16. (GBR) A sequence of real numbers $u_{1}, u_{2}, u_{3}, \ldots$ is determined by $u_{1}$ and the following recurrence relation for $n \geq 1$ :  $$ 4 u_{n+1}=\sqrt[3]{64 u_{n}+15} $$  Describe, with proof, the behavior of $u_{n}$ as $n \rightarrow \infty$.

Solution: 16. The sequence $\left\{u_{n}\right\}$ is bounded, whatever $u_{1}$ is. Indeed, assume the opposite, and let $u_{m}$ be the first member of the sequence such that $\left|u_{m}\right|>\max \left\{2,\left|u_{1}\right|\right\}$. Then $\left|u_{m-1}\right|=\left|u_{m}^{3}-15 / 64\right|>\left|u_{m}\right|$, which is impossible. Next, let us see for what values of $u_{m}, u_{m+1}$ is greater, equal, or smaller, respectively. If $u_{m+1}=u_{m}$, then $u_{m}=u_{m+1}^{3}-15 / 64=u_{m}^{3}-15 / 64$; i.e., $u_{m}$ is a root of $x^{3}-x-15 / 64=0$. This equation factors as $(x+1 / 4)\left(x^{2}-x / 4-\right.$ $15 / 16)=0$, and hence $u_{m}$ is equal to $x_{1}=(1-\sqrt{61}) / 8, x_{2}=-1 / 4$, or $x_{3}=(1+\sqrt{61}) / 8$, and these are the only possible limits of the sequence. Each of $u_{m+1}>u_{m}, u_{m+1}<u_{m}$ is equivalent to $u_{m}^{3}-u_{m}-15 / 64<0$ and $u_{m}^{3}-u_{m}-15 / 64>0$ respectively. Thus the former is satisfied for $u_{m}$ in the interval $I_{1}=\left(-\infty, x_{1}\right)$ or $I_{3}=\left(x_{2}, x_{3}\right)$, while the latter is satisfied for $u_{m}$ in $I_{2}=\left(x_{1}, x_{2}\right)$ or $I_{4}=\left(x_{3}, \infty\right)$. Moreover, since the function $f(x)=\sqrt[3]{x+15 / 64}$ is strictly increasing with fixed points $x_{1}, x_{2}, x_{3}$, it follows that $u_{m}$ will never escape from the interval $I_{1}, I_{2}, I_{3}$, or $I_{4}$ to which it belongs initially. Therefore: (1) if $u_{1}$ is one of $x_{1}, x_{2}, x_{3}$, the sequence $\left\{u_{m}\right\}$ is constant; (2) if $u_{1} \in I_{1}$, then the sequence is strictly increasing and tends to $x_{1}$; (3) if $u_{1} \in I_{2}$, then the sequence is strictly decreasing and tends to $x_{1}$; (4) if $u_{1} \in I_{3}$, then the sequence is strictly increasing and tends to $x_{3}$; (5) if $u_{1} \in I_{4}$, then the sequence is strictly decreasing and tends to $x_{3}$.


17. (USS) ${ }^{\text {IMO5 }}$ Three equal circles touch the sides of a triangle and have one common point $O$. Show that the center of the circle inscribed in and of the circle circumscribed about the triangle $A B C$ and the point $O$ are collinear.

Solution: 17. Let us denote by $S_{A}, S_{B}, S_{C}$ the centers of the given circles, where $S_{A}$ lies on the bisector of $\angle A$, etc. Then $S_{A} S_{B}\left\|A B, S_{B} S_{C}\right\| B C, S_{C} S_{A} \| C A$, so that the inner bisectors of the angles of triangle $A B C$ are also inner bisectors of the angles of $\triangle S_{A} S_{B} S_{C}$. These two triangles thus have a common incenter $S$, which is also the center of the homothety $\chi$ mapping $\triangle S_{A} S_{B} S_{C}$ onto $\triangle A B C$. The point $O$ is the circumcenter of triangle $S_{A} S_{B} S_{C}$, and so is mapped by $\chi$ onto the circumcenter $P$ of $A B C$. This means that $O, P$, and the center $S$ of $\chi$ are collinear.


18. (USS) Several equal spherical planets are given in outer space. On the surface of each planet there is a set of points that is invisible from any of the remaining planets. Prove that the sum of the areas of all these sets is equal to the area of the surface of one planet.

Solution: 18. Let $C$ be the convex hull of the set of the planets: its border consists of parts of planes, parts of cylinders, and parts of the surfaces of some planets. These parts of planets consist exactly of all the invisible points; any point on a planet that is inside $C$ is visible. Thus it remains to show that the areas of all the parts of planets lying on the border of $C$ add up to the area of one planet. As we have seen, an invisible part of a planet is bordered by some main spherical arcs, parallel two by two. Now fix any planet $P$, and translate these arcs onto arcs on the surface of $P$. All these arcs partition the surface of $P$ into several parts, each of which corresponds to the invisible part of one of the planets. This correspondence is bijective, and therefore the statement follows.


19. (YUG) A finite set of unit circles is given in a plane such that the area of their union $U$ is $S$. Prove that there exists a subset of mutually disjoint circles such that the area of their union is greater that $\frac{2 S}{9}$.

Solution: 19. Consider the partition of plane $\pi$ into regular hexagons, each having inradius 2. Fix one of these hexagons, denoted by $\gamma$. For any other hexagon $x$ in the partition, there exists a unique translation $\tau_{x}$ taking it onto $\gamma$. Define the mapping $\varphi: \pi \rightarrow \gamma$ as follows: If $A$ belongs to the interior of a hexagon $x$, then $\varphi(A)=\tau_{x}(A)$ (if $A$ is on the border of some hexagon, it does not actually matter where its image is). The total area of the images of the union of the given circles equals $S$, while the area of the hexagon $\gamma$ is $8 \sqrt{3}$. Thus there exists a point $B$ of $\gamma$ that is covered at least $\frac{S}{8 \sqrt{3}}$ times, i.e., such that $\varphi^{-1}(B)$ consists of at least $\frac{S}{8 \sqrt{3}}$ distinct points of the plane that belong to some of the circles. For any of these points, take a circle that contains it. All these circles are disjoint, with total area not less than $\frac{\pi}{8 \sqrt{3}} S \geq 2 S / 9$. Remark. The statement becomes false if the constant $2 / 9$ is replaced by any number greater than $1 / 4$. In that case a counterexample is, for example, a set of unit circles inside a circle of radius 2 covering a sufficiently large part of its area.


