5. (SPA 4) In the triangle $A B C$, with $\measuredangle A=60^{\circ}$, a parallel $I F$ to $A C$ is drawn through the incenter $I$ of the triangle, where $F$ lies on the side $A B$. The point $P$ on the side $B C$ is such that $3 B P=B C$. Show that $\measuredangle B F P=\measuredangle B / 2$.

Solution: 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. (USS 4) ${ }^{\mathrm{IMO}}$ Prove for each triangle $A B C$ the inequality  $$ \frac{1}{4}<\frac{I A \cdot I B \cdot I C}{l_{A} l_{B} l_{C}} \leq \frac{8}{27} $$  where $I$ is the incenter and $l_{A}, l_{B}, l_{C}$ are the lengths of the angle bisectors of $A B C$.

Solution: 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. (CHN 2) Let $O$ be the center of the circumsphere of a tetrahedron $A B C D$. Let $L, M, N$ be the midpoints of $B C, C A, A B$ respectively, and assume that $A B+B C=A D+C D, B C+C A=B D+A D$, and $C A+A B=$ $C D+B D$. Prove that $\angle L O M=\angle M O N=\angle N O L$.

Solution: 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 ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-556.jpg?height=317&width=410&top_left_y=248&top_left_x=888) $M N M_{1} N_{1}$ 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. (NET 1) Let $S$ be a set of $n$ points in the plane. No three points of $S$ are collinear. Prove that there exists a set $P$ containing $2 n-5$ points satisfying the following condition: In the interior of every triangle whose three vertices are elements of $S$ lies a point that is an element of $P$.

Solution: 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 $x_{1}<x_{2}<\cdots<x_{n}$ (choosing adequate axes and renumbering the points if necessary). Define $d$ to be half the minimum distance of $P_{i}$ from the line $P_{j} P_{k}$, where $i, j, k$ go through all possible combinations of mutually distinct indices. First we define a set $T$ containing $2 n-4$ points:  $$ T=\left\{\left(x_{i}, y_{i}-d\right),\left(x_{i}, y_{i}+d\right) \mid i=2,3, \ldots, n-1\right\} $$  Consider any triangle $P_{k} P_{l} P_{m}$, where $k<l<m$. Its interior contains at least one of the two points $\left(x_{l}, y_{l} \pm d\right)$, so $T$ is a set of $2 n-4$ points with the required property. However, at least one of the points of $T$ is useless. The convex hull of $S$ is a polygon with at least three points in $S$ as vertices. Let $P_{j}$ be a vertex of that hull distinct from $P_{1}$ and $P_{n}$. Clearly one of the points $\left(x_{j}, y_{j} \pm d\right)$ lies outside the convex hull, and thus can be left out. The remaining set of $2 n-5$ points satisfies the conditions.


9. (FRA 3) In the plane we are given a set $E$ of 1991 points, and certain pairs of these points are joined with a path. We suppose that for every point of $E$, there exist at least 1593 other points of $E$ to which it is joined by a path. Show that there exist six points of $E$ every pair of which are joined by a path. Alternative version. Is it possible to find a set $E$ of 1991 points in the plane and paths joining certain pairs of the points in $E$ such that every point of $E$ is joined with a path to at least 1592 other points of $E$, and in every subset of six points of $E$ there exist at least two points that are not joined?

Solution: 9. Let $A_{1}, A_{2}$ be two points of $E$ which are joined. In $E \backslash\left\{A_{1}, A_{2}\right\}$, there are at most 397 points to which $A_{1}$ is not joined, and at most as much to which $A_{2}$ is not joined. Consequently, there exists a point $A_{3}$ which is joined to both $A_{1}$ and $A_{2}$. There are at most $3 \cdot 397=1191$ points of $E \backslash\left\{A_{1}, A_{2}, A_{3}\right\}$ to which at least one of $A_{1}, A_{2}, A_{3}$ is not joined, hence it is possible to choose a point $A_{4}$ joined to $A_{1}, A_{2}, A_{3}$. Similarly, there exists a point $A_{5}$ which is joined to all $A_{1}, A_{2}, A_{3}, A_{4}$. Finally, among the remaining 1986 points, there are at most $5 \cdot 397=1985$ which are not joined to one of the points $A_{1}, \ldots, A_{5}$. Thus there is at least one point $A_{6}$ joined to all $A_{1}, \ldots, A_{5}$. It is clear that $A_{1}, \ldots, A_{6}$ are pairwise joined. Solution of the alternative version. Let be given 1991 points instead. Number the points from 1 to 1991, and join $i$ and $j$ if and only if $i-j$ is not a multiple of 5 . Then each $i$ is joined to 1592 or 1593 other points, and obviously among any six points there are two which are not joined.


1. (AUS 2) Prove that for any positive integer $m$ there exist an infinite number of pairs of integers $(x, y)$ such that (i) $x$ and $y$ are relatively prime; (ii) $y$ divides $x^{2}+m$; (iii) $x$ divides $y^{2}+m$.

Solution: 1. Assume that a pair $(x, y)$ with $x<y$ satisfies the required conditions. We claim that the pair $\left(y, x_{1}\right)$ also satisfies the conditions, where $x_{1}=\frac{y^{2}+m}{x}$ (note that $x_{1}>y$ 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 $\left(x_{0}, y_{0}\right)=(1,4)$.


10. (ITA 1) ${ }^{\mathrm{IMO5}}$ Let $V$ be a finite subset of Euclidean space consisting of points $(x, y, z)$ with integer coordinates. Let $S_{1}, S_{2}, S_{3}$ be the projections of $V$ onto the $y z, x z, x y$ planes, respectively. Prove that  $$ |V|^{2} \leq\left|S_{1}\right|\left|S_{2}\right|\left|S_{3}\right| $$  ( $|X|$ denotes the number of elements of $X$ ).

Solution: 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. (JAP 2) In a triangle $A B C$, let $D$ and $E$ be the intersections of the bisectors of $\angle A B C$ and $\angle A C B$ with the sides $A C, A B$, respectively. Determine the angles $\angle A, \angle B, \angle C$ if  $$ \measuredangle B D E=24^{\circ}, \quad \measuredangle C E D=18^{\circ} . $$

Solution: 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 $\angle A=96^{\circ}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-571.jpg?height=295&width=1043&top_left_y=1086&top_left_x=260)  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. (NET 1) Let $f, g$, and $a$ be polynomials with real coefficients, $f$ and $g$ in one variable and $a$ in two variables. Suppose  $$ f(x)-f(y)=a(x, y)(g(x)-g(y)) \quad \text { for all } x, y \in \mathbb{R} $$  Prove that there exists a polynomial $h$ with $f(x)=h(g(x))$ for all $x \in \mathbb{R}$.

Solution: 12. Let us set $\operatorname{deg} f=n$ and $\operatorname{deg} g=m$. We shall prove the result by induction on $n$. If $n<m$, then $\operatorname{deg}_{x}[f(x)-f(y)]<\operatorname{deg}_{x}[g(x)-g(y)]$, which implies that $f(x)-f(y)=0$, i.e., that $f$ is constant. The statement trivially holds. Assume now that $n \geq m$. Transition to $f_{1}(x)=f(x)-f(0)$ and $g_{1}(x)=$ $g(x)-g(0)$ allows us to suppose that $f(0)=g(0)=0$. Then the given condition for $y=0$ gives us $f(x)=f_{1}(x) g(x)$, where $f_{1}(x)=a(x, 0)$ and $\operatorname{deg} f_{1}=n-m$. We now have  $$ \begin{aligned} a(x, y)(g(x)-g(y)) & =f(x)-f(y)=f_{1}(x) g(x)-f_{1}(y) g(y) \\ & =\left[f_{1}(x)-f_{1}(y)\right] g(x)+f_{1}(y)[g(x)-g(y)] . \end{aligned} $$  Since $g(x)$ is relatively prime to $g(x)-g(y)$, it follows that $f_{1}(x)-f_{1}(y)=$ $b(x, y)(g(x)-g(y))$ for some polynomial $b(x, y)$. By the induction hypothesis there exists a polynomial $h_{1}$ such that $f_{1}(x)=h_{1}(g(x))$ and consequently $f(x)=g(x) \cdot h_{1}(g(x))=h(g(x))$ for $h(t)=t h_{1}(t)$. Thus the induction is complete.


