to both of the progressions $a+b_{1} t$ and $a+b_{2} t$. Hence $z^{6}$ is a member of the progression $a+b t$.
16. Let $d_{a}(X), d_{b}(X), d_{c}(X)$ denote the distances of a point $X$ interior to $\triangle A B C$ from the lines $B C, C A, A B$ respectively. We claim that $X \in P Q$ if and only if $d_{a}(X)+d_{b}(X)=d_{c}(X)$. Indeed, if $X \in P Q$ and $P X=$ $k P Q$ then $d_{a}(X)=k d_{a}(Q), d_{b}(X)=(1-k) d_{b}(P)$, and $d_{c}(X)=(1-$ $k) d_{c}(P)+k d_{c}(Q)$, and simple substitution yields $d_{a}(X)+d_{b}(X)=d_{c}(X)$. The converse follows easily. In particular, $O \in P Q$ if and only if $d_{a}(O)+$ $d_{b}(O)=d_{c}(O)$, i.e., $\cos \alpha+\cos \beta=\cos \gamma$.
We shall now show that $I \in D E$ if and only if $A E+B D=D E$. Let $K$ be the point on the segment $D E$ such that $A E=E K$. Then $\angle E K A=$ $\frac{1}{2} \angle D E C=\frac{1}{2} \angle C B A=\angle I B A$; hence the points $A, B, I, K$ are concyclic. The point $I$ lies on $D E$ if and only if $\angle B K D=\angle B A I=\frac{1}{2} \angle B A C=$ $\frac{1}{2} \angle C D E=\angle D B K$, which is equivalent to $K D=B D$, i.e., to $A E+B D=$ $D E$. But since $A E=A B \cos \alpha, B D=A B \cos \beta$, and $D E=A B \cos \gamma$, we have that $I \in D E \Leftrightarrow \cos \alpha+\cos \beta=\cos \gamma$. The conditions for $O \in P Q$ and $I \in D E$ are thus equivalent.
Second solution. We know that three points $X, Y, Z$ are collinear if and only if for some $\lambda, \mu \in \mathbb{R}$ with sum 1 , we have $\lambda \overrightarrow{C X}+\mu \overrightarrow{C Y}=\overrightarrow{C Z}$. Specially, if $\overrightarrow{C X}=p \overrightarrow{C A}$ and $\overrightarrow{C Y}=q \overrightarrow{C B}$ for some $p, q$, and $\overrightarrow{C Z}=k \overrightarrow{C A}+$ $l \overrightarrow{C B}$, then $Z$ lies on $X Y$ if and only if $k q+l p=p q$.
Using known relations in a triangle we directly obtain
$$
\begin{array}{rlrl}
\overrightarrow{C P} & =\frac{\sin \beta}{\sin \beta+\sin \gamma} \overrightarrow{C B}, & \overrightarrow{C Q}=\frac{\sin \alpha}{\sin \alpha+\sin \gamma} \overrightarrow{C A} \\
\overrightarrow{C O} & =\frac{\sin 2 \alpha \cdot \overrightarrow{C A}+\sin 2 \beta \cdot \overrightarrow{C B}}{\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma} ; & \overrightarrow{C D}=\frac{\tan \beta}{\tan \beta+\tan \gamma} \overrightarrow{C B} \\
\overrightarrow{C E} & =\frac{\tan \beta}{\tan \beta+\tan \gamma} \overrightarrow{C A}, & \overrightarrow{C I} & =\frac{\sin \alpha \cdot \overrightarrow{C A}+\sin \beta \cdot \overrightarrow{C B}}{\sin \alpha+\sin \beta+\sin \gamma}
\end{array}
$$
Now by the above considerations we get that the conditions (1) $P, Q, O$ are collinear and (2) $D, E, I$ are collinear are both equivalent to $\cos \alpha+\cos \beta=$ $\cos \gamma$.
17. We note first that $x$ and $y$ must be powers of the same positive integer. Indeed, if $x=p_{1}^{\alpha_{1}} \cdots p_{k}^{\alpha_{k}}$ and $y=p_{1}^{\beta_{1}} \cdots p_{k}^{\beta_{k}}$ (some of $\alpha_{i}$ and $\beta_{i}$ may be 0 , but not both for the same index $i$ ), then $x^{y^{2}}=y^{x}$ implies $\frac{\alpha_{i}}{\beta_{i}}=\frac{x}{y^{2}}=\frac{p}{q}$ for some $p, q>0$ with $\operatorname{gcd}(p, q)=1$, so for $a=p_{1}^{\alpha_{1} / p} \cdots p_{k}^{\alpha_{k} / p}$ we can take $x=a^{p}$ and $y=a^{q}$.
If $a=1$, then $(x, y)=(1,1)$ is the trivial solution. Let $a>1$. The given equation becomes $a^{p a^{2 q}}=a^{q a^{p}}$, which reduces to $p a^{2 q}=q a^{p}$. Hence $p \neq q$, so we distinguish two cases:
(i) $p>q$. Then from $a^{2 q}2 q$. We can rewrite the equation as $p=a^{p-2 q} q$, and putting $p=2 q+d, d>0$, we obtain $d=q\left(a^{d}-2\right)$. By induction, $2^{d}-2>d$ for each $d>2$, so we must have $d \leq 2$. For $d=1$ we get $q=1$ and $a=p=3$, and therefore $(x, y)=(27,3)$, which is indeed a solution. For $d=2$ we get $q=1$, $a=2$, and $p=4$, so $(x, y)=(16,2)$, which is another solution.
(ii) $p0$, this is transformed to $a^{d}=a^{\left(2 a^{d}-1\right) p}$, or equivalently to $d=\left(2 a^{d}-1\right) p$. However, this equality cannot hold, because $2 a^{d}-1>d$ for each $a \geq 2$, $d \geq 1$.
The only solutions are thus $(1,1),(16,2)$, and $(27,3)$.
18. By symmetry, assume that $A B>A C$. The point $D$ lies between $M$ and $P$ as well as between $Q$ and $R$, and if we show that $D M \cdot D P=D Q \cdot D R$, it will imply that $M, P, Q, R$ lie on a circle.
Since the triangles $A B C, A E F, A Q R$ are similar, the points $B, C, Q, R$ lie on a circle. Hence $D B \cdot D C=D Q \cdot D R$, and it remains to prove that
$$
D B \cdot D C=D M \cdot D P
$$
However, the points $B, C, E, F$ are concyclic, but so are the points $E, F, D, M$ (they lie on the nine-point circle), and we obtain $P B \cdot P C=$ $P E \cdot P F=P D \cdot P M$. Set $P B=x$ and $P C=y$. We have $P M=\frac{x+y}{2}$ and hence $P D=\frac{2 x y}{x+y}$. It follows that $D B=P B-P D=\frac{x(x-y)}{x+y}$, $D C=\frac{y(x-y)}{x+y}$, and $D M=\frac{(x-y)^{2}}{2(x+y)}$, from which we immediately obtain $D B \cdot D C=D M \cdot D P=\frac{x y(x-y)^{2}}{(x+y)^{2}}$, as needed.
19. Using that $a_{n+1}=0$ we can transform the desired inequality into
$$
\begin{aligned}
& \sqrt{a_{1}+}+a_{2}+\cdots+a_{n+1} \\
& \quad \leq \sqrt{1} \sqrt{a_{1}}+(\sqrt{2}-\sqrt{1}) \sqrt{a_{2}}+\cdots+(\sqrt{n+1}-\sqrt{n}) \sqrt{a_{n+1}}
\end{aligned}
$$
We shall prove by induction on $n$ that (1) holds for any $a_{1} \geq a_{2} \geq \cdots \geq$ $a_{n+1} \geq 0$, i.e., not only when $a_{n+1}=0$. For $n=0$ the inequality is
obvious. For the inductive step from $n-1$ to $n$, where $n \geq 1$, we need to prove the inequality
$$
\sqrt{a_{1}+\cdots+a_{n+1}}-\sqrt{a_{1}+\cdots+a_{n}} \leq(\sqrt{n+1}-\sqrt{n}) \sqrt{a_{n+1}}
$$
Putting $S=a_{1}+a_{2}+\cdots+a_{n}$, this simplifies to $\sqrt{S+a_{n+1}}-\sqrt{S} \leq$ $\sqrt{n a_{n+1}+a_{n+1}}-\sqrt{n a_{n+1}}$. For $a_{n+1}=0$ the inequality is obvious. For $a_{n+1}>0$ we have that the function $f(x)=\sqrt{x+a_{n+1}}-\sqrt{x}=$ $\frac{a_{n+1}}{\sqrt{x+a_{n+1}}+\sqrt{x}}$ is strictly decreasing on $\mathbb{R}^{+}$; hence (2) will follow if we show that $S \geq n a_{n+1}$. However, this last is true because $a_{1}, \ldots, a_{n} \geq a_{n+1}$.
Equality holds if and only if $a_{1}=a_{2}=\cdots=a_{k}$ and $a_{k+1}=\cdots=a_{n+1}=$ 0 for some $k$.
Second solution. Setting $b_{k}=\sqrt{a_{k}}-\sqrt{a_{k+1}}$ for $k=1, \ldots, n$ we have $a_{i}=\left(b_{i}+\cdots+b_{n}\right)^{2}$, so the desired inequality after squaring becomes
$$
\sum_{k=1}^{n} k b_{k}^{2}+2 \sum_{1 \leq k\frac{n+1}{2}$ for any $\pi$.
Further, we note that if $\pi^{\prime}$ is obtained from $\pi$ by interchanging two neighboring elements, say $y_{k}$ and $y_{k+1}$, then $S(\pi)$ and $S\left(\pi^{\prime}\right)$ differ by $\left|y_{k}+y_{k+1}\right| \leq n+1$, and consequently they must be of the same sign.