2. (BUL) A sphere $S$ is tangent to the edges $A B, B C, C D, D A$ of a tetrahedron $A B C D$ at the points $E, F, G, H$ respectively. The points $E, F, G, H$ are the vertices of a square. Prove that if the sphere is tangent to the edge $A C$, then it is also tangent to the edge $B D$.

Solution: 2. Lemma. Let $E, F, G, H, I$, and $K$ be points on edges $A B, B C, C D, D A$, $A C$, and $B D$ of a tetrahedron. Then there is a sphere that touches the edges at these points if and only if  $$ \begin{aligned} & A E=A H=A I, \quad B E=B F=B K, \\ & C F=C G=C I, \quad D G=D H=D K . \end{aligned} $$  Proof. The "only if" side of the equivalence is obvious. We now assume (*). Denote by $\epsilon, \phi, \gamma, \eta, \iota$, and $\kappa$ planes through $E, F, G, H, I, K$ perpendicular to $A B, B C, C D, D A, A C$ and $B D$ respectively. Since the three planes $\epsilon, \eta$, and $\iota$ are not mutually parallel, they intersect in a common point $O$. Clearly, $\triangle A E O \cong$ ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-451.jpg?height=324&width=496&top_left_y=1374&top_left_x=845) $\triangle A H O \cong \triangle A I O$; hence $O E=O H=O I=r$, and the sphere $\sigma(O, r)$ is tangent to $A B, A D, A C$. To prove that $\sigma$ is also tangent to $B C, C D, B D$ it suffices to show that planes $\phi, \gamma$, and $\kappa$ also pass through $O$. Without loss of generality we can prove this for just $\phi$. By the conditions for $E, F, I$, these are exactly the points of tangency of the incircle of $\triangle A B C$ and its sides, and if $S$ is the incenter, then $S E \perp A B, S F \perp B C, S I \perp A C$. Hence $\epsilon, \iota$, and $\phi$ all pass through $S$ and are perpendicular to the plane $A B C$, and consequently all share the line $l$ through $S$ perpendicular to $A B C$.  Since $l=\epsilon \cap \iota$, the point $O$ will be situated on $l$, and hence $\phi$ will also contain $O$. This completes our proof of the lemma. Let $A H=A E=x, B E=B F=y, C F=C G=z$, and $D G=D H=w$. If the sphere is also tangent to $A C$ at some point $I$, then $A I=x$ and $I C=z$. Using the stated lemma it suffices to prove that if $A C=x+z$, then $B D=y+w$. Let $E F=F G=G H=H I=t, \angle B A D=\alpha, \angle A B C=\beta, \angle B C D=\gamma$, and $\angle A D C=\delta$. We get  $$ t^{2}=E H^{2}=A E^{2}+A H^{2}-2 \cdot A E \cdot A H \cos \alpha=2 x^{2}(1-\cos \alpha) $$  We similarly conclude that $t^{2}=2 y^{2}(1-\cos \beta)=2 z^{2}(1-\cos \gamma)=2 w^{2}(1-$ $\cos \delta$ ). Further, using that $A B=x+y, B C=y+z, \cos \beta=1-t^{2} / 2 y^{2}$, we obtain  $$ A C^{2}=A B^{2}+B C^{2}-2 A B \cdot B C \cos \beta=(x-z)^{2}+t^{2}\left(\frac{x}{y}+1\right)\left(\frac{z}{y}+1\right) $$  Analogously, from the triangle $A D C$ we get $A C^{2}=(x-z)^{2}+t^{2}(x / w+$ $1)(z / w+1)$, which gives $(x / y+1)(z / y+1)=(x / w+1)(z / w+1)$. Since $f(s)=(x / s+1)(z / s+1)$ is a decreasing function in $s$, it follows that $y=w$; similarly $x=z$. Hence $C F=C G=x$ and $D G=D H=y$. Hence $A C \| E F$ and $A C: t=$ $A C: E F=A B: E B=(x+y): y$; i.e., $A C=t(x+y) / y$. Similarly, from the triangle $A B D$, we get that $B D=t(x+y) / x$. Hence if $A C=x+z=2 x$, it follows that $2 x=t(x+y) / y \Rightarrow 2 x y=t(x+y) \Rightarrow B D=t(x+y) / x=$ $2 y=y+w$. This completes the proof. Second solution. Without loss of generality, assume that $E F=2$. Consider the Cartesian system in which points $O, E, F, G, H$ respectively have coordinates $(0,0,0),(-1,-1, a),(1,-1, a),(1,1, a),(-1,1, a)$. Line $A H$ is perpendicular to $O H$ and $A E$ is perpendicular to $O E$; hence from Pythagoras's theorem $A O^{2}=A H^{2}+H O^{2}=A E^{2}+E O^{2}=A E^{2}+H O^{2}$, which implies $A H=A E$. Therefore the $y$-coordinate of $A$ is zero; analogously the $x$-coordinates of $B$ and $D$ and the $y$-coordinate of $C$ are 0 . Let $A$ have coordinates $\left(x_{0}, 0, z_{1}\right)$ : then $\overrightarrow{E A}\left(x_{0}+1,1, z_{1}-a\right) \perp \overrightarrow{E O}(1,1,-a)$, i.e., $\overrightarrow{E A} \cdot \overrightarrow{E O}=x_{0}+2+a\left(a-z_{1}\right)=0$. Similarly, for $B\left(0, y_{0}, z_{2}\right)$ we have $y_{0}+2+a\left(a-z_{2}\right)=0$. This gives us  $$ z_{1}=\frac{x_{0}+a^{2}+2}{a}, \quad z_{2}=\frac{y_{0}+a^{2}+2}{a} $$  We haven't used yet that $A\left(x_{0}, 0, z_{1}\right), E(-1,-1, a)$ and $B\left(0, y_{0}, z_{2}\right)$ are collinear, so let $A^{\prime}, B^{\prime}, E^{\prime}$ be the feet of perpendiculars from $A, B, E$ to the plane $x y$. The line $A^{\prime} B^{\prime}$, given by $y_{0} x+x_{0} y=x_{0} y_{0}, z=0$, contains the point $E^{\prime}(-1,-1,0)$, from which we obtain  $$ \left(x_{0}+1\right)\left(y_{0}+1\right)=1 $$  In the same way, from the points $B$ and $C$ we get relations similar to (1) and (2) and conclude that $C$ has the coordinates $C\left(-x_{0}, 0, z_{1}\right)$. Similarly we get $D\left(0,-y_{0}, z_{2}\right)$. The condition that $A C$ is tangent to the sphere $\sigma(O, O E)$ is equivalent to $z_{1}=\sqrt{a^{2}+2}$, i.e., to $x_{0}=a \sqrt{a^{2}+2}-\left(a^{2}+2\right)$. But then (2) implies that $y_{0}=-a \sqrt{a^{2}+2}-\left(a^{2}+2\right)$ and $z_{2}=-\sqrt{a^{2}+2}$, which means that the sphere $\sigma$ is tangent to $B D$ as well. This finishes the proof.