13. (NZL 1) ${ }^{\mathrm{IMO1}}$ Find all integer triples $(p, q, r)$ such that $1<p<q<r$ and $(p-1)(q-1)(r-1)$ is a divisor of $(p q r-1)$.

Solution: 13. Let us define  $$ \begin{aligned} F(p, q, r)= & \frac{(p q r-1)}{(p-1)(q-1)(r-1)} \\ = & 1+\frac{1}{p-1}+\frac{1}{q-1}+\frac{1}{r-1} \\ & +\frac{1}{(p-1)(q-1)}+\frac{1}{(q-1)(r-1)}+\frac{1}{(r-1)(p-1)} \end{aligned} $$  Obviously $F$ is a decreasing function of $p, q, r$. Suppose that $1<p<q<r$ are integers for which $F(p, q, r)$ is an integer. Observe that $p, q, r$ are either all even or all odd. Indeed, if for example $p$ is odd and $q$ is even, then $p q r-1$ is odd while $(p-1)(q-1)(r-1)$ is even, which is impossible. Also, if $p, q, r$ are even then $F(p, q, r)$ is odd. If $p \geq 4$, then $1<F(p, q, r) \leq F(4,6,8)=191 / 105<2$, which is impossible. Hence $p \leq 3$. Let $p=2$. Then $q, r$ are even and $1<F(2, q, r) \leq F(2,4,6)=47 / 15<4$. Therefore $F(2, q, r)=3$. This equality reduces to $(q-3)(r-3)=5$, with the unique solution $q=4, r=8$. Let $p=3$. Then $q, r$ are odd and $1<F(3, q, r) \leq F(3,5,7)=104 / 48<3$. Therefore $F(3, q, r)=2$. This equality reduces to $(q-4)(r-4)=11$, which leads to $q=5, r=15$. Hence the only solutions $(p, q, r)$ of the problem are $(2,4,8)$ and $(3,5,15)$.


14. (POL 2) For any positive integer $x$ define  $$ \begin{aligned} g(x) & =\text { greatest odd divisor of } x, \\ f(x) & =\left\{\begin{array}{l} x / 2+x / g(x), \text { if } x \text { is even; } \\ 2^{(x+1) / 2}, \text { if } x \text { is odd. } \end{array}\right. \end{aligned} $$  Construct the sequence $x_{1}=1, x_{n+1}=f\left(x_{n}\right)$. Show that the number 1992 appears in this sequence, determine the least $n$ such that $x_{n}=1992$, and determine whether $n$ is unique.

Solution: 14. We see that $x_{1}=2^{0}$. Suppose that for some $m, r \in \mathbb{N}$ we have $x_{m}=2^{r}$. Then inductively $x_{m+i}=2^{r-i}(2 i+1)$ for $i=1,2, \ldots, r$ and $x_{m+r+1}=$ $2^{r+1}$. Since every natural number can be uniquely represented as the product of an odd number and a power of two, we conclude that every natural number occurs in our sequence exactly once. Moreover, it follows that $2 k-1=x_{k(k+1) / 2}$. Thus $x_{n}=1992=2^{3} \cdot 249$ implies that $x_{n+3}=255=2 \cdot 128-1=x_{128 \cdot 129 / 2}=x_{8256}$. Hence $n=8253$.


15. (PRK 2) Does there exist a set $M$ with the following properties? (i) The set $M$ consists of 1992 natural numbers. (ii) Every element in $M$ and the sum of any number of elements have the form $m^{k}(m, k \in \mathbb{N}, k \geq 2)$.

Solution: 15. The result follows from the following lemma by taking $n=\frac{1992 \cdot 1993}{2}$ and $M=\{d, 2 d, \ldots, 1992 d\}$. Lemma. For every $n \in \mathbb{N}$ there exists a natural number $d$ such that all the numbers $d, 2 d, \ldots, n d$ are of the form $m^{k}(m, k \in \mathbb{N}, k \geq 2)$. Proof. Let $p_{1}, p_{2}, \ldots, p_{n}$ be distinct prime numbers. We shall find $d$ in the form $d=2^{\alpha_{2}} 3^{\alpha_{3}} \cdots n^{\alpha_{n}}$, where $\alpha_{i} \geq 0$ are integers such that $k d$ is a perfect $p_{k}$ th power. It is sufficient to find $\alpha_{i}, i=2,3, \ldots, n$, such that $\alpha_{i} \equiv 0\left(\bmod p_{j}\right)$ if $i \neq j$ and $\alpha_{i} \equiv-1\left(\bmod p_{j}\right)$ if $i=j$. But the existence of such $\alpha_{i}$ 's is an immediate consequence of the Chinese remainder theorem.


16. (KOR 2) Prove that $N=\frac{5^{125}-1}{5^{25}-1}$ is a composite number.

Solution: 16. Observe that $x^{4}+x^{3}+x^{2}+x+1=\left(x^{2}+3 x+1\right)^{2}-5 x(x+1)^{2}$. Thus for $x=5^{25}$ we have  $$ \begin{aligned} N & =x^{4}+x^{3}+x^{2}+x+1 \\ & =\left(x^{2}+3 x+1-5^{13}(x+1)\right)\left(x^{2}+3 x+1+5^{13}(x+1)\right)=A \cdot B . \end{aligned} $$  Clearly, both $A$ and $B$ are positive integers greater than 1.


17. (SWE 1) Let $\alpha(n)$ be the number of digits equal to one in the binary representation of a positive integer $n$. Prove that: (a) the inequality $\alpha\left(n^{2}\right) \leq \frac{1}{2} \alpha(n)(\alpha(n)+1)$ holds; (b) the above inequality is an equality for infinitely many positive integers; (c) there exists a sequence $\left(n_{i}\right)_{1}^{\infty}$ such that $\alpha\left(n_{i}^{2}\right) / \alpha\left(n_{i}\right) \rightarrow 0$ as $i \rightarrow \infty$.  Alternative parts: Prove that there exists a sequence $\left(n_{i}\right)_{1}^{\infty}$ such that $\alpha\left(n_{i}^{2}\right) / \alpha\left(n_{i}\right)$ tends to (d) $\infty$; (e) an arbitrary real number $\gamma \in(0,1)$; (f) an arbitrary real number $\gamma \geq 0$.

Solution: 17. (a) Let $n=\sum_{i=1}^{k} 2^{a_{i}}$, so that $\alpha(n)=k$. Then  $$ n^{2}=\sum_{i} 2^{2 a_{i}}+\sum_{i<j} 2^{a_{i}+a_{j}+1} $$  has at most $k+\binom{k}{2}=\frac{k(k+1)}{2}$ binary ones. (b) The above inequality is an equality for all numbers $n_{k}=2^{k}$. (c) Put $n_{m}=2^{2^{m}-1}-\sum_{j=1}^{m} 2^{2^{m}-2^{j}}$, where $m>1$. It is easy to see that $\alpha\left(n_{m}\right)=2^{m}-m$. On the other hand, squaring and simplifying yields $n_{m}^{2}=1+\sum_{i<j} 2^{2^{m+1}+1-2^{i}-2^{j}}$. Hence $\alpha\left(n_{m}^{2}\right)=1+\frac{m(m+1)}{2}$ and thus  $$ \frac{\alpha\left(n_{m}^{2}\right)}{\alpha\left(n_{m}\right)}=\frac{2+m(m+1)}{2\left(2^{m}-m\right)} \rightarrow 0 \quad \text { as } m \rightarrow \infty $$  Solution to the alternative parts. (1) Let $n=\sum_{i=1}^{n} 2^{2^{i}}$. Then $n^{2}=\sum_{i=1}^{n} 2^{2^{i+1}}+\sum_{i<j} 2^{2^{i}+2^{j}+1}$ has exactly $\frac{k(k+1)}{2}$ binary ones, and therefore $\frac{\alpha\left(n^{2}\right)}{\alpha(n)}=\frac{2 k}{k(k+1)} \rightarrow \infty$. (2) Consider the sequence $n_{i}$ constructed in part (c). Let $\theta>1$ 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. (USA 2) Let $[x]$ denote the greatest integer less than or equal to $x$. Pick any $x_{1}$ in $[0,1)$ and define the sequence $x_{1}, x_{2}, x_{3}, \ldots$ by $x_{n+1}=0$ if $x_{n}=0$ and $x_{n+1}=1 / x_{n}-\left[1 / x_{n}\right]$ otherwise. Prove that  $$ x_{1}+x_{2}+\cdots+x_{n}<\frac{F_{1}}{F_{2}}+\frac{F_{2}}{F_{3}}+\cdots+\frac{F_{n}}{F_{n+1}} $$  where $F_{1}=F_{2}=1$ and $F_{n+2}=F_{n+1}+F_{n}$ for $n \geq 1$.