Now consider the identity permutation $\pi_{0}=\left(x_{1}, \ldots, x_{n}\right)$ and the reverse permutation $\overline{\pi_{0}}=\left(x_{n}, \ldots, x_{1}\right)$. There is a sequence of permutations $\pi_{0}, \pi_{1}, \ldots, \pi_{m}=\overline{\pi_{0}}$ such that for each $i, \pi_{i+1}$ is obtained from $\pi_{i}$ by interchanging two neighboring elements. Indeed, by successive interchanges we can put $x_{n}$ in the first place, then $x_{n-1}$ in the second place, etc. Hence all $S\left(\pi_{0}\right), \ldots, S\left(\pi_{m}\right)$ are of the same sign. However, since $\left|S\left(\pi_{0}\right)+S\left(\pi_{m}\right)\right|=(n+1)\left|x_{1}+\cdots+x_{n}\right|=n+1$, this implies that one of
$S\left(\pi_{0}\right)$ and $S\left(\overline{\pi_{0}}\right)$ is smaller than $\frac{n+1}{2}$ in absolute value, contradicting the initial assumption.
22. (a) Suppose that $f$ and $g$ are such functions. From $g(f(x))=x^{3}$ we have $f\left(x_{1}\right) \neq f\left(x_{2}\right)$ whenever $x_{1} \neq x_{2}$. In particular, $f(-1), f(0)$, and $f(1)$ are three distinct numbers. However, since $f(x)^{2}=f(g(f(x)))=$ $f\left(x^{3}\right)$, each of the numbers $f(-1), f(0), f(1)$ is equal to its square, and so must be either 0 or 1 . This contradiction shows that no such $f, g$ exist.
(b) The answer is yes. We begin with constructing functions $F, G:(1, \infty)$ $\rightarrow(1, \infty)$ with the property $F(G(x))=x^{2}$ and $G(F(x))=x^{4}$ for $x>$ 1. Define the functions $\varphi, \psi$ by $F\left(2^{2^{t}}\right)=2^{2^{\varphi(t)}}$ and $G\left(2^{2^{t}}\right)=2^{2^{\psi(t)}}$. These functions determine $F$ and $G$ on the entire interval $(1, \infty)$, and satisfy $\varphi(\psi(t))=t+1$ and $\psi(\varphi(t))=t+2$. It is easy to find examples of $\varphi$ and $\psi$ : for example, $\varphi(t)=\frac{1}{2} t+1, \psi(t)=2 t$. Thus we also arrive at an example for $F, G$ :
$$
F(x)=2^{2^{\frac{1}{2} \log _{2} \log _{2} x+1}}=2^{2 \sqrt{\log _{2} x}}, \quad G(x)=2^{2^{2 \log _{2} \log _{2} x}}=2^{\log _{2}^{2} x}
$$
It remains only to extend these functions to the whole of $\mathbb{R}$. This can be done as follows:
$$
\widetilde{f}(x)= \begin{cases}F(x) & \text { for } x>1 \\
1 / F(1 / x) & \text { for } 0 1 } \\
{ x } & { \text { for } x \in \{ 0 , 1 \} ; }
\end{array} \left\{\begin{array}{cl}
1 / G(1 / x) & \text { for } 0\beta_{4}$. Then point $A$ lies inside the circle $B C D$, which is further equivalent to $\beta_{1}>\alpha_{2}$. On the other hand, from $\alpha_{1}+\beta_{2}=\alpha_{3}+\beta_{4}$ we deduce $\alpha_{3}>\beta_{2}$, and similarly $\beta_{3}>\alpha_{4}$. Therefore,
since the cosine is strictly decreasing on $(0, \pi)$, the left side of (1) is strictly negative, yielding a contradiction.
24. There is a bijective correspondence between representations in the given form of $2 k$ and $2 k+1$ for $k=0,1, \ldots$, since adding 1 to every representation of $2 k$, we obtain a representation of $2 k+1$, and conversely, every representation of $2 k+1$ contains at least one 1 , which can be removed. Hence, $f(2 k+1)=f(2 k)$.
Consider all representations of $2 k$. The number of those that contain at least one 1 equals $f(2 k-1)=f(2 k-2)$, while the number of those not containing a 1 equals $f(k)$ (the correspondence is given by division of summands by 2). Therefore
$$
f(2 k)=f(2 k-2)+f(k) .
$$
Summing these equalities over $k=1, \ldots, n$, we obtain
$$
f(2 n)=f(0)+f(1)+\cdots+f(n)
$$
We first prove the right-hand inequality. Since $f$ is increasing, and $f(0)+$ $f(1)=f(2)$, (2) yields $f(2 n) \leq n f(n)$ for $n \geq 2$. Now $f\left(2^{3}\right)=f(0)+$ $\cdots+f(4)=10<2^{3^{2} / 2}$, and one can easily conclude by induction that $f\left(2^{n+1}\right) \leq 2^{n} f\left(2^{n}\right)<2^{n} \cdot 2^{n^{2} / 2}<2^{(n+1)^{2} / 2}$ for each $n \geq 3$.
We now derive the lower estimate. It follows from (1) that $f(x+2)-f(x)$ is increasing. Consequently, for each $m$ and $k(2 m-1) f(2 m)$, which together with the above inequality gives
$$
f(8 m)=f(0)+f(1)+\cdots+f(4 m)>4 m f(2 m) .
$$
Finally, we have that the inequality $f\left(2^{n}\right)>2^{n^{2} / 4}$ holds for $n=2$ and $n=3$, while for larger $n$ we have by induction $f\left(2^{n}\right)>2^{n-1} f\left(2^{n-2}\right)>$ $2^{n-1+(n-2)^{2} / 4}=2^{n^{2} / 4}$. This completes the proof.
Remark. Despite the fact that the lower estimate is more difficult, it is much weaker than the upper estimate. It can be shown that $f\left(2^{n}\right)$ eventually (for large $n$ ) exceeds $2^{c n^{2}}$ for any $c<\frac{1}{2}$.
25. Let $M R$ meet the circumcircle of triangle $A B C$ again at a point $X$. We claim that $X$ is the common point of the lines $K P, L Q, M R$. By symmetry, it will be enough to show that $X$ lies on $K P$. It is easy to see that $X$ and $P$ lie on the same side of $A B$ as $K$. Let $I_{a}=A K \cap B P$ be the excenter of $\triangle A B C$ corresponding to $A$. It is easy to calculate that $\angle A I_{a} B=\gamma / 2$, from which we get $\angle R P B=\angle A I_{a} B=\angle M C B=\angle R X B$. Therefore $R, B, P, X$ are concyclic. Now if $P$ and $K$ are on distinct sides of $B X$ (the
other case is similar), we have $\angle R X P=180^{\circ}-\angle R B P=90^{\circ}-$ $\beta / 2=\angle M A K=180^{\circ}-\angle R X K$, from which it follows that $K, X, P$ are collinear, as claimed.
Remark. It is not essential for the statement of the problem that $R$ be an internal point of $A B$. Work with cases can be avoided using oriented
angles.
26. Let us first examine the case that all the inequalities in the problem are actually equalities. Then $a_{n-2}=a_{n-1}+a_{n}, a_{n-3}=2 a_{n-1}+a_{n}, \ldots, a_{0}=$ $F_{n} a_{n-1}+F_{n-1} a_{n}=1$, where $F_{n}$ is the $n$th Fibonacci number. Then it is easy to see (from $F_{1}+F_{2}+\cdots+F_{k}=F_{k+2}$ ) that $a_{0}+\cdots+a_{n}=$ $\left(F_{n+2}-1\right) a_{n-1}+F_{n+1} a_{n}=\frac{F_{n+2}-1}{F_{n}}+\left(F_{n+1}-\frac{F_{n-1}\left(F_{n+2}-1\right)}{F_{n}}\right) a_{n}$. Since $\frac{F_{n-1}\left(F_{n+2}-1\right)}{F_{n}} \leq F_{n+1}$, it follows that $a_{0}+a_{1}+\cdots+a_{n} \geq \frac{F_{n+2}-1}{F_{n}}$, with equality holding if and only if $a_{n}=0$ and $a_{n-1}=\frac{1}{F_{n}}$.
We denote by $M_{n}$ the required minimum in the general case. We shall prove by induction that $M_{n}=\frac{F_{n+2}-1}{F_{n}}$. For $M_{1}=1$ and $M_{2}=2$ it is easy to show that the formula holds; hence the inductive basis is true. Suppose that $n>2$. The sequences $1, \frac{a_{2}}{a_{1}}, \ldots, \frac{a_{n}}{a_{1}}$ and $1, \frac{a_{3}}{a_{2}}, \ldots, \frac{a_{n}}{a_{2}}$ also satisfy the conditions of the problem. Hence we have
$$
a_{0}+\cdots+a_{n}=a_{0}+a_{1}\left(1+\frac{a_{2}}{a_{1}}+\cdots+\frac{a_{n}}{a_{1}}\right) \geq 1+a_{1} M_{n-1}
$$
and
$$
a_{0}+\cdots+a_{n}=a_{0}+a_{1}+a_{2}\left(1+\frac{a_{3}}{a_{2}}+\cdots+\frac{a_{n}}{a_{2}}\right) \geq 1+a_{1}+a_{2} M_{n-2}
$$
Multiplying the first inequality by $M_{n-2}-1$ and the second one by $M_{n-1}$, adding the inequalities and using that $a_{1}+a_{2} \geq 1$, we obtain ( $M_{n-1}+$ $\left.M_{n-2}+1\right)\left(a_{0}+\cdots+a_{n}\right) \geq M_{n-1} M_{n-2}+M_{n-1}+M_{n-2}+1$, so
$$
M_{n} \geq \frac{M_{n-1} M_{n-2}+M_{n-1}+M_{n-2}+1}{M_{n-1}+M_{n-2}+1}
$$
Since $M_{n-1}=\frac{F_{n+1}-1}{F_{n-1}}$ and $M_{n-2}=\frac{F_{n}-1}{F_{n-2}}$, the above inequality easily yields $M_{n} \geq \frac{F_{n+2}-1}{F_{n}}$. However, we have shown above that equality can occur; hence $\frac{F_{n+2}-1}{F_{n}}$ is indeed the required minimum.