3. (CAN) Find the minimum value of  $$ \max (a+b+c, b+c+d, c+d+e, d+e+f, e+f+g) $$  subject to the constraints (i) $a, b, c, d, e, f, g \geq 0$, (ii) $a+b+c+d+e+f+g=1$.

Solution: 3. Denote $\max (a+b+c, b+c+d, c+d+e, d+e+f, e+f+g)$ by $p$. We have  $$ (a+b+c)+(c+d+e)+(e+f+g)=1+c+e \leq 3 p $$  which implies that $p \geq 1 / 3$. However, $p=1 / 3$ is achieved by taking $(a, b, c, d, e, f, g)=(1 / 3,0,0,1 / 3,0,0,1 / 3)$. Therefore the answer is $1 / 3$. Remark. In fact, one can prove a more general statement in the same way. Given positive integers $n, k, n \geq k$, if $a_{1}, a_{2}, \ldots, a_{n}$ are nonnegative real numbers with sum 1 , then the minimum value of $\max _{i=1, \ldots, n-k+1}\left\{a_{i}+\right.$ $\left.a_{i+1}+\cdots+a_{i+k-1}\right\}$ is $1 / r$, where $r$ is the integer with $k(r-1)<n \leq k r$.


4. (CAN) Let $\left\{f_{n}\right\}$ be the Fibonacci sequence $\{1,1,2,3,5, \ldots\}$. (a) Find all pairs $(a, b)$ of real numbers such that for each $n, a f_{n}+b f_{n+1}$ is a member of the sequence. (b) Find all pairs $(u, v)$ of positive real numbers such that for each $n$, $u f_{n}^{2}+v f_{n+1}^{2}$ is a member of the sequence.

Solution: 4. We shall use the known formula for the Fibonacci sequence  $$ f_{n}=\frac{1}{\sqrt{5}}\left(\alpha^{n}-(-1)^{n} \alpha^{-n}\right), \quad \text { where } \alpha=\frac{1+\sqrt{5}}{2} . $$  (a) Suppose that $a f_{n}+b f_{n+1}=f_{k_{n}}$ for all $n$, where $k_{n}>0$ is an integer depending on $n$. By (1), this is equivalent to $a\left(\alpha^{n}-(-1)^{n} \alpha^{-n}\right)+$ $b\left(\alpha^{n+1}+(-1)^{n} \alpha^{-n-1}\right)=\alpha^{k_{n}}-(-1)^{k_{n}} \alpha^{-k_{n}}$, i.e.,  $$ \alpha^{k_{n}-n}=a+b \alpha-\alpha^{-2 n}(-1)^{n}\left(a-b \alpha^{-1}-(-\alpha)^{n-k_{n}}\right) \rightarrow a+b \alpha $$  as $n \rightarrow \infty$. Hence, since $k_{n}$ is an integer, $k_{n}-n$ must be constant from some point on: $k_{n}=n+k$ and $\alpha^{k}=a+b \alpha$. Then it follows from (2) that $\alpha^{-k}=a-b \alpha^{-1}$, and from (1) we conclude that $a f_{n}+b f_{n+1}=$ $f_{k+n}$ holds for every $n$. Putting $n=1$ and $n=2$ in the previous relation and solving the obtained system of equations we get $a=f_{k-1}$, $b=f_{k}$. It is easy to verify that such $a$ and $b$ satisfy the conditions. (b) As in (a), suppose that $u f_{n}^{2}+v f_{n+1}^{2}=f_{l_{n}}$ for all $n$. This leads to  $$ \begin{aligned} u+v \alpha^{2}-\sqrt{5} \alpha^{l_{n}-2 n}= & 2(u-v)(-1)^{n} \alpha^{-2 n} \\ & -\left(u \alpha^{-4 n}+v \alpha^{-4 n-2}+(-1)^{l_{n}} \sqrt{5} \alpha^{-l_{n}-2 n}\right) \\ \rightarrow & 0 \end{aligned} $$  as $n \rightarrow \infty$. Thus $u+v \alpha^{2}=\sqrt{5} \alpha^{l_{n}-2 n}$, and $l_{n}-2 n=k$ is equal to a constant. Putting this into the above equation and multiplying by $\alpha^{2 n}$ we get $u-v \rightarrow 0$ as $n \rightarrow \infty$, i.e., $u=v$. Finally, substituting $n=1$ and $n=2$ in $u f_{n}^{2}+u f_{n+1}^{2}=f_{l_{n}}$ we easily get that the only possibility is $u=v=1$ and $k=1$. It is easy to verify that such $u$ and $v$ satisfy the conditions.