Solution: 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. (IRE 1) Let $f(x)=x^{8}+4 x^{6}+2 x^{4}+28 x^{2}+1$. Let $p>3$ be a prime and suppose there exists an integer $z$ such that $p$ divides $f(z)$. Prove that there exist integers $z_{1}, z_{2}, \ldots, z_{8}$ such that if  $$ g(x)=\left(x-z_{1}\right)\left(x-z_{2}\right) \cdots\left(x-z_{8}\right) $$  then all coefficients of $f(x)-g(x)$ are divisible by $p$.

Solution: 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.


2. (CHN 1) Let $\mathbb{R}^{+}$be the set of all nonnegative real numbers. Given two positive real numbers $a$ and $b$, suppose that a mapping $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ satisfies the functional equation  $$ f(f(x))+a f(x)=b(a+b) x $$  Prove that there exists a unique solution of this equation.

Solution: 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.


20. (FRA 1) ${ }^{\mathrm{IMO} 4}$ In the plane, let there be given a circle $C$, a line $l$ tangent to $C$, and a point $M$ on $l$. Find the locus of points $P$ that have the following property: There exist two points $Q$ and $R$ on $l$ such that $M$ is the midpoint of $Q R$ and $C$ is the incircle of $P Q R$.

Solution: 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 ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-575.jpg?height=543&width=539&top_left_y=908&top_left_x=799) the extension of $U^{\prime} X$ over $X$.


21. (GBR 2) ${ }^{\mathrm{IMO} 6}$ For each positive integer $n$, denote by $s(n)$ the greatest integer such that for all positive integers $k \leq s(n), n^{2}$ can be expressed as a sum of squares of $k$ positive integers. (a) Prove that $s(n) \leq n^{2}-14$ for all $n \geq 4$. (b) Find a number $n$ such that $s(n)=\overline{n^{2}}-14$. (c) Prove that there exist infinitely many positive integers $n$ such that $s(n)=n^{2}-14$.

Solution: 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 $n$ :  $$ f(n)= \begin{cases}1, & \text { if } n \text { has a prime divisor congruent to } 3 \bmod 4 \\ 2, & \text { if } n \text { is of the form } 5 \cdot 2^{k}, k \text { a positive integer; } \\ n^{2}-14, & \text { otherwise. }\end{cases} $$


3. (CHN 2) The diagonals of a quadrilateral $A B C D$ are perpendicular: $A C \perp B D$. Four squares, $A B E F, B C G H, C D I J, D A K L$, are erected externally on its sides. The intersection points of the pairs of straight lines $C L, D F ; D F, A H ; A H, B J ; B J, C L$ are denoted by $P_{1}, Q_{1}, R_{1}, S_{1}$, respectively, and the intersection points of the pairs of straight lines $A I, B K$; $B K, C E ; C E, D G ; D G, A I$ are denoted by $P_{2}, Q_{2}, R_{2}, S_{2}$, respectively. Prove that $P_{1} Q_{1} R_{1} S_{1} \cong P_{2} Q_{2} R_{2} S_{2}$.

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 $\angle F T A=$ ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-567.jpg?height=342&width=470&top_left_y=1659&top_left_x=847) $\angle A T K=45^{\circ}$. 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. $(\mathbf{C H N} 3)^{\mathrm{IMO}}$ Given nine points in space, no four of which are coplanar, find the minimal natural number $n$ such that for any coloring with red or blue of $n$ edges drawn between these nine points there always exists a triangle having all edges of the same color.

Solution: 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. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-568.jpg?height=338&width=396&top_left_y=1069&top_left_x=326)  Fig. 1 ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-568.jpg?height=326&width=396&top_left_y=1079&top_left_x=868)  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 $r(k, l)<n$. Every graph with $n$ vertices and $t(n, r(k, l))$ +1 edges contains a complete subgraph with $r(k, l)$ vertices, and this subgraph contains either a red complete $k$-graph or a blue complete $l$ graph. We claim that $t(n, r(k, l))+1$ is the smallest number of edges with the above property. By the definition of $r(k, l)$ there exists a coloring of the complete graph $H$ with $r(k, l)-1$ vertices in two colors such that no red complete $k$-graph or blue complete $l$-graph exists. Let $c_{i j}$ be the color in which the edge $(i, j)$ of $H$ is colored, $1 \leq i<j \leq r(k, l)-1$. Consider a complete $r(k, l)$-1-partite graph $G$ with $n$ vertices and exactly $t(n, r(k, l))$ edges and denote its partitions by $P_{i}, i=1, \ldots, r(k, l)-1$. If we color each edge of $H$ between $P_{i}$ and $P_{j}(j<i)$ in the color $c_{i j}$, we obviously obtain a graph with $n$ vertices and $t(n, r(k, l))$ edges in two colors that contains neither a red complete $k$-graph nor a blue complete $l$-graph. Therefore the answer to our problem is $t(9, r(3,3))+1=t(9,6)+1=33$.


5. (COL 3) Let $A B C D$ be a convex quadrilateral such that $A C=$ $B D$. Equilateral triangles are constructed on the sides of the quadrilateral. Let $O_{1}, O_{2}, O_{3}, O_{4}$ be the centers of the triangles constructed on $A B, B C, C D, D A$ respectively. Show that $O_{1} O_{3}$ is perpendicular to $O_{2} O_{4}$.

Solution: 5. Denote by $K, L, M$, and $N$ the midpoints of the sides $A B, B C, C D$, and $D A$, respectively. The quadrilateral $K L M N$ is a rhombus. We shall prove that $O_{1} O_{3} \| K M$. Similarly, $O_{2} O_{4} \| L N$, and the desired result follows immediately. We have $\overrightarrow{O_{1} O_{3}}=\overrightarrow{K M}+\left(\overrightarrow{O_{1} K}+\overrightarrow{M O_{3}}\right)$. Assume that $A B C D$ is positively oriented. A rotational homothety $\mathcal{R}$ with angle $-90^{\circ}$ and coefficient $1 / \sqrt{3}$ takes the vectors $\overrightarrow{B K}$ and $\overrightarrow{C M}$ into $\overrightarrow{O_{1} K}$ and $\overrightarrow{M O_{3}}$ respectively. Therefore  $$ \begin{aligned} \overrightarrow{O_{1} O_{3}} & =\overrightarrow{K M}+\left(\overrightarrow{O_{1} K}+\overrightarrow{M O_{3}}\right)=\overrightarrow{K M}+\mathcal{R}(\overrightarrow{B K}+\overrightarrow{C M}) \\ & =\overrightarrow{K M}+\frac{1}{2} \mathcal{R}(\overrightarrow{B A}+\overrightarrow{C D})=\overrightarrow{K M}+\mathcal{R}(\overrightarrow{L N}) \end{aligned} $$  Since $L N \perp K M$, it follows that $\mathcal{R}(L N)$ is parallel to $K M$ and so is $\mathrm{O}_{1} \mathrm{O}_{3}$.