### 4.39 Solutions to the Shortlisted Problems of IMO 1998
1. We begin with the following observation: Suppose that $P$ lies in $\triangle A E B$, where $E$ is the intersection of $A C$ and $B D$ (the other cases are similar). Let $M, N$ be the feet of the perpendiculars from $P$ to $A C$ and $B D$ respectively. We have $S_{A B P}=S_{A B E}-S_{A E P}-S_{B E P}=\frac{1}{2}(A E \cdot B E-A E \cdot E N-B E$. $E M)=\frac{1}{2}(A M \cdot B N-E M \cdot E N)$. Similarly, $S_{C D P}=\frac{1}{2}(C M \cdot D N-E M$. $E N)$. Therefore, we obtain
$$
S_{A B P}-S_{C D P}=\frac{A M \cdot B N-C M \cdot D N}{2}
$$
Now suppose that $A B C D$ is cyclic. Then $P$ is the circumcenter of $A B C D$; hence $M$ and $N$ are the midpoints of $A C$ and $B D$. Hence $A M=C M$ and $B N=D N$; thus (1) gives us $S_{A B P}=S_{C D P}$. On the other hand, suppose that
$A B C D$ is not cyclic and let w.l.o.g. $P A=P B>P C=P D$. Then we must have $A M>C M$ and $B N>$ $D N$, and consequently by (1), $S_{A B P}>S_{C D P}$. This proves the other implication.
Second solution. Let $F$ and $G$ denote the midpoints of $A B$ and $C D$, and assume that $P$ is on the same side of $F G$ as $B$ and $C$. Since $P F \perp A B$, $P G \perp C D$, and $\angle F E B=\angle A B E, \angle G E C=\angle D C E$, a direct computation yields $\angle F P G=\angle F E G=90^{\circ}+\angle A B E+\angle D C E$.
Taking into account that $S_{A B P}=\frac{1}{2} A B \cdot F P=F E \cdot F P$, we note that $S_{A B P}=S_{C D P}$ is equivalent to $F E \cdot F P=G E \cdot G P$, i.e., to $F E / E G=$ $G P / P F$. But this last is equivalent to triangles $E F G$ and $P G F$ being similar, which holds if and only if $E F P G$ is a parallelogram. This last is equivalent to $\angle E F P=\angle E G P$, or $2 \angle A B E=2 \angle D C E$. Thus $S_{A B P}=$ $S_{C D P}$ is equivalent to $A B C D$ being cyclic.
Remark. The problems also allows an analytic solution, for example putting the $x$ and $y$ axes along the diagonals $A C$ and $B D$.
2. If $A D$ and $B C$ are parallel, then $A B C D$ is an isosceles trapezoid with $A B=C D$, so $P$ is the midpoint of $E F$. Let $M$ and $N$ be the midpoints of $A B$ and $C D$. Then $M N \| B C$, and the distance $d(E, M N)$ equals the distance $d(F, M N)$ because $B$ and $D$ are the same distance from $M N$ and $E M / B M=F N / D N$. It follows that the midpoint $P$ of $E F$ lies on $M N$, and consequently $S_{A P D}: S_{B P C}=A D: B C$.
If $A D$ and $B C$ are not parallel, then they meet at some point $Q$. It is plain that $\triangle Q A B \sim \triangle Q C D$, and since $A E / A B=C F / C D$, we also deduce that $\triangle Q A E \sim \triangle Q C F$. Therefore $\angle A Q E=\angle C Q F$. Further, from these similarities we obtain $Q E / Q F=Q A / Q C=A B / C D=P E / P F$,
which in turn means that $Q P$ is the internal bisector of $\angle E Q F$. But since $\angle A Q E=\angle C Q F$, this is also the internal bisector of $\angle A Q B$. Hence $P$ is at equal distances from $A D$ and $B C$, so again $S_{A P D}: S_{B P C}=A D: B C$. Remark. The part $A B \| C D$ could also be regarded as a limiting case of the other part.
Second solution. Denote $\lambda=\frac{A E}{A B}, A B=a, B C=b, C D=c, D A=d$, $\angle D A B=\alpha, \angle A B C=\beta$. Since $d(P, A D)=\frac{c \cdot d(E, A D)+a \cdot d(F, A D)}{a+c}$, we have $S_{A P D}=\frac{c S_{E A D}+a S_{F A D}}{a+c}=\frac{\lambda c S_{A B D}+(1-\lambda) a S_{A C D}}{a+c}$. Since $S_{A B D}=\frac{1}{2} a d \sin \alpha$ and $S_{A C D}=\frac{1}{2} c d \sin \beta$, we are led to $S_{A P D}=\frac{a c d}{a+c}[\lambda \sin \alpha+(1-\lambda) \sin \beta]$, and analogously $S_{B P C}=\frac{a b c}{a+c}[\lambda \sin \alpha+(1-\lambda) \sin \beta]$. Thus we obtain $S_{A P D}: S_{B P C}=d: b$.
3. Lemma. If $U, W, V$ are three points on a line $l$ in this order, and $X$ a point in the plane with $X W \perp U V$, then $\angle U X V<90^{\circ}$ if and only if $X W^{2}>U W \cdot V W$.
Proof. Let $X W^{2}>U W \cdot V W$, and let $X_{0}$ be a point on the segment $X W$ such that $X_{0} W^{2} \geq U W \cdot V W$. Then $X_{0} W / U W=V W / X_{0} W$, so that triangles $X_{0} W U$ and $V W X_{0}$ are similar. Thus $\angle U X_{0} V=\angle U X_{0} W+$ $\angle W U X_{0}=90^{\circ}$, which immediately implies that $\angle U X V<90^{\circ}$.
Similarly, if $X W^{2} \leq U W \cdot V W$, then $\angle U X V \geq 90^{\circ}$.
Since $B I \perp R S$, it will be enough by the lemma to show that $B I^{2}>$ $B R \cdot B S$. Note that $\triangle B K R \sim \triangle B S L$ : in fact, we have $\angle K B R=\angle S B L=$ $90^{\circ}-\beta / 2$ and $\angle B K R=\angle A K M=\angle K L M=\angle B S L=90^{\circ}-\alpha / 2$. In particular, we obtain $B R / B K=B L / B S=B K / B S$, so that $B R \cdot B S=$ $B K^{2}B K$ and $B M>B L$. We conclude that $\angle M B E<\frac{1}{2} \angle M B K$ and $\angle M B F<\frac{1}{2} \angle M B L$. Adding these two inequalities gives $\angle E B F<$ $\beta / 2$. Therefore $\angle R I S<90^{\circ}$.
Remark. It can be shown (using vectors) that the statement remains true for an arbitrary line $t$ passing through $B$.
4. Let $K$ be the point on the ray $B N$ with $\angle B C K=\angle B M A$. Since $\angle K B C=\angle A B M$, we get $\triangle B C K \sim \triangle B M A$. It follows that $B C / B M=$ $B K / B A$, which implies that also $\triangle B A K \sim \triangle B M C$. The quadrilateral $A N C K$ is cyclic, because $\angle B K C=\angle B A M=\angle N A C$. Then by Ptolemy's theorem we obtain
$$
A C \cdot B K=A C \cdot B N+A N \cdot C K+C N \cdot A K
$$
On the other hand, from the similarities noted above we get
$$
C K=\frac{B C \cdot A M}{B M}, A K=\frac{A B \cdot C M}{B M} \text { and } B K=\frac{A B \cdot B C}{B M} .
$$
After substitution of these values, the equality (1) becomes
$$
\frac{A B \cdot B C \cdot A C}{B M}=A C \cdot B N+\frac{B C \cdot A M \cdot A N}{B M}+\frac{A B \cdot C M \cdot C N}{B M},
$$
which is exactly the equality we must prove multiplied by $\frac{A B \cdot B C \cdot C A}{B M}$.
5. Let $G$ be the centroid of $\triangle A B C$ and $\mathcal{H}$ the homothety with center $G$ and ratio $-\frac{1}{2}$. It is well-known that $\mathcal{H}$ maps $H$ into $O$. For every other point $X$, let us denote by $X^{\prime}$ its image under $\mathcal{H}$. Also, let $A_{2} B_{2} C_{2}$ be the triangle in which $A, B, C$ are the midpoints of $B_{2} C_{2}, C_{2} A_{2}$, and $A_{2} B_{2}$, respectively.
It is clear that $A^{\prime}, B^{\prime}, C^{\prime}$ are the midpoints of sides $B C, C A, A B$ respectively. Furthermore, $D^{\prime}$ is the reflection of $A^{\prime}$ across $B^{\prime} C^{\prime}$. Thus $D^{\prime}$ must lie on $B_{2} C_{2}$ and $A^{\prime} D^{\prime} \perp$
$B_{2} C_{2}$. However, it also holds that $O A^{\prime} \perp B_{2} C_{2}$, so we conclude that $O, D^{\prime}, A^{\prime}$ are collinear and $D^{\prime}$ is the projection of $O$ on $B_{2} C_{2}$. Analogously, $E^{\prime}, F^{\prime}$ are the projections of $O$ on $C_{2} A_{2}$ and $A_{2} B_{2}$.
Now we apply Simson's theorem. It claims that $D^{\prime}, E^{\prime}, F^{\prime}$ are collinear (which is equivalent to $D, E, F$ being collinear) if and only if $O$ lies on the circumcircle of $A_{2} B_{2} C_{2}$. However, this circumcircle is centered at $H$ with radius $2 R$, so the last condition is equivalent to $H O=2 R$.
6. Let $P$ be the point such that $\triangle C D P$ and $\triangle C B A$ are similar and equally oriented. Since then $\angle D C P=\angle B C A$ and $\frac{B C}{C A}=\frac{D C}{C P}$, it follows that $\angle A C P=\angle B C D$ and $\frac{A C}{C P}=\frac{B C}{C D}$, so $\triangle A C P \sim \triangle B C D$. In particular, $\frac{B C}{C A}=\frac{D B}{P A}$.
Furthermore, by the conditions of the problem we have $\angle E D P=360^{\circ}-$ $\angle B-\angle D=\angle F$ and $\frac{P D}{D E}=\frac{P D}{C D} \cdot \frac{C D}{D E}=\frac{A B}{B C} \cdot \frac{C D}{D E}=\frac{A F}{F E}$. Therefore $\triangle E D P \sim \triangle E F A$ as well, so that similarly as above we conclude that $\triangle A E P \sim \triangle F E D$ and consequently $\frac{A E}{E F}=\frac{P A}{F D}$.