5. (COL) A cube is assembled with 27 white cubes. The larger cube is then painted black on the outside and disassembled. A blind man reassembles it. What is the probability that the cube is now completely black on the outside? Give an approximation of the size of your answer.

Solution: 5. There are four types of small cubes upon disassembling: (1) 8 cubes with three faces, painted black, at one corner; (2) 12 cubes with two black faces, both at one edge; (3) 6 cubes with one black face; (4) 1 completely white cube.  All cubes of type (1) must go to corners, and be placed in a correct way (one of three): for this step we have $3^{8} \cdot 8$ ! possibilities. Further, all cubes of type (2) must go in a correct way (one of two) to edges, admitting $2^{12} \cdot 12$ ! possibilities; similarly, there are $4^{6} \cdot 6$ ! ways for cubes of type (3), and 24 ways for the cube of type (4). Thus the total number of good reassemblings is $3^{8} 8!\cdot 2^{12} 12!\cdot 4^{6} 6!\cdot 24$, while the number of all possible reassemblings is $24^{27} \cdot 27!$. The desired probability is $\frac{3^{8} 8!\cdot 2^{12} 12!\cdot 4^{6} 6!\cdot 24}{24^{27} \cdot 27!}$. It is not necessary to calculate these numbers to find out that the blind man practically has no chance to reassemble the cube in a right way: in fact, the probability is of order $1.8 \cdot 10^{-37}$.


6. (CUB) Let $P(z)$ and $Q(z)$ be complex-variable polynomials, with degree not less than 1. Let  $$ P_{k}=\{z \in \mathbb{C} \mid P(z)=k\}, \quad Q_{k}=\{z \in \mathbb{C} \mid Q(z)=k\} $$  Let also $P_{0}=Q_{0}$ and $P_{1}=Q_{1}$. Prove that $P(z) \equiv Q(z)$.

Solution: 6. Assume w.l.o.g. that $n=\operatorname{deg} P \geq \operatorname{deg} Q$, and let $P_{0}=\left\{z_{1}, z_{2}, \ldots, z_{k}\right\}$, $P_{1}=\left\{z_{k+1}, z_{k+2}, \ldots z_{k+m}\right\}$. The polynomials $P$ and $Q$ match at $k+m$ points $z_{1}, z_{2}, \ldots, z_{k+m}$; hence if we prove that $k+m>n$, the result will follow. By the assumption, $P(x)=\left(x-z_{1}\right)^{\alpha_{1}} \cdots\left(x-z_{k}\right)^{\alpha_{k}}=\left(x-z_{k+1}\right)^{\alpha_{k+1}} \cdots\left(x-z_{k+m}\right)^{\alpha_{k+m}}+1$ for some positive integers $\alpha_{1}, \ldots, \alpha_{k+m}$. Let us consider $P^{\prime}(x)$. As we know, it is divisible by $\left(x-z_{i}\right)^{\alpha_{i}-1}$ for $i=1,2, \ldots, k+m$; i.e.,  $$ \prod_{i=1}^{k+m}\left(x-z_{i}\right)^{\alpha_{i}-1} \mid P^{\prime}(x) $$  Therefore $2 n-k-m=\operatorname{deg} \prod_{i=1}^{k+m}\left(x-z_{i}\right)^{\alpha_{i}-1} \leq \operatorname{deg} P^{\prime}=n-1$, i.e., $k+m \geq n+1$, as we claimed.


7. (FIN) ${ }^{\mathrm{IMO} 6}$ Assume that $f(x, y)$ is defined for all positive integers $x$ and $y$, and that the following equations are satisfied:  $$ \begin{aligned} f(0, y) & =y+1 \\ f(x+1,0) & =f(x, 1) \\ f(x+1, y+1) & =f(x, f(x+1, y)) \end{aligned} $$  Determine $f(2,2), f(3,3)$ and $f(4,4)$. Alternative version: Determine $f(4,1981)$.

Solution: 7. We immediately find that $f(1,0)=f(0,1)=2$. Then $f(1, y+1)=$ $f(0, f(1, y))=f(1, y)+1$; hence $f(1, y)=y+2$ for $y \geq 0$. Next we find that $f(2,0)=f(1,1)=3$ and $f(2, y+1)=f(1, f(2, y))=f(2, y)+2$, from which $f(2, y)=2 y+3$. Particularly, $f(2,2)=7$. Further, $f(3,0)=$ $f(2,1)=5$ and $f(3, y+1)=f(2, f(3, y))=2 f(3, y)+3$. This gives by induction $f(3, y)=2^{y+3}-3$. For $y=3, f(3,3)=61$. Finally, from $f(4,0)=f(3,1)=13$ and $f(4, y+1)=f(3, f(4, y))=2^{f(4, y)+3}-3$, we conclude that  $$ f(4, y)=2^{2 y^{2}}-3 \quad(y+3 \text { twos }) $$


8. (FRG) ${ }^{\mathrm{IMO} 2}$ Let $f(n, r)$ be the arithmetic mean of the minima of all $r$ subsets of the set $\{1,2, \ldots, n\}$. Prove that $f(n, r)=\frac{n+1}{r+1}$.