6. (IND 2) ${ }^{\mathrm{IMO} 2}$ Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that  $$ f\left(x^{2}+f(y)\right)=y+f(x)^{2} \quad \text { for all } x, y \text { in } \mathbb{R} $$

Solution: 6. It is easy to see that $f$ is injective and surjective. From $f\left(x^{2}+f(y)\right)=$ $f\left((-x)^{2}+f(y)\right)$ it follows that $f(x)^{2}=(f(-x))^{2}$, which implies $f(-x)=$ $-f(x)$ because $f$ is injective. Furthermore, there exists $z \in \mathbb{R}$ such that $f(z)=0$. From $f(-z)=-f(z)=0$ we deduce that $z=0$. Now we have $f\left(x^{2}\right)=f\left(x^{2}+f(0)\right)=0+(f(x))^{2}=f(x)^{2}$, and consequently $f(x)=f(\sqrt{x})^{2}>0$ 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. (IND 4) Circles $G, G_{1}, G_{2}$ are three circles related to each other as follows: Circles $G_{1}$ and $G_{2}$ are externally tangent to one another at a point $W$ and both these circles are internally tangent to the circle $G$. Points $A, B, C$ are located on the circle $G$ as follows: Line $B C$ is a direct common tangent to the pair of circles $G_{1}$ and $G_{2}$, and line $W A$ is the transverse common tangent at $W$ to $G_{1}$ and $G_{2}$, with $W$ and $A$ lying on the same side of the line $B C$. Prove that $W$ is the incenter of the triangle $A B C$.

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. (IND 5) Show that in the plane there exists a convex polygon of 1992 sides satisfying the following conditions: (i) its side lengths are $1,2,3, \ldots, 1992$ in some order; (ii) the polygon is circumscribable about a circle.  Alternative formulation. Does there exist a 1992-gon with side lengths $1,2,3, \ldots, 1992$ circumscribed about a circle? Answer the same question for a 1990-gon.

Solution: 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 $f(r) \rightarrow 0$ ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-570.jpg?height=331&width=444&top_left_y=1156&top_left_x=887) 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. (IRN 1) Let $f(x)$ be a polynomial with rational coefficients and $\alpha$ be a real number such that $\alpha^{3}-\alpha=f(\alpha)^{3}-f(\alpha)=33^{1992}$. Prove that for each $n \geq 1$,  $$ \left(f^{(n)}(\alpha)\right)^{3}-f^{(n)}(\alpha)=33^{1992} $$  where $f^{(n)}(x)=f(f(\ldots f(x)))$, and $n$ is a positive integer.

Solution: 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 $g\left(f^{(n)}(\alpha)\right)=0$.


1. (BRA 1) Show that there exists a finite set $A \subset \mathbb{R}^{2}$ such that for every $X \in A$ there are points $Y_{1}, Y_{2}, \ldots, Y_{1993}$ in $A$ such that the distance between $X$ and $Y_{i}$ is equal to 1 , for every $i$.

Solution: 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 $A$.


10. (IND 5) A natural number $n$ is said to have the property $P$ if whenever $n$ divides $a^{n}-1$ for some integer $a, n^{2}$ also necessarily divides $a^{n}-1$. (a) Show that every prime number has property $P$. (b) Show that there are infinitely many composite numbers $n$ that possess property $P$.

Solution: 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. (IRE 1) ${ }^{\mathrm{IMO1}}$ Let $n>1$ be an integer and let $f(x)=x^{n}+5 x^{n-1}+3$. Prove that there do not exist polynomials $g(x), h(x)$, each having integer coefficients and degree at least one, such that $f(x)=g(x) h(x)$.

Solution: 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 $p<a-1$ is a prime, then $F(x)=x^{m}(x-a)^{n}+p$ is irreducible. The details are left to the reader.


12. (IRE 2) Let $n, k$ be positive integers with $k \leq n$ and let $S$ be a set containing $n$ distinct real numbers. Let $T$ be the set of all real numbers of the form $x_{1}+x_{2}+\cdots+x_{k}$, where $x_{1}, x_{2}, \ldots, x_{k}$ are distinct elements of $S$. Prove that $T$ contains at least $k(n-k)+1$ distinct elements.

Solution: 12. Let $x_{1}<x_{2}<\cdots<x_{n}$ be the elements of $S$. We use induction on $n$. The result is trivial for $k=1$ or $n=k$, so assume that it is true for $n-1$ numbers. Then there exist $m=(k-1)(n-k)+1$ distinct sums of $k-1$ numbers among $x_{2}, \ldots, x_{n}$; call these sums $S_{i}, S_{1}<S_{2}<\cdots<S_{m}$. Then $x_{1}+S_{1}, x_{1}+S_{2}, \ldots, x_{1}+S_{m}$ are distinct sums of $k$ of the numbers $x_{1}, x_{2}, \ldots, x_{n}$. However, the biggest of these sums is  $$ x_{1}+S_{m} \leq x_{1}+x_{n-k+2}+x_{n-k+3}+\cdots+x_{n} $$  hence we can find $n-k$ sums that are greater and thus not included here: $x_{2}+x_{n-k+2}+\cdots+x_{n}, x_{3}+x_{n-k+2}+\cdots+x_{n}, \ldots, x_{n-k+1}+x_{n-k+2}+\cdots+x_{n}$. This counts for $k(n-k)+1$ sums in total. Remark. Equality occurs if $S$ is an arithmetic progression.


13. (IRE 3) Let $S$ be the set of all pairs $(m, n)$ of relatively prime positive integers $m, n$ with $n$ even and $m<n$. For $s=(m, n) \in S$ write $n=2^{k} n_{0}$, where $k, n_{0}$ are positive integers with $n_{0}$ odd and define $f(s)=\left(n_{0}, m+\right.$ $n-n_{0}$ ). Prove that $f$ is a function from $S$ to $S$ and that for each $s=(m, n) \in S$, there exists a positive integer $t \leq \frac{m+n+1}{4}$ such that $f^{t}(s)=s$, where  $$ f^{t}(s)=\underbrace{(f \circ f \circ \cdots \circ f)}_{t \text { times }}(s) . $$  If $m+n$ is a prime number that does not divide $2^{k}-1$ for $k=1,2, \ldots, m+$ $n-2$, prove that the smallest value of $t$ that satisfies the above conditions is $\left[\frac{m+n+1}{4}\right]$, where $[x]$ denotes the greatest integer less than or equal to $x$.