Finally, $\frac{B C}{C A} \cdot \frac{A E}{E F} \cdot \frac{F D}{D B}=\frac{D B}{P A} \cdot \frac{P A}{F D} \cdot \frac{F D}{D B}=1$.
Second solution. Let $a, b, c, d, e, f$ be the complex coordinates of $A, B$, $C, D, E, F$, respectively. The condition of the problem implies that $\frac{a-b}{b-c}$. $\frac{c-d}{d-e} \cdot \frac{e-f}{f-a}=-1$.
On the other hand, since $(a-b)(c-d)(e-f)+(b-c)(d-e)(f-a)=$ $(b-c)(a-e)(f-d)+(c-a)(e-f)(d-b)$ holds identically, we immediately deduce that $\frac{b-c}{c-a} \cdot \frac{a-e}{e-f} \cdot \frac{f-d}{d-b}=-1$. Taking absolute values gives $\frac{B C}{C A} \cdot \frac{A E}{E F}$. $\frac{F D}{D B}=1$.
7. We shall use the following result.
Lemma. In a triangle $A B C$ with $B C=a, C A=b$, and $A B=c$,
i. $\angle C=2 \angle B$ if and only if $c^{2}=b^{2}+a b$;
ii. $\angle C+180^{\circ}=2 \angle B$ if and only if $c^{2}=b^{2}-a b$.
Proof.
i. Take a point $D$ on the extension of $B C$ over $C$ such that $C D=b$. The condition $\angle C=2 \angle B$ is equivalent to $\angle A D C=\frac{1}{2} \angle C=\angle B$, and thus to $A D=A B=c$. This is further equivalent to triangles $C A D$ and $A B D$ being similar, so $C A / A D=A B / B D$, i.e., $c^{2}=$ $b(a+b)$.
ii. Take a point $E$ on the ray $C B$ such that $C E=b$. As above, $\angle C+180^{\circ}=2 \angle B$ if and only if $\triangle C A E \sim \triangle A B E$, which is equivalent to $E B / B A=E A / A C$, or $c^{2}=b(b-a)$.
Let $F, G$ be points on the ray $C B$ such that $C F=\frac{1}{3} a$ and $C G=\frac{4}{3} a$. Set $B C=a, C A=b, A B=c, E C=b_{1}$, and $E B=c_{1}$. By the lemma it follows that $c^{2}=b^{2}+a b$. Also $b_{1}=A G$ and $c_{1}=A F$, so Stewart's theorem gives us $c_{1}^{2}=\frac{2}{3} b^{2}+\frac{1}{3} c^{2}-\frac{2}{9} a^{2}=b^{2}+\frac{1}{3} a b-\frac{2}{9} a^{2}$ and $b_{1}^{2}=$ $-\frac{1}{3} b^{2}+\frac{4}{3} c^{2}+\frac{4}{9} a^{2}=b^{2}+\frac{4}{3} a b+\frac{4}{9} a^{2}$. It follows that $b_{1}=\frac{2}{3} a+b$ and $c_{1}^{2}=b_{1}^{2}-\left(a b+\frac{2}{3} a^{2}\right)=b_{1}^{2}-a b_{1}$. The statement of the problem follows immediately by the lemma.
8. Let $M$ be the point of intersection of $A E$ and $B C$, and let $N$ be the point on $\omega$ diametrically opposite $A$.
Since $\angle B<\angle C$, points $N$ and $B$ are on the same side of $A E$. Furthermore, $\angle N A E=\angle B A X=$ $90^{\circ}-\angle A B E$; hence the triangles $N A E$ and $B A X$ are similar. Consequently, $\triangle B A Y$ and $\triangle N A M$ are also similar, since $M$ is the midpoint
of $A E$. Thus $\angle A N Z=\angle A B Z=\angle A B Y=\angle A N M$, implying that $N, M, Z$ are collinear. Now we have $\angle Z M D=90^{\circ}-\angle Z M A=\angle E A Z=$ $\angle Z E D$ (the last equality because $E D$ is tangent to $\omega$ ); hence $Z M E D$ is a cyclic quadrilateral. It follows that $\angle Z D M=\angle Z E A=\angle Z A D$, which is enough to conclude that $M D$ is tangent to the circumcircle of $A Z D$.
Remark. The statement remains valid if $\angle B \geq \angle C$.
9. Set $a_{n+1}=1-\left(a_{1}+\cdots+a_{n}\right)$. Then $a_{n+1}>0$, and the desired inequality becomes
$$
\frac{a_{1} a_{2} \cdots a_{n+1}}{\left(1-a_{1}\right)\left(1-a_{2}\right) \cdots\left(1-a_{n+1}\right)} \leq \frac{1}{n^{n+1}}
$$
To prove it, we observe that
$1-a_{i}=a_{1}+\cdots+a_{i-1}+a_{i+1}+\cdots+a_{n+1} \geq n \sqrt[n]{a_{1} \cdots a_{i-1} a_{i+1} \cdots a_{n+1}}$.
Multiplying these inequalities for $i=1,2, \ldots, n+1$, we get exactly the inequality we need.
10. We shall first prove the inequality for $n$ of the form $2^{k}, k=0,1,2, \ldots$ The case $k=0$ is clear. For $k=1$, we have
$$
\frac{1}{r_{1}+1}+\frac{1}{r_{2}+1}-\frac{2}{\sqrt{r_{1} r_{2}}+1}=\frac{\left(\sqrt{r_{1} r_{2}}-1\right)\left(\sqrt{r_{1}}-\sqrt{r_{2}}\right)^{2}}{\left(r_{1}+1\right)\left(r_{2}+1\right)\left(\sqrt{r_{1} r_{2}}+1\right)} \geq 0
$$
For the inductive step it suffices to show that the claim for $k$ and 2 implies that for $k+1$. Indeed,
$$
\begin{aligned}
\sum_{i=1}^{2^{k+1}} \frac{1}{r_{i}+1} & \geq \frac{2^{k}}{\sqrt[2^{k}]{r_{1} r_{2} \cdots r_{2^{k}}}+1}+\frac{2^{k}}{\sqrt[2^{k}]{r_{2^{k}+1^{\prime} r_{2^{k}}+2^{\cdots r_{2^{k+1}}}+1}}} \\
& \geq \frac{2^{k+1}}{\sqrt[2^{k+1}]{r_{1} r_{2} \cdots r_{2^{k+1}}}+1}
\end{aligned}
$$
and the induction is complete.
We now show that if the statement holds for $2^{k}$, then it holds for every $n<2^{k}$ as well. Put $r_{n+1}=r_{n+2}=\cdots=r_{2^{k}}=\sqrt[n]{r_{1} r_{2} \ldots r_{n}}$. Then (1) becomes
$$
\frac{1}{r_{1}+1}+\cdots+\frac{1}{r_{n}+1}+\frac{2^{k}-n}{\sqrt[n]{r_{1} \cdots r_{n}}+1} \geq \frac{2^{k}}{\sqrt[n]{r_{1} \cdots r_{n}}+1}
$$
This proves the claim.
Second solution. Define $r_{i}=e^{x_{i}}$, where $x_{i}>0$. The function $f(x)=\frac{1}{1+e^{x}}$ is convex for $x>0$ : indeed, $f^{\prime \prime}(x)=\frac{e^{x}\left(e^{x}-1\right)}{\left(e^{x}+1\right)^{3}}>0$. Thus by Jensen's inequality applied to $f\left(x_{1}\right), \ldots, f\left(x_{n}\right)$, we get $\frac{1}{r_{1}+1}+\cdots+\frac{1}{r_{n}+1} \geq \frac{n}{\sqrt[n]{r_{1} \cdots r_{n}}+1}$.
11. The given inequality is equivalent to $x^{3}(x+1)+y^{3}(y+1)+z^{3}(z+1) \geq$ $\frac{3}{4}(x+1)(y+1)(z+1)$. By the A-G mean inequality, it will be enough to prove a stronger inequality:
$$
x^{4}+x^{3}+y^{4}+y^{3}+z^{4}+z^{3} \geq \frac{1}{4}\left[(x+1)^{3}+(y+1)^{3}+(z+1)^{3}\right] .
$$
If we set $S_{k}=x^{k}+y^{k}+z^{k}$, (1) takes the form $S_{4}+S_{3} \geq \frac{1}{4} S_{3}+\frac{3}{4} S_{2}+\frac{3}{4} S_{1}+\frac{3}{4}$. Note that by the A-G mean inequality, $S_{1}=x+y+z \geq 3$. Thus it suffices to prove the following:
$$
\text { If } S_{1} \geq 3 \text { and } m>n \text { are positive integers, then } S_{m} \geq S_{n} \text {. }
$$
This can be shown in many ways. For example, by Hölder's inequality,
$$
\left(x^{m}+y^{m}+z^{m}\right)^{n / m}(1+1+1)^{(m-n) / m} \geq x^{n}+y^{n}+z^{n} .
$$
(Another way is using the Chebyshev inequality: if $x \geq y \geq z$ then $x^{k-1} \geq$ $y^{k-1} \geq z^{k-1}$; hence $S_{k}=x \cdot x^{k-1}+y \cdot y^{k-1}+z \cdot z^{k-1} \geq \frac{1}{3} S_{1} S_{k-1}$, and the claim follows by induction.)
Second solution. Assume that $x \geq y \geq z$. Then also $\frac{1}{(y+1)(z+1)} \geq$ $\frac{1}{(x+1)(z+1)} \geq \frac{1}{(x+1)(y+1)}$. Hence Chebyshev's inequality gives that
$$
\begin{aligned}
& \frac{x^{3}}{(1+y)(1+z)}+\frac{y^{3}}{(1+x)(1+z)}+\frac{z^{3}}{(1+x)(1+y)} \\
\geq & \frac{1}{3} \frac{\left(x^{3}+y^{3}+z^{3}\right) \cdot(3+x+y+z)}{(1+x)(1+y)(1+z)}
\end{aligned}
$$
Now if we put $x+y+z=3 S$, we have $x^{3}+y^{3}+z^{3} \geq 3 S$ and $(1+$ $x)(1+y)(1+z) \leq(1+a)^{3}$ by the A-G mean inequality. Thus the needed inequality reduces to $\frac{6 S^{3}}{(1+S)^{3}} \geq \frac{3}{4}$, which is obviously true because $S \geq 1$.