Solution: 8. Since the number $k, k=1,2, \ldots, n-r+1$, is the minimum in exactly $\binom{n-k}{r-1} r$-element subsets of $\{1,2, \ldots, n\}$, it follows that  $$ f(n, r)=\frac{1}{\binom{n}{r}} \sum_{k=1}^{n-r+1} k\binom{n-k}{r-1} $$  To calculate the sum in the above expression, using the equality $\binom{r+j}{j}=$ $\sum_{i=0}^{j}\binom{r+i-1}{r-1}$, we note that  $$ \begin{aligned} \sum_{k=1}^{n-r+1} k\binom{n-k}{r-1} & =\sum_{j=0}^{n-r}\left(\sum_{i=0}^{j}\binom{r+i-1}{r-1}\right) \\ & =\sum_{j=0}^{n-r}\binom{r+j}{r}=\binom{n+1}{r+1}=\frac{n+1}{r+1}\binom{n}{r} . \end{aligned} $$  Therefore $f(n, r)=(n+1) /(r+1)$.


9. (FRG) A sequence $\left(a_{n}\right)$ is defined by means of the recursion  $$ a_{1}=1, \quad a_{n+1}=\frac{1+4 a_{n}+\sqrt{1+24 a_{n}}}{16} $$  Find an explicit formula for $a_{n}$.

Solution: 9. If we put $1+24 a_{n}=b_{n}^{2}$, the given recurrent relation becomes  $$ \frac{2}{3} b_{n+1}^{2}=\frac{3}{2}+\frac{b_{n}^{2}}{6}+b_{n}=\frac{2}{3}\left(\frac{3}{2}+\frac{b_{n}}{2}\right)^{2}, \quad \text { i.e., } \quad b_{n+1}=\frac{3+b_{n}}{2} $$  where $b_{1}=5$. To solve this recurrent equation, we set $c_{n}=2^{n-1} b_{n}$. From (1) we obtain  $$ \begin{aligned} c_{n+1} & =c_{n}+3 \cdot 2^{n-1}=\cdots=c_{1}+3\left(1+2+2^{2}+\cdots+2^{n-1}\right) \\ & =5+3\left(2^{n}-1\right)=3 \cdot 2^{n}+2 \end{aligned} $$  Therefore $b_{n}=3+2^{-n+2}$ and consequently  $$ a_{n}=\frac{b_{n}^{2}-1}{24}=\frac{1}{3}\left(1+\frac{3}{2^{n}}+\frac{1}{2^{2 n-1}}\right)=\frac{1}{3}\left(1+\frac{1}{2^{n-1}}\right)\left(1+\frac{1}{2^{n}}\right) . $$


1. A1 (GBR 3) ${ }^{\mathrm{IMO}}$ The function $f(n)$ is defined for all positive integers $n$ and takes on nonnegative integer values. Also, for all $m, n$,  $$ \begin{gathered} f(m+n)-f(m)-f(n)=0 \text { or } 1 \\ f(2)=0, \quad f(3)>0, \quad \text { and } \quad f(9999)=3333 \end{gathered} $$  Determine $f(1982)$.

Solution: 1. From $f(1)+f(1) \leq f(2)=0$ we obtain $f(1)=0$. Since $0<f(3) \leq$ $f(1)+f(2)+1$, it follows that $f(3)=1$. Note that if $f(3 n) \geq n$, then $f(3 n+3) \geq f(3 n)+f(3) \geq n+1$. Hence by induction $f(3 n) \geq n$ holds for all $n \in \mathbb{N}$. Moreover, if the inequality is strict for some $n$, then it is so for all integers greater than $n$ as well. Since $f(9999)=3333$, we deduce that $f(3 n)=n$ for all $n \leq 3333$. By the given condition, we have $3 f(n) \leq f(3 n) \leq 3 f(n)+2$. Therefore $f(n)=[f(3 n) / 3]=[n / 3]$ for $n \leq 3333$. In particular, $f(1982)=$ $[1982 / 3]=660$.


10. B4 (BRA 1) A box contains $p$ white balls and $q$ black balls. Beside the box there is a pile of black balls. Two balls are taken out of the box. If they have the same color, a black ball from the pile is put into the box. If they have different colors, the white ball is put back into the box. This procedure is repeated until the last two balls are removed from the box and one last ball is put in. What is the probability that this last ball is white?

Solution: 10. If the two balls taken from the box are both white, then the number of white balls decreases by two; otherwise, it remains unchanged. Hence the parity of the number of white balls does not change during the procedure. Therefore if $p$ is even, the last ball cannot be white; the probability is 0 . If $p$ is odd, the last ball has to be white; the probability is 1 .


11. B5 (CAN 3) (a) Find the rearrangement $\left\{a_{1}, \ldots, a_{n}\right\}$ of $\{1,2, \ldots, n\}$ that maximizes  $$ a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n} a_{1}=Q $$  (b) Find the rearrangement that minimizes $Q$.

Solution: 11. (a) Suppose $\left\{a_{1}, a_{2}, \ldots, a_{n}\right\}$ is the arrangement that yields the maximal value $Q_{\max }$ of $Q$. Note that the value of $Q$ for the rearrangement $\left\{a_{1}, \ldots, a_{i-1}, a_{j}, a_{j-1}, \ldots, a_{i}, a_{j+1}, \ldots, a_{n}\right\}$ equals $Q_{\text {max }}-\left(a_{i}-\right.$ $\left.a_{j}\right)\left(a_{i-1}-a_{j+1}\right)$, where $1<i<j<n$. Hence $\left(a_{i}-a_{j}\right)\left(a_{i-1}-a_{j+1}\right) \geq 0$ for all $1<i<j<n$.  We may suppose w.l.o.g. that $a_{1}=1$. Let $a_{i}=2$. If $2<i<$ $n$, then $\left(a_{2}-a_{i}\right)\left(a_{1}-a_{i+1}\right)<0$, which is impossible. Therefore $i$ is either 2 or $n$; let w.l.o.g. $a_{n}=2$. Further, if $a_{j}=3$ for $2<j<n$, then $\left(a_{1}-a_{j+1}\right)\left(a_{2}-a_{j}\right)<0$, which is impossible; therefore $a_{2}=3$. Continuing this argument we obtain that $A=\{1,3,5, \ldots, 2[(n-1) / 2]+1,2[n / 2], \ldots, 4,2\}$. (b) A similar argument leads to the minimizing rearrangement $\{1, n, 2$, $n-1, \ldots,[n / 2]+1\}$.