Solution: 13. For an odd integer $N>1$, 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 $m_{1} \leq \frac{n}{2}<$ $\frac{N}{2}<n_{1}$, so $f$ maps $S_{N}$ to $S_{N}$. Also $f$ is bijective, since if $f(m, n)=$ $\left(m_{1}, n_{1}\right)$, then $n$ is uniquely determined as the even number of the form $2^{k} m_{1}$ that belongs to the interval $\left[\frac{N+1}{2}, N\right]$, and this also determines $m$. Note that $S_{N}$ has at most $\left[\frac{N+1}{4}\right]$ elements, with equality if and only if $N$ is prime. Thus if $(m, n) \in S_{N}$, there exist $s, r$ with $1 \leq s<r \leq\left[\frac{N+5}{4}\right]$ such that $f^{s}(m, n)=f^{r}(m, n)$. Consequently $f^{t}(m, n)=(m, n)$, where $t=r-s, 0<t \leq\left[\frac{N+1}{4}\right]=\left[\frac{m+n+1}{4}\right]$. Suppose that $(m, n) \in S_{N}$ and $t$ is the least positive integer with $f^{t}(m, n)=(m, n)$. We write $(m, n)=\left(m_{0}, n_{0}\right)$ and $f^{i}(m, n)=\left(m_{i}, n_{i}\right)$ for $i=1, \ldots, t$. Then there exist positive integers $a_{i}$ such that $2^{a_{i}} m_{i}=n_{i-1}$, $i=1, \ldots, t$. Since $m_{t}=m_{0}$, multiplying these equalities gives  $$ \begin{aligned} 2^{a_{1}+a_{2}+\cdots+a_{t}} m_{0} m_{1} \cdots m_{t-1} & =n_{0} n_{1} \cdots n_{t-1} \\ & \equiv(-1)^{t} m_{0} m_{1} \cdots m_{t-1}(\bmod N) . \end{aligned} $$  It follows that $N \mid 2^{k} \pm 1$ and consequently $N \mid 2^{2 k}-1$, where $k=$ $a_{1}+\cdots+a_{t}$. On the other hand, it also follows that $2^{k}\left|n_{0} n_{1} \cdots n_{t-1}\right|$ $(N-1)(N-3) \cdots(N-2[N / 4])$. But since  $$ \frac{(N-1)(N-3) \cdots\left(N-2\left[\frac{N}{4}\right]\right)}{1 \cdot 3 \cdots\left(2\left[\frac{N-2}{4}\right]+1\right)}=\frac{2 \cdot 4 \cdots(N-1)}{1 \cdot 2 \cdots \frac{N-1}{2}}=2^{\frac{N-1}{2}} $$  we conclude that $0<k \leq \frac{N-1}{2}$, where equality holds if and only if $\left\{n_{1}, \ldots, n_{t}\right\}$ is the set of all even integers from $\frac{N+1}{2}$ to $N-1$, and consequently $t=\frac{N+1}{4}$. Now if $N \nmid 2^{h}-1$ for $1 \leq h<N-1$, we must have $2 k=N-1$. Therefore $t=\frac{N+1}{4}$.


14. (ISR 1) The vertices $D, E, F$ of an equilateral triangle lie on the sides $B C, C A, A B$ respectively of a triangle $A B C$. If $a, b, c$ are the respective lengths of these sides, and $S$ the area of $A B C$, prove that  $$ D E \geq \frac{2 \sqrt{2} S}{\sqrt{a^{2}+b^{2}+c^{2}+4 \sqrt{3} S}} $$

Solution: 14. Consider any point $T$ inside the triangle $A B C$ or on its boundary. Since  $$ \begin{aligned} 2 S & =2\left(S_{A E T F}+S_{B F T D}+S_{C D T E}\right) \\ & \leq A T \cdot E F+B T \cdot F D+C T \cdot D E=(A T+B T+C T) D E \end{aligned} $$  it suffices to find a point $T$ such that  $$ (A T+B T+C T)^{2} \geq \frac{a^{2}+b^{2}+c^{2}+4 S \sqrt{3}}{2} . $$  We distinguish two cases: (i) If all angles of $\triangle A B C$ are less than $120^{\circ}$, then the sum $A T+B T+C T$ attains its minimum when $T$ is the Torricelli point, i.e., such that $\angle A T B=\angle B T C=\angle C T A=120^{\circ}$. In this case, by the cosine theorem we get  $$ \begin{aligned} A T^{2}+A T \cdot B T+B T^{2} & =c^{2}, \\ B T^{2}+B T \cdot C T+C T^{2} & =a^{2}, \\ C T^{2}+C T \cdot A T+A T^{2} & =b^{2}, \\ 3(A T \cdot B T+B T \cdot C T+C T \cdot A T) & =4 \sqrt{3}\left(S_{A T B}+S_{B T C}+S_{C T A}\right) \\ & =4 \sqrt{3} S . \end{aligned} $$  Adding these four equalities, we obtain $2(A T+B T+C T)^{2}=a^{2}+$ $b^{2}+c^{2}+4 \sqrt{3} S$. (ii) Let $\angle A C B \geq 120^{\circ}$. We claim that $T=C$ satisfies the requirements. Indeed, $a^{2}+b^{2}+c^{2}+4 \sqrt{3} S=a^{2}+b^{2}+\left(a^{2}+b^{2}-2 a b \cos \angle C\right)+$ $2 \sqrt{3} a b \sin \angle C=2\left(a^{2}+b^{2}\right)+2 a b(\sqrt{3} \sin \angle C-\cos \angle C)=2\left(a^{2}+b^{2}\right)+$ $4 a b \sin \left(\angle C-30^{\circ}\right) \leq 2(a+b)^{2}$, which proves the desired inequality.


15. (MCD 1) ${ }^{\mathrm{IMO} 4}$ For three points $A, B, C$ in the plane we define $m(A B C)$ to be the smallest length of the three altitudes of the triangle $A B C$, where in the case of $A, B, C$ collinear, $m(A B C)=0$. Let $A, B, C$ be given points in the plane. Prove that for any point $X$ in the plane,  $$ m(A B C) \leq m(A B X)+m(A X C)+m(X B C) $$

Solution: 15. Denote by $d(P Q R)$ the diameter of a triangle $P Q R$. It is clear that $d(P Q R) \cdot m(P Q R)=2 S_{P Q R}$. So if the point $X$ lies inside the triangle $A B C$ or on its boundary, we have $d(A B X), d(B C X), d(C A X) \leq d(A B C)$, which implies  $$ \begin{aligned} m(A B X)+m(B C X)+m(C A X) & =\frac{2 S_{A B X}}{d(A B X)}+\frac{2 S_{B C X}}{d(B C X)}+\frac{2 S_{C A X}}{d(C A X)} \\ & \geq \frac{2 S_{A B X}+2 S_{B C X}+2 S_{C A X}}{d(A B C)} \\ & =\frac{2 S_{A B C}}{d(A B C)}=m(A B C) \end{aligned} $$  If $X$ is outside $\triangle A B C$ but inside the angle $B A C$, consider the point $Y$ of intersection of $A X$ and $B C$. Then $m(A B X)+m(B C X)+m(C A X) \geq$ $m(A B Y)+m(B C Y)+m(C A Y) \geq m(A B C)$. Also, if $X$ is inside the opposite angle of $\angle B A C$ (i.e., $\angle D A E$, where $\mathcal{B}(D, A, B)$ and $\mathcal{B}(E, A, C)$ ), then $m(A B X)+m(B C X)+m(C A X) \geq m(B C X) \geq m(A B C)$. Since these are substantially all possible different positions of point $X$, we have finished the proof.


16. (MCD 3) Let $n \in \mathbb{N}, n \geq 2$, and $A_{0}=\left(a_{01}, a_{02}, \ldots, a_{0 n}\right)$ be any $n$-tuple of natural numbers such that $0 \leq a_{0 i} \leq i-1$, for $i=1, \ldots, n$. The $n$-tuples $A_{1}=\left(a_{11}, a_{12}, \ldots, a_{1 n}\right), A_{2}=\left(a_{21}, a_{22}, \ldots, a_{2 n}\right), \ldots$ are defined by $a_{i+1, j}=\operatorname{Card}\left\{a_{i, l} \mid 1 \leq l \leq j-1, a_{i, l} \geq a_{i, j}\right\}, \quad$ for $i \in \mathbb{N}$ and $j=1, \ldots, n$. Prove that there exists $k \in \mathbb{N}$, such that $A_{k+2}=A_{k}$.

Solution: 16. Let $S_{n}=\left\{A=\left(a_{1}, \ldots, a_{n}\right) \mid 0 \leq a_{i}<i\right\}$. For $A=\left(a_{1}, \ldots, a_{n}\right)$, let $A^{\prime}=\left(a_{1}, \ldots, a_{n-1}\right)$, so that we can write $A=\left(A^{\prime}, a_{n}\right)$. The proof of the statement from the problem will be given by induction on $n$. For $n=2$ there are two possibilities for $A_{0}$, so one directly checks that $A_{2}=A_{0}$. Now assume that $n \geq 3$ and that $A_{0}=\left(A_{0}^{\prime}, a_{0 n}\right) \in S_{n}$. It is clear that then any $A_{i}$ is in $S_{n}$ too. By the induction hypothesis there exists $k \in \mathbb{N}$ such that $A_{k}^{\prime}=A_{k+2}^{\prime}=A_{k+4}^{\prime}=\cdots$ and $A_{k+1}^{\prime}=A_{k+3}^{\prime}=\cdots$. Observe that if we increase (decrease) $a_{k n}, a_{k+1, n}$ will decrease (respectively increase), and this will also increase (respectively decrease) $a_{k+2, n}$. Hence $a_{k n}, a_{k+2, n}, a_{k+4, n}, \ldots$ is monotonically increasing or decreasing, and since it is bounded (by 0 and $n-1$ ), it follows that we will eventually have $a_{k+2 i, n}=a_{k+2 i+2, n}=\cdots$. Consequently $A_{k+2 i}=A_{k+2 i+2}$.