Remark. Both these solutions use only that $x+y+z \geq 3$.
12. The assertion is clear for $n=0$. We shall prove the general case by induction on $n$. Suppose that $c(m, i)=c(m, m-i)$ for all $i$ and $m \leq n$. Then by the induction hypothesis and the recurrence formula we have $c(n+1, k)=2^{k} c(n, k)+c(n, k-1)$ and $c(n+1, n+1-k)=$ $2^{n+1-k} c(n, n+1-k)+c(n, n-k)=2^{n+1-k} c(n, k-1)+c(n, k)$. Thus it remains only to show that
$$
\left(2^{k}-1\right) c(n, k)=\left(2^{n+1-k}-1\right) c(n, k-1)
$$
We prove this also by induction on $n$. By the induction hypothesis,
$$
c(n-1, k)=\frac{2^{n-k}-1}{2^{k}-1} c(n-1, k-1)
$$
and
$$
c(n-1, k-2)=\frac{2^{k-1}-1}{2^{n+1-k}-1} c(n-1, k-1)
$$
Using these formulas and the recurrence formula we obtain $\left(2^{k}-1\right) c(n, k)-$ $\left(2^{n+1-k}-1\right) c(n, k-1)=\left(2^{2 k}-2^{k}\right) c(n-1, k)-\left(2^{n}-3 \cdot 2^{k-1}+1\right) c(n-$ $1, k-1)-\left(2^{n+1-k}-1\right) c(n-1, k-2)=\left(2^{n}-2^{k}\right) c(n-1, k-1)-\left(2^{n}-\right.$ $\left.3 \cdot 2^{k-1}+1\right) c(n-1, k-1)-\left(2^{k-1}-1\right) c(n-1, k-1)=0$. This completes the proof.
Second solution. The given recurrence formula resembles that of binomial coefficients, so it is natural to search for an explicit formula of the form $c(n, k)=\frac{F(n)}{F(k) F(n-k)}$, where $F(m)=f(1) f(2) \cdots f(m)($ with $F(0)=1$ ) and $f$ is a certain function from the natural numbers to the real numbers. If there is such an $f$, then $c(n, k)=c(n, n-k)$ follows immediately.
After substitution of the above relation, the recurrence equivalently reduces to $f(n+1)=2^{k} f(n-k+1)+f(k)$. It is easy to see that $f(m)=2^{m}-1$ satisfies this relation.
Remark. If we introduce the polynomial $P_{n}(x)=\sum_{k=0}^{n} c(n, k) x^{k}$, the recurrence relation gives $P_{0}(x)=1$ and $P_{n+1}(x)=x P_{n}(x)+P_{n}(2 x)$. As a consequence of the problem, all polynomials in this sequence are symmetric, i.e., $P_{n}(x)=x^{n} P_{n}\left(x^{-1}\right)$.
13. Denote by $\mathcal{F}$ the set of functions considered. Let $f \in \mathcal{F}$, and let $f(1)=a$. Putting $n=1$ and $m=1$ we obtain $f(f(z))=a^{2} z$ and $f\left(a z^{2}\right)=f(z)^{2}$ for all $z \in \mathbb{N}$. These equations, together with the original one, imply $f(x)^{2} f(y)^{2}=f(x)^{2} f\left(a y^{2}\right)=f\left(x^{2} f\left(f\left(a y^{2}\right)\right)\right)=f\left(x^{2} a^{3} y^{2}\right)=$ $f\left(a(a x y)^{2}\right)=f(a x y)^{2}$, or $f(a x y)=f(x) f(y)$ for all $x, y \in \mathbb{N}$. Thus $f(a x)=a f(x)$, and we conclude that
$$
a f(x y)=f(x) f(y) \quad \text { for all } x, y \in \mathbb{N}
$$
We now prove that $f(x)$ is divisible by $a$ for each $x \in \mathbb{N}$. In fact, we inductively get that $f(x)^{k}=a^{k-1} f\left(x^{k}\right)$ is divisible by $a^{k-1}$ for every $k$. If $p^{\alpha}$ and $p^{\beta}$ are the exact powers of a prime $p$ that divide $f(x)$ and $a$ respectively, we deduce that $k \alpha \geq(k-1) \beta$ for all $k$, so we must have $\alpha \geq \beta$ for any $p$. Therefore $a \mid f(x)$.
Now we consider the function on natural numbers $g(x)=f(x) / a$. The above relations imply
$$
g(1)=1, \quad g(x y)=g(x) g(y), \quad g(g(x))=x \quad \text { for all } x, y \in \mathbb{N}
$$
Since $g \in \mathcal{F}$ and $g(x) \leq f(x)$ for all $x$, we may restrict attention to the functions $g$ only.
Clearly $g$ is bijective. We observe that $g$ maps a prime to a prime. Assume to the contrary that $g(p)=u v, u, v>1$. Then $g(u v)=p$, so either $g(u)=1$ and $g(v)=1$. Thus either $g(1)=u$ or $g(1)=v$, which is impossible.
We return to the problem of determining the least possible value of $g(1998)$. Since $g(1998)=g\left(2 \cdot 3^{3} \cdot 37\right)=g(2) \cdot g(3)^{3} \cdot g(37)$, and $g(2)$, $g(3), g(37)$ are distinct primes, $g(1998)$ is not smaller than $2^{3} \cdot 3 \cdot 5=120$. On the other hand, the value of 120 is attained for any function $g$ satisfying (2) and $g(2)=3, g(3)=2, g(5)=37, g(37)=5$. Hence the answer is 120 .
14. If $x^{2} y+x+y$ is divisible by $x y^{2}+y+7$, then so is the number $y\left(x^{2} y+\right.$ $x+y)-x\left(x y^{2}+y+7\right)=y^{2}-7 x$.
If $y^{2}-7 x \geq 0$, then since $y^{2}-7 x0$ is divisible by $x y^{2}+y+7$. But then $x y^{2}+y+7 \leq 7 x-y^{2}<7 x$, from which we obtain $y \leq 2$. For $y=1$, we are led to $x+8 \mid 7 x-1$, and hence $x+8 \mid 7(x+8)-(7 x-1)=57$. Thus the only possibilities are $x=11$ and $x=49$, and the obtained pairs $(11,1),(49,1)$ are indeed solutions. For $y=2$, we have $4 x+9 \mid 7 x-4$, so that $7(4 x+9)-4(7 x-4)=79$ is divisible by $4 x+9$. We do not get any new solutions in this case.
Therefore all required pairs $(x, y)$ are $\left(7 t^{2}, 7 t\right)(t \in \mathbb{N}),(11,1)$, and $(49,1)$.
15. The condition is obviously satisfied if $a=0$ or $b=0$ or $a=b$ or $a, b$ are both integers. We claim that these are the only solutions.
Suppose that $a, b$ belong to none of the above categories. The quotient $a / b=\lfloor a\rfloor /\lfloor b\rfloor$ is a nonzero rational number: let $a / b=p / q$, where $p$ and $q$ are coprime nonzero integers.
Suppose that $p \notin\{-1,1\}$. Then $p$ divides $\lfloor a n\rfloor$ for all $n$, so in particular $p$ divides $\lfloor a\rfloor$ and thus $a=k p+\varepsilon$ for some $k \in \mathbb{N}$ and $0 \leq \varepsilon<1$. Note that $\varepsilon \neq 0$, since otherwise $b=k q$ would also be an integer. It follows that there exists an $n \in \mathbb{N}$ such that $1 \leq n \varepsilon<2$. But then $\lfloor n a\rfloor=\lfloor k n p+n \varepsilon\rfloor=k n p+1$ is not divisible by $p$, a contradiction. Similarly, $q \notin\{-1,1\}$ is not possible. Therefore we must have $p, q= \pm 1$, and since $a \neq b$, the only possibility is $b=-a$. However, this leads to $\lfloor-a\rfloor=-\lfloor a\rfloor$, which is not valid if $a$ is not an integer.
16. Let $S$ be a set of integers such that for no four distinct elements $a, b, c, d \in$ $S$, it holds that $20 \mid a+b-c-d$. It is easily seen that there cannot exist distinct elements $a, b, c, d$ with $a \equiv b$ and $c \equiv d(\bmod 20)$. Consequently, if the elements of $S$ give $k$ different residues modulo 20, then $S$ itself has at most $k+2$ elements.
Next, consider these $k$ elements of $S$ with different residues modulo 20. They give $\frac{k(k-1)}{2}$ different sums of two elements. For $k \geq 7$ there are at least 21 such sums, and two of them, say $a+b$ and $c+d$, are equal modulo 20 ; it is easy to see that $a, b, c, d$ are discinct. It follows that $k$ cannot exceed 6 , and consequently $S$ has at most 8 elements.
An example of a set $S$ with 8 elements is $\{0,20,40,1,2,4,7,12\}$. Hence the answer is $n=9$.
17. Initially, we determine that the first few values for $a_{n}$ are $1,3,4,7,10$, $12,13,16,19,21,22,25$. Since these are exactly the numbers of the forms $3 k+1$ and $9 k+3$, we conjecture that this is the general pattern. In fact, it is easy to see that the equation $x+y=3 z$ has no solution in the set $K=\{3 k+1,9 k+3 \mid k \in \mathbb{N}\}$. We shall prove that the sequence $\left\{a_{n}\right\}$ is actually this set ordered increasingly.
Suppose $a_{n}>25$ is the first member of the sequence not belonging to $K$. We have several cases:
(i) $a_{n}=3 r+2, r \in \mathbb{N}$. By the assumption, one of $r+1, r+2, r+3$ is of the form $3 k+1$ (and smaller than $a_{n}$ ), and therefore is a member $a_{i}$ of the sequence. Then $3 a_{i}$ equals $a_{n}+1$, $a_{n}+4$, or $a_{n}+7$, which is a contradiction because $1,4,7$ are in the sequence.
(ii) $a_{n}=9 r, r \in \mathbb{N}$. Then $a_{n}+a_{2}=3(3 r+1)$, although $3 r+1$ is in the sequence, a contradiction.