12. B6 (FIN 3) Four distinct circles $C, C_{1}, C_{2}, C_{3}$ and a line $L$ are given in the plane such that $C$ and $L$ are disjoint and each of the circles $C_{1}, C_{2}, C_{3}$ touches the other two, as well as $C$ and $L$. Assuming the radius of $C$ to be 1 , determine the distance between its center and $L$.

Solution: 12. Let $y$ be the line perpendicular to $L$ passing through the center of $C$. It can be shown by a continuity argument that there exists a point $Y \in y$ such that an inversion $\Psi$ centered at $Y$ maps $C$ and $L$ onto two concentric circles $\widehat{C}$ and $\widehat{L}$. Let $\widehat{X}$ denote the image of an object $X$ under $\Psi$. Then the circles $\widehat{C_{i}}$ touch $\widehat{C}$ externally and $\widehat{L}$ internally, and all have the same radius. Let us now rotate the picture around the common center $Z$ of $\widehat{C}$ and $\widehat{L}$ so that $\widehat{C_{3}}$ passes through $Y$. Applying the inversion $\Psi$ again on the picture thus obtained, $\widehat{C}$ and $\widehat{L}$ go back to $C$ and $L$, but $\widehat{C_{3}}$ goes to a line $C_{3}^{\prime}$ parallel to $L$, while the images of $\widehat{C_{1}}$ and $\widehat{C_{2}}$ go to two equal circles $C_{1}^{\prime}$ and $C_{2}^{\prime}$ touching $L, C_{3}^{\prime}$, and $C$. This way we have achieved that $C_{3}$ becomes a line. Denote by $O_{1}, O_{2}, O$ respectively the centers of the circles $C_{1}^{\prime}, C_{2}^{\prime}, C$ and by $T$ the point of tangency of the circles $C_{1}^{\prime}$ and $C_{2}^{\prime}$. If $x$ is the common radius of the circles $C_{1}^{\prime}$ and $C_{2}^{\prime}$, then from $\triangle O_{1} T O$ we obtain ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-463.jpg?height=266&width=496&top_left_y=1028&top_left_x=832) that $(x-1)^{2}+x^{2}=(x+1)^{2}$, and thus $x=4$. Hence the distance of $O$ from $L$ equals $2 x-1=7$.


13. C1 (NET 1) ${ }^{\mathrm{IMO} 2} \mathrm{~A}$ scalene triangle $A_{1} A_{2} A_{3}$ is given with sides $a_{1}, a_{2}, a_{3}$ ( $a_{i}$ is the side opposite to $A_{i}$ ). For all $i=1,2,3, M_{i}$ is the midpoint of side $a_{i}, T_{i}$ is the point where the incircle touches side $a_{i}$, and the reflection of $T_{i}$ in the interior bisector of $A_{i}$ yields the point $S_{i}$. Prove that the lines $M_{1} S_{1}, M_{2} S_{2}$, and $M_{3} S_{3}$ are concurrent.

Solution: 13. Points $S_{1}, S_{2}, S_{3}$ clearly lie on the inscribed circle. Let $\widehat{X Y}$ denote the oriented arc $X Y$. The $\operatorname{arcs} \widehat{T_{2} S_{1}}$ and $\widehat{T_{1} T_{3}}$ are equal, since they are symmetric with respect to the bisector of $\angle A_{1}$. Similarly, $\widehat{T_{3} T_{2}}=\widehat{S_{2} T_{1}}$. Therefore $\widehat{T_{3} S_{1}}=$ $\widehat{T_{3} T_{2}}+\widehat{T_{2} S_{1}}=\widehat{S_{2} T_{1}}+\widehat{T_{1} T_{3}}=$ $S_{2} T_{3}$. It follows that $S_{1} S_{2}$ is parallel to $A_{1} A_{2}$, and consequently $S_{1} S_{2} \|$ $M_{1} M_{2}$. Analogously $S_{1} S_{3} \| M_{1} M_{3}$ ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-463.jpg?height=358&width=505&top_left_y=1450&top_left_x=827) and $S_{2} S_{3} \| M_{2} M_{3}$. Since the circumcircles of $\triangle M_{1} M_{2} M_{3}$ and $\triangle S_{1} S_{2} S_{3}$ are not equal, these triangles are not congruent and hence they must be homothetic. Then all the lines $M_{i} S_{i}$ pass through the center of homothety. Second solution. Set the complex plane so that the incenter of $\triangle A_{1} A_{2} A_{3}$ is the unit circle centered at the origin. Let $t_{i}, s_{i}$ respectively denote the complex numbers of modulus 1 corresponding to $T_{i}, S_{i}$. Clearly $t_{1} \overline{t_{1}}=$ $t_{2} \overline{t_{2}}=t_{3} \overline{t_{3}}=1$. Since $T_{2} T_{3}$ and $T_{1} S_{1}$ are parallel, we obtain $t_{2} t_{3}=t_{1} s_{1}$, or $s_{1}=t_{2} t_{3} \overline{t_{1}}$. Similarly $s_{2}=t_{1} t_{3} \overline{t_{2}}, s_{3}=t_{1} t_{2} \overline{t_{3}}$, from which it follows that $s_{2}-s_{3}=t_{1}\left(t_{3} \overline{t_{2}}-t_{2} \overline{t_{3}}\right)$. Since the number in parentheses is strictly imaginary, we conclude that $O T_{1} \perp S_{2} S_{3}$ and consequently $S_{2} S_{3} \| A_{2} A_{3}$. We proceed as in the first solution.