17. (NET 2) ${ }^{\mathrm{IMO} 6}$ Let $n$ be an integer greater than 1 . In a circular arrangement of $n$ lamps $L_{0}, \ldots, L_{n-1}$, each one of that can be either ON or OFF, we start with the situation where all lamps are ON, and then carry out a sequence of steps, $S_{t e p_{0}}, S t e p_{1}, \ldots$. If $L_{j-1}(j$ is taken $\bmod n)$ is ON, then $S_{t e p}^{j}$ changes the status of $L_{j}$ (it goes from ON to OFF or from OFF to ON) but does not change the status of any of the other lamps. If $L_{j-1}$ is OFF, then $S_{t e p}^{j}$ does not change anything at all. Show that: (a) There is a positive integer $M(n)$ such that after $M(n)$ steps all lamps are ON again. (b) If $n$ has the form $2^{k}$, then all lamps are ON after $n^{2}-1$ steps. (c) If $n$ has the form $2^{k}+1$, then all lamps are ON after $n^{2}-n+1$ steps.

Solution: 17. We introduce the rotation operation Rot to the left by one, so that $S_{t e p_{j}}=$ Rot $^{-j} \circ$ Step $_{0} \circ$ Rot $^{j}$. Now writing Step ${ }^{*}={\text { Rot } \circ \text { Step }_{0} \text {, the problem is }}^{-20}$ transformed into the question whether there is an $M(n)$ such that all lamps are $O N$ again after $M(n)$ successive applications of Step*. We operate in the field $\mathbb{Z}_{2}$, representing $O F F$ by 0 and $O N$ by 1 . So if the status of $L_{j}$ at some moment is given by $v_{j} \in \mathbb{Z}_{2}$, the effect of $S t e p_{j}$ is that $v_{j}$ is replaced by $v_{j}+v_{j-1}$. With the $n$-tuple $v_{0}, \ldots, v_{n-1}$ we associate the polynomial  $$ P(x)=v_{n-1} x^{n-1}+v_{0} x^{n-2}+v_{1} x^{n-3}+\cdots+v_{n-2} . $$  By means of Step*, this polynomial is transformed into the polynomial $Q(x)$ over $\mathbb{Z}$ of degree less than $n$ that satisfies $Q(x) \equiv x P(x)(\bmod$ $\left.x^{n}+x^{n-1}+1\right)$. From now on, the sign $\equiv$ always stands for congruence with this modulus. (i) It suffices to show the existence of $M(n)$ with $x^{M(n)} \equiv 1$. Because the number of residue classes is finite, there are $r, q, r<q$ such that $x^{q} \equiv x^{r}$, i.e., $x^{r}\left(x^{q-r}-1\right)=0$. One can take $M(n)=q-r$. (Or simply note that there are only finitely many possible configurations; since each operation is bijective, the configuration that reappears first must be $O N, O N, \ldots, O N$.) (ii) We shall prove that if $n=2^{k}$, then $x^{n^{2}-1} \equiv 1$. We have $x^{n^{2}} \equiv$ $\left(x^{n-1}+1\right)^{n} \equiv x^{n^{2}-n}+1$, because all binomial coefficients of order $n=2^{k}$ are even, apart from the first one and the last one. Since also $x^{n^{2}} \equiv x^{n^{2}-1}+x^{n^{2}-n}$, this is what we wanted. (iii) Now if $n=2^{k}+1$, we prove that $x^{n^{2}-n+1} \equiv 1$. We have $x^{n^{2}-1} \equiv$ $\left(x^{n+1}\right)^{n-1} \equiv\left(x+x^{n}\right)^{n-1} \equiv x^{n-1}+x^{n^{2}-n}$ (again by evenness of binomial coefficients of order $n-1=2^{k}$ ). Together with $x^{n^{2}} \equiv x^{n^{2}-1}+$ $x^{n^{2}-n}$, this leads to $x^{n^{2}} \equiv x^{n-1}$.


18. (POL 1) Let $S_{n}$ be the number of sequences $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$, where $a_{i} \in$ $\{0,1\}$, in which no six consecutive blocks are equal. Prove that $S_{n} \rightarrow \infty$ as $n \rightarrow \infty$.

Solution: 18. Let $B_{n}$ be the set of sequences with the stated property $\left(S_{n}=\left|B_{n}\right|\right)$. We shall prove by induction on $n$ that $S_{n} \geq \frac{3}{2} S_{n-1}$ for every $n$. Suppose that for every $i \leq n, S_{i} \geq \frac{3}{2} S_{i-1}$, and consequently $S_{i} \leq$ $\left(\frac{2}{3}\right)^{n-i} S_{n}$. Let us consider the $2 S_{n}$ sequences obtained by putting 0 or 1 at the end of any sequence from $B_{n}$. If some sequence among them does not belong to $B_{n+1}$, then for some $k \geq 1$ it can be obtained by extending some sequence from $B_{n+1-6 k}$ by a sequence of $k$ terms repeated six times. The number of such sequences is $2^{k} S_{n+1-6 k}$. Hence the number of sequences not satisfying our condition is not greater than  $$ \sum_{k \geq 1} 2^{k} S_{n+1-6 k} \leq \sum_{k \geq 1} 2^{k}\left(\frac{2}{3}\right)^{6 k-1} S_{n}=\frac{3}{2} S_{n} \frac{2(2 / 3)^{6}}{1-2(2 / 3)^{6}}=\frac{192}{601} S_{n}<\frac{1}{2} S_{n} $$  Therefore $S_{n+1}$ is not smaller than $2 S_{n}-\frac{1}{2} S_{n}=\frac{3}{2} S_{n}$. Thus we have $S_{n} \geq\left(\frac{3}{2}\right)^{n}$.


19. (ROM 2) Let $a, b, n$ be positive integers, $b>1$ and $b^{n}-1 \mid a$. Show that the representation of the number $a$ in the base $b$ contains at least $n$ digits different from zero.

Solution: 19. Let $s$ be the minimum number of nonzero digits that can appear in the $b$ adic representation of any number divisible by $b^{n}-1$. Among all numbers divisible by $b^{n}-1$ and having $s$ nonzero digits in base $b$, we choose the number $A$ with the minimum sum of digits. Let $A=a_{1} b^{n_{1}}+\cdots+a_{s} b^{n_{s}}$, where $0<a_{i} \leq b-1$ and $n_{1}>n_{2}>\cdots>n_{s}$. 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 $a_{i}+a_{j}<b$, then $B$ has $s-1$ digits in base $b$, which is impossible; on the other hand, $a_{i}+a_{j} \geq b$ is also impossible, for otherwise $B$ would have sum of digits less for $b-1$ than that of $A$ (because $B$ would have digits 1 and $a_{i}+a_{j}-b$ in the positions $\left.n_{j}+k n+1, n_{j}+k n\right)$. Therefore $n_{i} \not \equiv n_{j}$ if $i \neq j$. Let $n_{i} \equiv r_{i}$, where $r_{i} \in\{0,1, \ldots, n-1\}$ are distinct. The number $C=$ $a_{1} b^{r_{1}}+\cdots+a_{s} b^{r_{s}}$ also has $s$ digits and is divisible by $b^{n}-1$. But since $C<b^{n}$, the only possibility is $C=b^{n}-1$ which has exactly $n$ digits in base $b$. It follows that $s=n$.