(iii) $a_{n}=9 r+6, r \in \mathbb{N}$. Then one of the numbers $3 r+3,3 r+6,3 r+9$ is a member $a_{j}$ of the sequence, and thus $3 a_{j}$ is equal to $a_{n}+3$, $a_{n}+12$, or $a_{n}+21$, where $3,12,21$ are members of the sequence, again a contradiction.
Once we have revealed the structure of the sequence, it is easy to compute $a_{1998}$. We have $1998=4 \cdot 499+2$, which implies $a_{1998}=9 \cdot 499+a_{2}=4494$.
18. We claim that, if $2^{n}-1$ divides $m^{2}+9$ for some $m \in \mathbb{N}$, then $n$ must be a power of 2 . Suppose otherwise that $n$ has an odd divisor $d>1$. Then $2^{d}-1 \mid 2^{n}-1$ is also a divisor of $m^{2}+9=m^{2}+3^{2}$. However, $2^{d}-1$ has some prime divisor $p$ of the form $4 k-1$, and by a well-known fact, $p$ divides both $m$ and 3 . Hence $p=3$ divides $2^{d}-1$, which is impossible, because for $d$ odd, $2^{d} \equiv 2(\bmod 3)$. Hence $n=2^{r}$ for some $r \in \mathbb{N}$.
Now let $n=2^{r}$. We prove the existence of $m$ by induction on $r$. The case $r=1$ is trivial. Now for any $r>1$ note that $2^{2^{r}}-1=\left(2^{2^{r-1}}-1\right)\left(2^{2^{r-1}}+\right.$ 1). The induction hypothesis claims that there exists an $m_{1}$ such that $2^{2^{r-1}}-1 \mid m_{1}^{2}+9$. We also observe that $2^{2^{r-1}}+1 \mid m_{2}^{2}+9$ for simple $m_{2}=3 \cdot 2^{2^{r-2}}$. By the Chinese remainder theorem, there is an $m \in \mathbb{N}$ that satisfies $m \equiv m_{1}\left(\bmod 2^{2^{r-1}}-1\right)$ and $m \equiv m_{2}\left(\bmod 2^{2^{r-1}}+1\right)$. It is easy to see that this $m^{2}+9$ will be divisible by both $2^{2^{r-1}}-1$ and $2^{2^{r-1}}+1$, i.e., that $2^{2^{r}}-1 \mid m^{2}+9$. This completes the induction.
19. For $n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{r}^{\alpha_{r}}$, where $p_{i}$ are distinct primes and $\alpha_{i}$ natural numbers, we have $\tau(n)=\left(\alpha_{1}+1\right) \cdots\left(\alpha_{r}+1\right)$ and $\tau\left(n^{2}\right)=\left(2 \alpha_{1}+1\right) \ldots\left(2 \alpha_{r}+1\right)$. Putting $k_{i}=\alpha_{i}+1$, the problem reduces to determining all natural values of $m$ that can be represented as
$$
m=\frac{2 k_{1}-1}{k_{1}} \cdot \frac{2 k_{2}-1}{k_{2}} \cdots \frac{2 k_{r}-1}{k_{r}}
$$
Since the numerator $\tau\left(n^{2}\right)$ is odd, $m$ must be odd too. We claim that every odd $m$ has a representation of the form (1). The proof will be done by induction.
This is clear for $m=1$. Now for every $m=2 k-1$ with $k$ odd the result follows easily, since $m=\frac{2 k-1}{k} \cdot k$, and $k$ can be written as (1). We cannot do the same if $k$ is even; however, in the case $m=4 k-1$ with $k$ odd, we can write it as $m=\frac{12 k-3}{6 k-1} \cdot \frac{6 k-1}{3 k} \cdot k$, and this works.
In general, suppose that $m=2^{t} k-1$, with $k$ odd. Following the same pattern, we can write $m$ as
$$
m=\frac{2^{t}\left(2^{t}-1\right) k-\left(2^{t}-1\right)}{2^{t-1}\left(2^{t}-1\right) k-\left(2^{t-1}-1\right)} \cdots \frac{4\left(2^{t}-1\right) k-3}{2\left(2^{t}-1\right) k-1} \cdot \frac{2\left(2^{t}-1\right) k-1}{\left(2^{t}-1\right) k} \cdot k
$$
The induction is finished. Hence $m$ can be represented as $\frac{\tau\left(n^{2}\right)}{\tau(n)}$ if and only if it is odd.
20. We first consider the special case $n=3^{r}$. Then the simplest choice $\frac{10^{n}-1}{9}=$ $11 \ldots 1$ ( $n$ digits) works. This can be shown by induction: it is true for $r=$ 1, while the inductive step follows from $10^{3^{r}}-1=\left(10^{3^{r-1}}-1\right)\left(10^{2 \cdot 3^{r-1}}+\right.$ $10^{3^{r-1}}+1$ ), because the second factor is divisible by 3 .
In the general case, let $k \geq n / 2$ be a positive integer and $a_{1}, \ldots, a_{n-k}$ be nonzero digits. We have
$$
\begin{aligned}
A & =\left(10^{k}-1\right) \overline{a_{1} a_{2} \ldots a_{n-k}} \\
& =\overline{a_{1} a_{2} \ldots a_{n-k-1} a_{n-k}^{\prime} \underbrace{99 \ldots 99}_{2 k-n}} b_{1} b_{2} \ldots b_{n-k-1} b_{n-k}^{\prime}
\end{aligned}
$$
where $a_{n-k}^{\prime}=a_{n-k}-1, b_{i}=9-a_{i}$, and $b_{n-k}^{\prime}=9-a_{n-k}^{\prime}$. The sum of digits of $A$ equals $9 k$ independently of the choice of digits $a_{1}, \ldots, a_{n-k}$. Thus we need only choose $k \geq \frac{n}{2}$ and digits $a_{1}, \ldots, a_{n-k-1} \notin\{0,9\}$ and $a_{n-k} \in\{0,1\}$ in order for the conditions to be fulfilled. Let us choose
$$
k=\left\{\begin{array}{l}
3^{r}, \quad \text { if } 3^{r}1$ we have $N_{s} \equiv 1(\bmod 4)$. Now we prove $3 \mid d\left(p_{s}\right)$. We have $p_{s}\left|N_{s}\right| 10^{3 s}-1$ and hence $d\left(p_{s}\right) \mid 3 s$. We cannot have $d\left(p_{s}\right) \mid s$, for otherwise $p_{s}\left|10^{s}-1 \Rightarrow p_{s}\right|\left(10^{2 s}+\right.$ $\left.10^{s}+1,10^{s}-1\right)=3$; and we cannot have $d\left(p_{s}\right) \mid 3$, for otherwise $p_{s} \mid 10^{3}-1=999=3^{3} \cdot 37$, both of which contradict $p_{s} \neq 3,37$. It follows that $d\left(p_{s}\right)=3 s$. Hence for every prime $s$ there exists a prime $p_{s}$ such that $d\left(p_{s}\right)=3 s$. It follows that the cardinality of $S$ is infinite.
(b) Let $r=r(s)$ be the fundamental period of $p \in S$. Then $p \mid 10^{3 r}-1$, $p \nmid 10^{r}-1 \Rightarrow p \mid 10^{2 r}+10^{r}+1$. Let $x_{j}=\frac{10^{j-1}}{p}$ and $y_{j}=\left\{x_{j}\right\}=$ $0 . a_{j} a_{j+1} a_{j+2} \ldots$. Then $a_{j}<10 y_{j}$, and hence
$$
f(k, p)=a_{k}+a_{k+r}+a_{k+2 r}<10\left(y_{k}+y_{k+r}+y_{k+2 r}\right) .
$$
We note that $x_{k}+x_{k+s(p)}+x_{k+2 s(p)}=\frac{10^{k-1} N_{p}}{p}$ is an integer, from which it follows that $y_{k}+y_{k+s(p)}+y_{k+2 s(p)} \in \mathbb{N}$. Hence $y_{k}+y_{k+s(p)}+$ $y_{k+2 s(p)} \leq 2$. It follows that $f(k, p)<20$. We note that $f(2,7)=$ $4+8+7=19$. Hence 19 is the greatest possible value of $f(k, p)$.
5. Since one can arbitrarily add zeros at the end of $m$, which increases divisibility by 2 and 5 to an arbitrary exponent, it suffices to assume $2,5 \nmid n$. If $(n, 10)=1$, there exists an integer $w \geq 2$ such that $10^{w} \equiv 1(\bmod n)$. We also note that $10^{i w} \equiv 1(\bmod n)$ and $10^{j w+1} \equiv 10(\bmod n)$ for all integers $i$ and $j$. Let us assume that $m$ is of the form $m=\sum_{i=1}^{u} 10^{i w}+\sum_{j=1}^{v} 10^{j w+1}$ for integers $u, v \geq 0$ (where if $u$ or $v$ is 0 , the corresponding sum is 0 ). Obviously, the sum of the digits of $m$ is equal to $u+v$, and also $m \equiv u+10 v(\bmod n)$. Hence our problem reduces to finding integers $u, v \geq 0$ such that $u+v=k$ and $n \mid u+10 v=k+9 v$. Since $(n, 9)=1$, it follows that there exists some $v_{0}$ such that $0 \leq v_{0}b)$ are respectively their smallest elements.
Lemma 1. Numbers $x, y$, and $x+y$ cannot belong to three different sets. Proof. The number $f(x, x+y)=f(y, x+y)$ must belong to both the set containing $y$ and the set containing $x$, a contradiction.
Lemma 2. The subset $C$ contains a multiple of $b$. Moreover, if $k b$ is the smallest such multiple, then $(k-1) b \in B$ and $(k-1) b+1, k b+1 \in A$.
Proof. Let $r$ be the residue of $c$ modulo $b$. If $r=0$, the first statement automatically holds. Let $02$ that for all $s$ we have $f(n, s)=\frac{1}{s}\binom{n-1}{s-1}\binom{n}{s-1}$. We shall prove that the given formula holds also for all $f(n+1, s)$, where $s \geq 2$.