14. C2 (AUS 4) Let $A B C D$ be a convex plane quadrilateral and let $A_{1}$ denote the circumcenter of $\triangle B C D$. Define $B_{1}, C_{1}, D_{1}$ in a corresponding way. (a) Prove that either all of $A_{1}, B_{1}, C_{1}, D_{1}$ coincide in one point, or they are all distinct. Assuming the latter case, show that $A_{1}, C_{1}$ are on opposite sides of the line $B_{1} D_{1}$, and similarly, $B_{1}, D_{1}$ are on opposite sides of the line $A_{1} C_{1}$. (This establishes the convexity of the quadrilateral $A_{1} B_{1} C_{1} D_{1}$.) (b) Denote by $A_{2}$ the circumcenter of $B_{1} C_{1} D_{1}$, and define $B_{2}, C_{2}, D_{2}$ in an analogous way. Show that the quadrilateral $A_{2} B_{2} C_{2} D_{2}$ is similar to the quadrilateral $A B C D$.

Solution: 14. (a) If any two of $A_{1}, B_{1}, C_{1}, D_{1}$ coincide, say $A_{1} \equiv B_{1}$, then $A B C D$ is inscribed in a circle centered at $A_{1}$ and hence all $A_{1}, B_{1}, C_{1}, D_{1}$ coincide. Assume now the opposite, and let w.l.o.g. $\angle D A B+\angle D C B<180^{\circ}$. Then $A$ is outside the circumcircle of $\triangle B C D$, so $A_{1} A>A_{1} C$. Similarly, $C_{1} C>C_{1} A$. Hence the perpendicular bisector $l_{A C}$ of $A C$ separates points $A_{1}$ and $C_{1}$. Since $B_{1}, D_{1}$ lie on $l_{A C}$, this means that $A_{1}$ and $C_{1}$ are on opposite sides $B_{1} D_{1}$. Similarly one can show that $B_{1}$ and $D_{1}$ are on opposite sides of $A_{1} C_{1}$. (b) Since $A_{2} B_{2} \perp C_{1} D_{1}$ and $C_{1} D_{1} \perp A B$, it follows that $A_{2} B_{2} \| A B$. Similarly $A_{2} C_{2}\left\|A C, A_{2} D_{2}\right\| A D, B_{2} C_{2}\left\|B C, B_{2} D_{2}\right\| B D$, and $C_{2} D_{2} \| C D$. Hence $\triangle A_{2} B_{2} C_{2} \sim \triangle A B C$ and $\triangle A_{2} D_{2} C_{2} \sim \triangle A D C$, and the result follows.


15. C3 (CAN 5) Show that  $$ \frac{1-s^{a}}{1-s} \leq(1+s)^{a-1} $$  holds for every $1 \neq s>0$ real and $0<a \leq 1$ rational.

Solution: 15. Let $a=k / n$, where $n, k \in \mathbb{N}, n \geq k$. Putting $t^{n}=s$, the given inequality becomes $\frac{1-t^{k}}{1-t^{n}} \leq\left(1+t^{n}\right)^{k / n-1}$, or equivalently  $$ \left(1+t+\cdots+t^{k-1}\right)^{n}\left(1+t^{n}\right)^{n-k} \leq\left(1+t+\cdots+t^{n-1}\right)^{n} $$  This is clearly true for $k=n$. Therefore it is enough to prove that the lefthand side of the above inequality is an increasing function of $k$. We are led to show that $\left(1+t+\cdots+t^{k-1}\right)^{n}\left(1+t^{n}\right)^{n-k} \leq\left(1+t+\cdots+t^{k}\right)^{n}\left(1+t^{n}\right)^{n-k-1}$. This is equivalent to $1+t^{n} \leq A^{n}$, where $A=\frac{1+t+\cdots+t^{k}}{1+t+\cdots+t^{k-1}}$. But this easily follows, since  $$ \begin{aligned} A^{n}-t^{n} & =(A-t)\left(A^{n-1}+A^{n-2} t+\cdots+t^{n-1}\right) \\ & \geq(A-t)\left(1+t+\cdots+t^{n-1}\right)=\frac{1+t+\cdots+t^{n-1}}{1+t+\cdots+t^{k-1}} \geq 1 \end{aligned} $$  Remark. The original problem asked to prove the inequality for real $a$.


16. C4 (GBR 2) ${ }^{\mathrm{IMO}}$ Prove that if $n$ is a positive integer such that the equation $x^{3}-3 x y^{2}+y^{3}=n$ has a solution in integers $(x, y)$, then it has at least three such solutions. Show that the equation has no solution in integers when $n=2891$.

Solution: 16. It is easy to verify that whenever $(x, y)$ is a solution of the equation $x^{3}-3 x y^{2}+y^{3}=n$, so are the pairs $(y-x,-x)$ and $(-y, x-y)$. No two of these three solutions are equal unless $x=y=n=0$. Observe that $2981 \equiv 2(\bmod 9)$. Since $x^{3}, y^{3} \equiv 0, \pm 1(\bmod 9), x^{3}-$ $3 x y^{2}+y^{3}$ cannot give the remainder 2 when divided by 9 . Hence the above equation for $n=2981$ has no integer solutions.


17. C5 (USS 5) The right triangles $A B C$ and $A B_{1} C_{1}$ are similar and have opposite orientation. The right angles are at $C$ and $C_{1}$, and we also have $\measuredangle C A B=\measuredangle C_{1} A B_{1}$. Let $M$ be the point of intersection of the lines $B C_{1}$ and $B_{1} C$. Prove that if the lines $A M$ and $C C_{1}$ exist, they are perpendicular.