2. (CAN 2) Let triangle $A B C$ be such that its circumradius $R$ is equal to 1. Let $r$ be the inradius of $A B C$ and let $p$ be the inradius of the orthic triangle $A^{\prime} B^{\prime} C^{\prime}$ of triangle $A B C$. Prove that $p \leq 1-\frac{1}{3}(1+r)^{2}$. Remark. The orthic triangle is the triangle whose vertices are the feet of the altitudes of $A B C$.

Solution: 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  $$ \begin{aligned} p+\frac{1}{3}(1+r)^{2} & =2 \cos A \cos B \cos C+\frac{1}{3}(\cos A+\cos B+\cos C)^{2} \\ & \leq 2 \cos A \cos B \cos C+\cos ^{2} A+\cos ^{2} B+\cos ^{2} C=1 \end{aligned} $$


20. (ROM 3) Let $c_{1}, \ldots, c_{n} \in \mathbb{R}(n \geq 2)$ such that $0 \leq \sum_{i=1}^{n} c_{i} \leq n$. Show that we can find integers $k_{1}, \ldots, k_{n}$ such that $\sum_{i=1}^{n} k_{i}=0$ and  $$ 1-n \leq c_{i}+n k_{i} \leq n \quad \text { for every } i=1, \ldots, n $$

Solution: 20. For every real $x$ we shall denote by $\lfloor x\rfloor$ and $\lceil x\rceil$ the greatest integer less than or equal to $x$ and the smallest integer greater than or equal to $x$ respectively. The condition $c_{i}+n k_{i} \in[1-n, n]$ is equivalent to $k_{i} \in I_{i}=$ $\left[\frac{1-c_{i}}{n}-1,1-\frac{c_{i}}{n}\right]$. For every $c_{i}$, this interval contains two integers (not necessarily distinct), namely $p_{i}=\left\lceil\frac{1-c_{i}}{n}-1\right\rceil \leq q_{i}=\left\lfloor 1-\frac{c_{i}}{n}\right\rfloor$. In order to show that there exist integers $k_{i} \in I_{i}$ with $\sum_{i=1}^{n} k_{i}=0$, it is sufficient to show that $\sum_{i=1}^{n} p_{i} \leq 0 \leq \sum_{i=1}^{n} q_{i}$. Since $p_{i}<\frac{1-c_{i}}{n}$, we have  $$ \sum_{i=1}^{n} p_{i}<1-\sum_{i=1}^{n} \frac{c_{i}}{n} \leq 1 $$  and consequently $\sum_{i=1}^{n} p_{i} \leq 0$ because the $p_{i}$ 's are integers. On the other hand, $q_{i}>-\frac{c_{i}}{n}$ 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. (GBR 1) A circle $S$ is said to cut a circle $\Sigma$ diametrally if their common chord is a diameter of $\Sigma$. Let $S_{A}, S_{B}, S_{C}$ be three circles with distinct centers $A, B, C$ respectively. Prove that $A, B, C$ are collinear if and only if there is no unique circle $S$ that cuts each of $S_{A}, S_{B}, S_{C}$ diametrally. Prove further that if there exists more than one circle $S$ that cuts each of $S_{A}, S_{B}, S_{C}$ diametrally, then all such circles pass through two fixed points. Locate these points in relation to the circles $S_{A}, S_{B}, S_{C}$.

Solution: 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 ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-586.jpg?height=480&width=544&top_left_y=1409&top_left_x=810) 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. (GBR 2) ${ }^{\mathrm{IMO} 2} A, B, C, D$ are four points in the plane, with $C, D$ on the same side of the line $A B$, such that $A C \cdot B D=A D \cdot B C$ and $\measuredangle A D B=$ $90^{\circ}+\measuredangle A C B$. Find the ratio  $$ \frac{A B \cdot C D}{A C \cdot B D} $$  and prove that circles $A C D, B C D$ are orthogonal. (Intersecting circles are said to be orthogonal if at either common point their tangents are perpendicular.)

Solution: 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 $\frac{A B \cdot C D}{A C \cdot B D}=\sqrt{2}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-587.jpg?height=380&width=391&top_left_y=239&top_left_x=882)  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. (GBR 3) A finite set of (distinct) positive integers is called a " $D S$-set" if each of the integers divides the sum of them all. Prove that every finite set of positive integers is a subset of some $D S$-set.

Solution: 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 $0=k_{0}<\cdots<k_{t}$. Adjoin to the set $\left\{a_{1}, \ldots, a_{n}\right\}$ the numbers $2^{k_{i}} s, i=1,2, \ldots, t$. The sum of the enlarged set is $r s$. Finally, adjoin $r s, 2 r s, 2^{2} r s, \ldots, 2^{l-1} r s$ for $l=$ $\max \left\{k, k_{t}\right\}$. The resulting set has sum $2^{l} r s$, which is divisible by $m$ and so by each of $a_{j}$, and also by the $2^{i} s$ above and by $r s, 2 r s, \ldots, 2^{l-1} r s$. Therefore this is a $D S$-set. Second solution. We show by induction that there is a $D S$-set containing 1 and $n$. For $n=2,3$, take $\{1,2,3\}$. Assume that $\left\{1, n, b_{1}, \ldots, b_{k}\right\}$ is a $D S$ set. Then $\left\{1, n+1, n, 2(n+1) n, 2(n+1) b_{1}, \ldots, 2(n+1) b_{k}\right\}$ is a $D S$-set too. For given $a_{1}, \ldots, a_{n}$ let $m$ be a sufficiently large common multiple of the $a_{i}$ 's such that $u=m-\left(a_{1}+\cdots+a_{n}\right) \neq a_{i}$ for all $i$. There exist $b_{1}, \ldots, b_{k}$ such that $\left\{1, u, b_{1}, \ldots, b_{k}\right\}$ is a $D S$-set. It is clear that $\left\{a_{1}, \ldots, a_{n}, u, m u, m b_{1}, \ldots, m b_{k}\right\}$ is a $D S$-set containing $a_{1}, \ldots, a_{n}$.


24. (USA 3) Prove that  $$ \frac{a}{b+2 c+3 d}+\frac{b}{c+2 d+3 a}+\frac{c}{d+2 a+3 b}+\frac{d}{a+2 b+3 c} \geq \frac{2}{3} $$  for all positive real numbers $a, b, c, d$.

Solution: 24. By the Cauchy-Schwarz inequality, if $x_{1}, x_{2}, \ldots, x_{n}$ and $y_{1}, y_{2}, \ldots, y_{n}$ are positive numbers, then  $$ \left(\sum_{i=1}^{n} \frac{x_{i}}{y_{i}}\right)\left(\sum_{i=1}^{n} x_{i} y_{i}\right) \geq\left(\sum_{i=1}^{n} x_{i}\right)^{2} $$  Applying this to the numbers $a, b, c, d$ and $b+2 c+3 d, c+2 d+3 a, d+2 a+$ $3 b, a+2 b+3 c$ (here $n=4$ ), we obtain  $$ \begin{gathered} \frac{a}{b+2 c+3 d}+\frac{b}{c+2 d+3 a}+\frac{c}{d+2 a+3 b}+\frac{d}{a+2 b+3 c} \\ \geq \frac{(a+b+c+d)^{2}}{4(a b+a c+a d+b c+b d+c d)} \geq \frac{2}{3} \end{gathered} $$  The last inequality follows, for example, from $(a-b)^{2}+(a-c)^{2}+\cdots+$ $(c-d)^{2} \geq 0$. Equality holds if and only if $a=b=c=d$. Second solution. Putting $A=b+2 c+3 d, B=c+2 d+3 a, C=d+2 a+3 b$, $D=a+2 b+3 c$, our inequality transforms into  $$ \begin{aligned} & \frac{-5 A+7 B+C+D}{24 A}+\frac{-5 B+7 C+D+A}{24 B} \\ & +\frac{-5 C+7 D+A+B}{24 C}+\frac{-5 D+7 A+B+C}{24 D} \geq \frac{2}{3} \end{aligned} $$  This follows from the arithmetic-geometric mean inequality, since $\frac{B}{A}+\frac{C}{B}+$ $\frac{D}{C}+\frac{A}{D} \geq 4$, etc.