We say that an $(n+1, s)$ - or $(n+1, s+1)$-path is related to a given $(n, s)$ path if it is obtained from the given path by inserting a step $E N$ between two moves or at the beginning or the end of the path. We note that by inserting the step between two moves that form a step one obtains an $(n+1, s)$-path; in all other cases one obtains an $(n+1, s+1)$-path. For each $(n, s)$-path there are exactly $2 n+1-s$ related $(n+1, s+1)$-paths, and for each $(n, s+1)$-path there are $s+1$ related $(n+1, s+1)$-paths. Also, each $(n+1, s+1)$-path is related to exactly $s+1$ different $(n, s)$ - or $(n, s+1)$-paths. Thus:
$$
\begin{aligned}
(s+1) f(n+1, s+1) & =(2 n+1-s) f(n, s)+(s+1) f(n, s+1) \\
& =\frac{2 n+1-s}{s}\binom{n-1}{s-1}\binom{n}{s-1}+\binom{n-1}{s}\binom{n}{s} \\
& =\binom{n}{s}\binom{n+1}{s},
\end{aligned}
$$
i.e., $f(n+1, s+1)=\frac{1}{s+1}\binom{n}{s}\binom{n+1}{s}$. This completes the proof.
22. (a) Color the first, third, and fifth row red, and the remaining squares white. There in total $n$ pieces and $3 n$ red squares. Since each piece can cover at most three red squares, it follows that each piece colors exactly three red squares. Then it follows that the two white squares it covers must be on the same row; otherwise, the piece has to cover
at least three. Hence, each white row can be partitioned into pairs of squares belonging to the same piece. Thus it follows that the number of white squares in a row, which is $n$, must be even.
(b) Let $a_{k}$ denote the number of different tilings of a $5 \times 2 k$ rectangle. Let $b_{k}$ be the number of tilings that cannot be partitioned into two smaller tilings along a vertical line (without cutting any pieces). It is easy to see that $a_{1}=b_{1}=2, b_{2}=2, a_{2}=6=2 \cdot 3, b_{3}=4$, and subsequently, by induction, $b_{3 k} \geq 4, b_{3 k+1} \geq 2$, and $b_{3 k+2} \geq 2$. We also have $a_{k}=b_{k}+\sum_{i=1}^{k-1} b_{i} a_{k-i}$. For $k \geq 3$ we now have inductively
$$
a_{k}>2+\sum_{i=1}^{k-1} 2 a_{k-i} \geq 2 \cdot 3^{k-1}+2 a_{k-1} \geq 2 \cdot 3^{k}
$$
23. Let $r(m)$ denote the rest period before the $m$ th catch, $t(m)$ the number of minutes before the $m$ th catch, and $f(n)$ as the number of flies caught in $n$ minutes. We have $r(1)=1, r(2 m)=r(m)$, and $r(2 m+1)=f(m)+1$. We then have by induction that $r(m)$ is the number of ones in the binary representation of $m$. We also have $t(m)=\sum_{i=1}^{m} r(i)$ and $f(t(m))=m$. From the recursive relations for $r$ we easily derive $t(2 m+1)=2 t(m)+m+1$ and consequently $t(2 m)=2 t(m)+m-r(m)$. We then have, by induction on $p, t\left(2^{p} m\right)=2^{p} t(m)+p \cdot m \cdot 2^{p-1}-\left(2^{p}-1\right) r(m)$.
(a) We must find the smallest number $m$ such that $r(m+1)=9$. The smallest number with nine binary digits is $\overline{111111111}_{2}=511$; hence the required $m$ is 510 .
(b) We must calculate $t(98)$. Using the recursive formulas we have $t(98)=$ $2 t(49)+49-r(49), t(49)=2 t(24)+25$, and $t(24)=8 t(3)+36-7 r(3)$. Since we have $t(3)=4, r(3)=2$ and $r(49)=r\left(\overline{110001}_{2}\right)=3$, it follows $t(24)=54 \Rightarrow t(49)=133 \Rightarrow t(98)=312$.
(c) We must find $m_{c}$ such that $t\left(m_{c}\right) \leq 1999e>2$, it follows that $\left|\bigcup_{j=1}^{N} A_{i_{j}}\right| \geq \frac{1}{2}|S|$; hence the chosen $i_{1}<\cdots0$. Hence $F_{B}(w) \geq 1$.
It remains to show that $F_{B}(w)=1$ if and only if $p=5$. We have the formula $(x-w)\left(x-w^{2}\right) \cdots\left(x-w^{p-1}\right)=x^{p-1}+x^{p-2}+\cdots+x+1=\frac{x^{p}-1}{x-1}$. Let $f_{B}(x)=x^{3}+x+1=(x-\lambda)(x-\mu)(x-\nu)$, where $\lambda$, $\mu$, and $\nu$ are the three zeros of the polynomial $f_{B}(x)$. It follows that
$F_{B}(w)=\left(\frac{\lambda^{p}-1}{\lambda-1}\right)\left(\frac{\mu^{p}-1}{\mu-1}\right)\left(\frac{\nu^{p}-1}{\nu-1}\right)=-\frac{1}{3}\left(\lambda^{p}-1\right)\left(\mu^{p}-1\right)\left(\nu^{p}-1\right)$,
since $(\lambda-1)(\mu-1)(\nu-1)=-f_{B}(1)=-3$. We also have $\lambda+\mu+\nu=0$, $\lambda \mu \nu=-1, \lambda \mu+\lambda \nu+\mu \nu=1$, and $\lambda^{2}+\mu^{2}+\nu^{2}=(\lambda+\mu+\nu)^{2}-2(\lambda \mu+$ $\lambda \nu+\mu \nu)=-2$. By induction (using that $\left(\lambda^{r}+\mu^{r}+\nu^{r}\right)+\left(\lambda^{r-2}+\mu^{r-2}+\right.$ $\left.\nu^{r-2}\right)+\left(\lambda^{r-3}+\mu^{r-3}+\nu^{r-3}\right)=0$ ), it follows that $\lambda^{r}+\mu^{r}+\nu^{r}$ is an integer for all $r \in \mathbb{N}$.
Let us assume $F_{B}(x)=1$. It follows that $\left(\lambda^{p}-1\right)\left(\mu^{p}-1\right)\left(\nu^{p}-1\right)=-3$. Hence $\lambda^{p}, \mu^{p}, \nu^{p}$ are roots of the polynomial $p(x)=x^{3}-q x^{2}+(1+q) x+1$, where $q=\lambda^{p}+\mu^{p}+\nu^{p}$. Since $f_{B}(x)$ is an increasing function in real numbers, it follows that it has only one real root (w.l.o.g.) $\lambda$, the other two roots being complex conjugates. From $f_{B}(-1)<00$; otherwise, $q\left(x^{2}-x\right)$ intersects $x^{3}+x+1$ at a value smaller than $\lambda$. Additionally, as $p$ increases, $\lambda^{p}$ approaches 0 , and hence $q$ must increase.
For $p=5$ we have $1+w+w^{3}=-w^{2}\left(1+w^{2}\right)$ and hence $G(w)=\prod_{i=1}^{p-1}(1+$ $\left.w^{2 j}\right)=1$. For a zero of $f_{B}(x)$ we have $x^{5}=-x^{3}-x^{2}=-x^{2}+x+1$ and hence $q=\lambda^{5}+\mu^{5}+\nu^{5}=-\left(\lambda^{2}+\mu^{2}+\nu^{2}\right)+(\lambda+\mu+\nu)+3=5$.
For $p>5$ we also have $q \geq 6$. Assuming again $F_{B}(x)=1$ and defining $p(x)$ as before, we have $p(-1)<0, p(0)>0, p(2)<0$, and $p(x)>0$ for a sufficiently large $x>2$. It follows that $p(x)$ must have three distinct real roots. However, since $\mu^{p}, \nu^{p} \in \mathbb{R} \Rightarrow \nu^{p}=\overline{\mu^{p}}=\mu^{p}$, it follows that $p(x)$ has at most two real roots, which is a contradiction. Hence, it follows that $F_{B}(x)>1$ for $p>5$ and thus $E(A) \leq E(B)$, where equality holds only for $p=5$.
### 4.41 Solutions to the Shortlisted Problems of IMO 2000
1. In order for the trick to work, whenever $x+y=z+t$ and the cards $x, y$ are placed in different boxes, either $z, t$ are in these boxes as well or they are both in the remaining box.
Case 1. The cards $i, i+1, i+2$ are in different boxes for some $i$. Since $i+(i+3)=(i+1)+(i+2)$, the cards $i$ and $i+3$ must be in the same box; moreover, $i-1$ must be in the same box as $i+2$, etc. Hence the cards $1,4,7, \ldots, 100$ are placed in one box, the cards $2,5, \ldots, 98$ are in the second, while $3,6, \ldots, 99$ are in the third box. The number of different arrangements of the cards is 6 in this case.
Case 2. No three successive cards are all placed in different boxes. Suppose that 1 is in the blue box, and denote by $w$ and $r$ the smallest numbers on cards lying in the white and red boxes; assume w.l.o.g. that $w$ $w+1$. Now suppose that $r<100$. Since $w+r=(w-1)+(r+1), r+1$ must be in the blue box. But then $(r+1)+w=r+(w+1)$ implies that $w+1$ must be red, which is a contradiction. Hence the red box contains only the card 100 . Since $99+w=100+(w-1)$, we deduce that the card 99 is in the white box. Moreover, if any of the cards $k$, $2 \leq k \leq 99$, were in the blue box, then since $k+99=(k-1)+100$, the card $k-1$ should be in the red box, which is impossible. Hence the blue box contains only the card 1 , whereas the cards $2,3, \ldots, 99$ are all in the white box.
In general, one box contains 1 , another box only 100 , while the remaining contains all the other cards. There are exactly 6 such arrangements, and the trick works in each of them.
Therefore the answer is 12.
2. Since the volume of each brick is 12 , the side of any such cube must be divisible by 6 .
Suppose that a cube of side $n=6 k$ can be built using $\frac{n^{3}}{12}=18 k^{3}$ bricks. Set a coordinate system in which the cube is given as $[0, n] \times[0, n] \times[0, n]$ and color in black each unit cube $[2 p, 2 p+1] \times[2 q, 2 q+1] \times[2 r, 2 r+1]$. There are exactly $\frac{n^{3}}{9}=27 k^{3}$ black cubes. Each brick covers either one or three black cubes, which is in any case an odd number. It follows that the total number of black cubes must be even, which implies that $k$ is even. Hence $12 \mid n$.
On the other hand, two bricks can be fitted together to give a $2 \times 3 \times 4$ box. Using such boxes one can easily build a cube of side 12 , and consequently any cube of side divisible by 12 .
3. Clearly $m(S)$ is the number of pairs of point and triangle $\left(P_{t}, P_{i} P_{j} P_{k}\right)$ such that $P_{t}$ lies inside the circle $P_{i} P_{j} P_{k}$. Consider any four-element set $S_{i j k l}=\left\{P_{i}, P_{j}, P_{k}, P_{l}\right\}$. If the convex hull of $S_{i j k l}$ is the triangle $P_{i} P_{j} P_{k}$, then we have $a_{i}=a_{j}=a_{k}=0, a_{l}=1$. Suppose that the convex hull is
the quadrilateral $P_{i} P_{j} P_{k} P_{l}$. Since this quadrilateral is not cyclic, we may suppose that $\angle P_{i}+\angle P_{k}<180^{\circ}<\angle P_{j}+\angle P_{l}$. In this case $a_{i}=a_{k}=0$ and $a_{j}=a_{l}=1$. Therefore $m\left(S_{i j k l}\right)$ is 2 if $P_{i}, P_{j}, P_{k}, P_{l}$ are vertices of a convex quadrilateral, and 1 otherwise.
There are $\binom{n}{4}$ four-element subsets $S_{i j k l}$. If $a(S)$ is the number of such subsets whose points determine a convex quadrilateral, we have $m(S)=$ $2 a(S)+\left(\binom{n}{4}-a(S)\right)=\binom{n}{4}+a(S) \leq 2\binom{n}{4}$. Equality holds if and only if every four distinct points of $S$ determine a convex quadrilateral, i.e. if and only if the points of $S$ determine a convex polygon. Hence $f(n)=2\binom{n}{4}$ has the desired property.
4. By a good placement of pawns we mean the placement in which there is no block of $k$ adjacent unoccupied squares in a row or column.
We can make a good placement as follows: Label the rows and columns with $0,1, \ldots, n-1$ and place a pawn on a square $(i, j)$ if and only if $k$ divides $i+j+1$. This is obviously a good placement in which the pawns are placed on three lines with $k, 2 n-2 k$, and $2 n-3 k$ squares, which adds up to $4 n-4 k$ pawns in total.
Now we shall prove that a good placement must contain at least $4 n-4 k$ pawns. Suppose we have a good placement of $m$ pawns. Partition the board into nine rectangular regions as shown in the picture. Let $a, b, \ldots, h$ be the numbers of pawns in the rectangles $A, B, \ldots, H$ respectively. Note that each row that
passes through $A, B$, and $C$ either
contains a pawn inside $B$, or contains a pawn in both $A$ and $C$. It follows that $a+c+2 b \geq 2(n-k)$. We similarly obtain that $c+e+2 d, e+g+2 f$, and $g+a+2 h$ are all at least $2(n-k)$. Adding and dividing by 2 yields $a+b+\cdots+h \geq 4(n-k)$, which proves the statement.
5. We say that a vertex of a nice region is convex if the angle of the region at that vertex equals $90^{\circ}$; otherwise (if the angle is $270^{\circ}$ ), we say that a vertex is concave.
For a simple broken line $C$ contained in the boundary of a nice region $R$ we call the pair $(R, C)$ a boundary pair. Such a pair is called outer if the region $R$ is inside the broken line $C$, and inner otherwise. Let $\mathcal{B}_{i}, \mathcal{B}_{o}$ be the sets of inner and outer boundary pairs of nice regions respectively, and let $\mathcal{B}=\mathcal{B}_{i} \cup \mathcal{B}_{o}$. For a boundary pair $b=(R, C)$ denote by $c_{b}$ and $v_{b}$ respectively the number of convex and concave vertices of $R$ that belong to $C$. We have the following facts:
(1) Each vertex of a rectangle corresponds to one concave angle of a nice region and vice versa. This correspondence is bijective, so $\sum_{b \in \mathcal{B}} v_{b}=$ $4 n$.
(2) For a boundary pair $b=(R, C)$ the sum of angles of $R$ that are on $C$ equals $\left(c_{b}+v_{b}-2\right) 180^{\circ}$ if $b$ is outer, and $\left(c_{b}+v_{b}+2\right) 180^{\circ}$ if $b$ is inner. On the other hand the sum of angles is obviously equal to $c_{b} \cdot 90^{\circ}+v_{b} \cdot 270^{\circ}$. It immediately follows that $c_{b}-v_{b}=\left\{\begin{array}{r}4 \text { if } b \in \mathcal{B}_{o}, \\ -4 \text { if } b \in \mathcal{B}_{i} .\end{array}\right.$
(3) Since every vertex of a rectangle appears in exactly two boundary pairs and each boundary pair contains at least one vertex of a rectangle, the number $K$ of boundary pairs is less than or equal to $8 n$.
(4) The set $\mathcal{B}_{i}$ is nonempty, because every boundary of the infinite region is inner.
Consequently, the sum of the numbers of the vertices of all nice regions is equal to
$$
\sum_{b \in \mathcal{B}}\left(c_{b}+v_{b}\right)=\sum_{b \in \mathcal{B}}\left(2 v_{b}+\left(c_{b}-v_{b}\right)\right) \leq 2 \cdot 4 n+4(K-1)-4 \leq 40 n-8
$$
6. Every integer $z$ has a unique representation $z=p x+q y$, where $x, y \in \mathbb{Z}$, $0 \leq x \leq q-1$. Consider the region $T$ in the $x y$-plane defined by the last inequality and $p x+q y \geq 0$. There is a bijective correspondence between lattice points of this region and nonnegative integers given by $(x, y) \mapsto$ $z=p x+q y$. Let us mark all lattice points of $T$ whose corresponding integers belong to $S$ and color in black the unit squares whose left-bottom vertices are at marked points. Due to the condition for $S$, this coloring has the property that all points lying on the right or above a colored point are colored as well. In particular, since the point $(0,0)$ is colored, all points above or on the line $y=0$ are colored. What we need is the number of such colorings of $T$.
The border of the colored subregion $C$ of $T$ determines a path from $(0,0)$ to $(q,-p)$ consisting of consecutive unit moves either to the right or downwards. There are $\binom{p+q}{p}$ such paths in total. We must find the number of such paths not going below the line $l: p x+q y=0$.
Consider any path $\gamma=A_{0} A_{1} \ldots A_{p+q}$ from $A_{0}=(0,0)$ to $A_{p+q}=(q,-p)$. We shall see the path $\gamma$ as a sequence $G_{1} G_{2} \ldots G_{p+q}$ of moves to the right $(R)$ or downwards ( $D$ ) with exactly $p D$ 's and $q R$ 's.
Two paths are said to be equivalent if one is obtained from the other by a circular shift of the corresponding sequence $G_{1} G_{2} \ldots G_{p+q}$. We note that all the $p+q$ circular shifts of a path are distinct. Indeed, $G_{1} \ldots G_{p+q} \equiv$ $G_{i+1} \ldots G_{i+p+q}$ would imply $G_{1}=G_{i+1}=G_{2 i+1}=\cdots$ (where $G_{j+p+q}=$ $G_{j}$ ), so $G_{1}=\cdots=G_{p+q}$, which is impossible. Hence each equivalence class contains exactly $p+q$ paths.
Let $l_{i}, 0 \leq i0$, this proves the result.
12. Since $D(A)=D(B)$, we can define $f(i)>g(i) \geq 0$ that satisfy $b_{i}-b_{i-1}=$ $a_{f(i)}-a_{g(i)}$ for all $i$.
The number $b_{i+1}-b_{i-1} \in D(B)=D(A)$ can be written in the form $a_{u}-a_{v}, u>v \geq 0$. Then $b_{i+1}-b_{i-1}=b_{i+1}-b_{i}+b_{i}-b_{i-1}$ implies
$a_{f(i+1)}+a_{f(i)}+a_{v}=a_{g(i+1)}+a_{g(i)}+a_{u}$, so the $B_{3}$ property of $A$ implies that $(f(i+1), f(i), v)$ and $(g(i+1), g(i), u)$ coincide up to a permutation. It follows that either $f(i+1)=g(i)$ or $f(i)=g(i+1)$. Hence if we define $R=\left\{i \in \mathbb{N}_{0} \mid f(i+1)=g(i)\right\}$ and $S=\left\{i \in \mathbb{N}_{0} \mid f(i)=g(i+1)\right\}$ it holds that $R \cup S=\mathbb{N}_{0}$.
Lemma. If $i \in R$, then also $i+1 \in R$.
Proof. Suppose to the contrary that $i \in R$ and $i+1 \in S$, i.e., $g(i)=$ $f(i+1)=g(i+2)$. There are integers $x$ and $y$ such that $b_{i+2}-b_{i-1}=$ $a_{x}-a_{y}$. Then $a_{x}-a_{y}=a_{f(i+2)}-a_{g(i+2)}+a_{f(i+1)}-a_{g(i+1)}+a_{f(i)}-$ $a_{g(i)}=a_{f(i+2)}+a_{f(i)}-a_{g(i+1)}-a_{g(i)}$, so by the $B_{3}$ property $(x, g(i+$ $1), g(i))$ and $(y, f(i+2), f(i))$ coincide up to a permutation. But this is impossible, since $f(i+2), f(i)>g(i+2)=g(i)=f(i+1)>g(i+1)$. This proves the lemma.
Therefore if $i \in R \neq \emptyset$, then it follows that every $j>i$ belongs to $R$. Consequently $g(i)=f(i+1)>g(i+1)=f(i+2)>g(i+2)=f(i+3)>$ $\cdots$ is an infinite decreasing sequence of nonnegative integers, which is impossible. Hence $S=\mathbb{N}_{0}$, i.e.,
$$
b_{i+1}-b_{i}=a_{f(i+1)}-a_{f(i)} \quad \text { for all } i \in \mathbb{N}_{0}
$$
Thus $f(0)=g(1)