25. (VIE 1) Solve the following system of equations, in which $a$ is a given number satisfying $|a|>1$ :  $$ \begin{aligned} x_{1}^{2} & =a x_{2}+1, \\ x_{2}^{2} & =a x_{3}+1 \\ \cdots & \cdots \\ x_{999}^{2} & =a x_{1000}+1 \\ x_{1000}^{2} & =a x_{1}+1 \end{aligned} $$

Solution: 25. We need only consider the case $a>1$ (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. (VIE 2) Let $a, b, c, d$ be four nonnegative numbers satisfying $a+b+c+d=$ 1. Prove the inequality  $$ a b c+b c d+c d a+d a b \leq \frac{1}{27}+\frac{176}{27} a b c d $$

Solution: 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.


3. (SPA 1) Consider the triangle $A B C$, its circumcircle $k$ with center $O$ and radius $R$, and its incircle with center $I$ and radius $r$. Another circle $k_{c}$ is tangent to the sides $C A, C B$ at $D, E$, respectively, and it is internally tangent to $k$. Show that the incenter $I$ is the midpoint of $D E$.

Solution: 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. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-578.jpg?height=496&width=523&top_left_y=967&top_left_x=827)  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. (SPA 2) In the triangle $A B C$, let $D, E$ be points on the side $B C$ such that $\angle B A D=\angle C A E$. If $M, N$ are, respectively, the points of tangency with $B C$ of the incircles of the triangles $A B D$ and $A C E$, show that  $$ \frac{1}{M B}+\frac{1}{M D}=\frac{1}{N C}+\frac{1}{N E} $$

Solution: 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. (FIN 3) ${ }^{\mathrm{IMO} 3}$ On an infinite chessboard, a solitaire game is played as follows: At the start, we have $n^{2}$ pieces occupying $n^{2}$ squares that form a square of side $n$. The only allowed move is a jump horizontally or vertically over an occupied square to an unoccupied one, and the piece that has been jumped over is removed. For what positive integers $n$ can the game end with only one piece remaining on the board?

Solution: 5. For $n=1$ the game is trivially over. If $n=2$, it can end, for example, in the following way: ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-579.jpg?height=144&width=534&top_left_y=1229&top_left_x=510)  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 . ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-579.jpg?height=123&width=595&top_left_y=1716&top_left_x=265)  Fig. 2 ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-579.jpg?height=272&width=272&top_left_y=1642&top_left_x=1050)  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. (GER 1) ${ }^{\mathrm{IMO} 5}$ Let $\mathbb{N}=\{1,2,3, \ldots\}$. Determine whether there exists a strictly increasing function $f: \mathbb{N} \rightarrow \mathbb{N}$ with the following properties:  $$ \begin{aligned} f(1) & =2 \\ f(f(n)) & =f(n)+n \quad(n \in \mathbb{N}) \end{aligned} $$

Solution: 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. (GEO 3) Let $a, b, c$ be given integers $a>0, a c-b^{2}=P=P_{1} \cdots P_{m}$ where $P_{1}, \ldots, P_{m}$ are (distinct) prime numbers. Let $M(n)$ denote the number of pairs of integers $(x, y)$ for which  $$ a x^{2}+2 b x y+c y^{2}=n $$  Prove that $M(n)$ is finite and $M(n)=M\left(P^{k} \cdot n\right)$ for every integer $k \geq 0$.

Solution: 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. (IND 1) Define a sequence $\langle f(n)\rangle_{n=1}^{\infty}$ of positive integers by $f(1)=1$ and  $$ f(n)= \begin{cases}f(n-1)-n, & \text { if } f(n-1)>n ; \\ f(n-1)+n, & \text { if } f(n-1) \leq n,\end{cases} $$  for $n \geq 2$. Let $S=\{n \in \mathbb{N} \mid f(n)=1993\}$. (a) Prove that $S$ is an infinite set. (b) Find the least positive integer in $S$. (c) If all the elements of $S$ are written in ascending order as $n_{1}<n_{2}<$ $n_{3}<\cdots$, show that  $$ \lim _{i \rightarrow \infty} \frac{n_{i+1}}{n_{i}}=3 . $$

Solution: 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. (IND 4) (a) Show that the set $\mathbb{Q}^{+}$of all positive rational numbers can be partitioned into three disjoint subsets $A, B, C$ satisfying the following conditions:  $$ B A=B, \quad B^{2}=C, \quad B C=A, $$  where $H K$ stands for the set $\{h k \mid h \in H, k \in K\}$ for any two subsets $H, K$ of $\mathbb{Q}^{+}$and $H^{2}$ stands for $H H$. (b) Show that all positive rational cubes are in $A$ for such a partition of $\mathbb{Q}^{+}$. (c) Find such a partition $\mathbb{Q}^{+}=A \cup B \cup C$ with the property that for no positive integer $n \leq 34$ are both $n$ and $n+1$ in $A$; that is,  $$ \min \{n \in \mathbb{N} \mid n \in A, n+1 \in A\}>34 $$

Solution: 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 $\min \{n \in \mathbb{N} \mid n \in A, n+1 \in A\} \leq 77$.


1. A1 (USA) Let $a_{0}=1994$ and $a_{n+1}=\frac{a_{n}^{2}}{a_{n}+1}$ for each nonnegative integer $n$. Prove that $1994-n$ is the greatest integer less than or equal to $a_{n}$, $0 \leq n \leq 998$.

Solution: 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 $\left\lfloor a_{n}\right\rfloor=1994-n$.


10. C5 (SWE) At a round table are 1994 girls, playing a game with a deck of $n$ cards. Initially, one girl holds all the cards. In each turn, if at least one girl holds at least two cards, one of these girls must pass a card to each of her two neighbors. The game ends when and only when each girl is holding at most one card. (a) Prove that if $n \geq 1994$, then the game cannot end. (b) Prove that if $n<1994$, then the game must end.

Solution: 10. (a) The case $n>1994$ 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. C6 (FIN) On an infinite square grid, two players alternately mark symbols on empty cells. The first player always marks $X$ 's, the second $O$ 's. One symbol is marked per turn. The first player wins if there are 11 consecutive $X$ 's in a row, column, or diagonal. Prove that the second player can prevent the first from winning.

Solution: 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. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-593.jpg?height=385&width=399&top_left_y=1025&top_left_x=851)


12. C7 (BRA) Prove that for any integer $n \geq 2$, there exists a set of $2^{n-1}$ points in the plane such that no 3 lie on a line and no $2 n$ are the vertices of a convex $2 n$-gon.

Solution: 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 $x_{1}<\cdots<x_{n+1}$. It follows that $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}>\cdots>\frac{y_{n+1}-y_{n}}{x_{n+1}-x_{n}}$. 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. G1 (FRA) A semicircle $\Gamma$ is drawn on one side of a straight line $l$. $C$ and $D$ are points on $\Gamma$. The tangents to $\Gamma$ at $C$ and $D$ meet $l$ at $B$ and $A$ respectively, with the center of the semicircle between them. Let $E$ be the point of intersection of $A C$ and $B D$, and $F$ the point on $l$ such that $E F$ is perpendicular to $l$. Prove that $E F$ bisects $\angle C F D$.

Solution: 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. G2 (UKR) $A B C D$ is a quadrilateral with $B C$ parallel to $A D . M$ is the midpoint of $C D, P$ that of $M A$ and $Q$ that of $M B$. The lines $D P$ and $C Q$ meet at $N$. Prove that $N$ is not outside triangle $A B M .{ }^{8}$

Solution: 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 $H L<K L$, and the statement follows.