Solution: 17. Let $A$ be the origin of the Cartesian plane. Suppose that $B C: A C=k$ and that $(a, b)$ and $\left(a_{1}, b_{1}\right)$ are coordinates of the points $C$ and $C_{1}$, respectively. Then the coordinates of the point $B$ are $(a, b)+k(-b, a)=(a-k b, b+k a)$, while the coordinates of $B_{1}$ are $\left(a_{1}, b_{1}\right)+k\left(b_{1},-a_{1}\right)=\left(a+k b_{1}, b_{1}-k a_{1}\right)$. Thus the lines $B C_{1}$ and $C B_{1}$ are given by the equations $\frac{x-a_{1}}{y-b_{1}}=\frac{x-(a-k b)}{y-(b+k a)}$ and $\frac{x-a}{y-b}=\frac{x-\left(a_{1}+k b_{1}\right)}{y-\left(b_{1}-k a_{1}\right)}$ respectively. After multiplying, these equations transform into the forms  $$ \begin{aligned} B C_{1}: & k a x+k b y & =k a a_{1}+k b b_{1}+b a_{1}-a b_{1}-\left(b-b_{1}\right) x+\left(a-a_{1}\right) y \\ C B_{1}: & k a_{1} x+k b_{1} y & =k a a_{1}+k b b_{1}+b a_{1}-a b_{1}-\left(b-b_{1}\right) x+\left(a-a_{1}\right) y . \end{aligned} $$  The coordinates $\left(x_{0}, y_{0}\right)$ of the point $M$ satisfy these equations, from which we deduce that $k a x_{0}+k b y_{0}=k a_{1} x_{0}+k b_{1} y_{0}$. This yields $\frac{x_{0}}{y_{0}}=-\frac{b_{1}-b}{a_{1}-a}$, implying that the lines $C C_{1}$ and $A M$ are perpendicular.


18. C6 (FRA 2) Let $O$ be a point of three-dimensional space and let $l_{1}, l_{2}, l_{3}$ be mutually perpendicular straight lines passing through $O$. Let $S$ denote the sphere with center $O$ and radius $R$, and for every point $M$ of $S$, let $S_{M}$ denote the sphere with center $M$ and radius $R$. We denote by $P_{1}, P_{2}, P_{3}$ the intersection of $S_{M}$ with the straight lines $l_{1}, l_{2}, l_{3}$, respectively, where we put $P_{i} \neq O$ if $l_{i}$ meets $S_{M}$ at two distinct points and $P_{i}=O$ otherwise ( $i=$ $1,2,3)$. What is the set of centers of gravity of the (possibly degenerate) triangles $P_{1} P_{2} P_{3}$ as $M$ runs through the points of $S$ ?

Solution: 18. Set the coordinate system with the axes $x, y, z$ along the lines $l_{1}, l_{2}, l_{3}$ respectively. The coordinates $(a, b, c)$ of $M$ satisfy $a^{2}+b^{2}+c^{2}=R^{2}$, and so $S_{M}$ is given by the equation $(x-a)^{2}+(y-b)^{2}+(z-c)^{2}=R^{2}$. Hence the coordinates of $P_{1}$ are $(x, 0,0)$ with $(x-a)^{2}+b^{2}+c^{2}=R^{2}$, implying that either $x=2 a$ or $x=0$. Thus by the definition we obtain $x=2 a$. Similarly, the coordinates of $P_{2}$ and $P_{3}$ are $(0,2 b, 0)$ and $(0,0,2 c)$ respectively. Now, the centroid of $\triangle P_{1} P_{2} P_{3}$ has the coordinates $(2 a / 3,2 b / 3,2 c / 3)$. Therefore the required locus of points is the sphere with center $O$ and radius $2 R / 3$.


19. C7 (CZS 3) Let $M$ be the set of real numbers of the form $\frac{m+n}{\sqrt{m^{2}+n^{2}}}$, where $m$ and $n$ are positive integers. Prove that for every pair $x \in M$, $y \in M$ with $x<y$, there exists an element $z \in M$ such that $x<z<y$.

Solution: 19. Let us set $x=m / n$. Since $f(x)=(m+n) / \sqrt{m^{2}+n^{2}}=(x+1) / \sqrt{1+x^{2}}$ is a continuous function of $x, f(x)$ takes all values between any two values of $f$; moreover, the corresponding $x$ can be rational. This completes the proof.  Remark. Since $f$ is increasing for $x \geq 1,1 \leq x<z<y$ implies $f(x)<$ $f(z)<f(y)$.


2. A2 (YUG 1) Let $K$ be a convex polygon in the plane and suppose that $K$ is positioned in the coordinate system in such a way that  $$ \operatorname{area}\left(K \cap Q_{i}\right)=\frac{1}{4} \text { area } K(i=1,2,3,4,), $$  where the $Q_{i}$ denote the quadrants of the plane. Prove that if $K$ contains no nonzero lattice point, then the area of $K$ is less than 4.

Solution: 2. Since $K$ does not contain a lattice point other than $O(0,0)$, it is bounded by four lines $u, v, w, x$ that pass through the points $U(1,0), V(0,1)$, $W(-1,0), X(0,-1)$ respectively. Let $P Q R S$ be the quadrilateral formed by these lines, where $U \in S P, V \in P Q, W \in Q R, X \in R S$. If one of the quadrants, say $Q_{1}$, contains no vertices of $P Q R S$, then $K \cap Q_{1}$ is contained in $\triangle O U V$ and hence has area less than $1 / 2$. Consequently the area of $K$ is less than 2 . Let us now suppose that $P, Q, R, S$ lie in different quadrants. One of the angles of $P Q R S$ is at least $90^{\circ}$ : let it be $\angle P$. Then $S_{U P V} \leq P U \cdot P V / 2 \leq$ $\left(P U^{2}+P V^{2}\right) / 4 \leq U V^{2} / 4=1 / 2$, which implies that $S_{K \cap Q_{1}}<S_{O U P V} \leq$ 1. Hence the area of $K$ is less than 4.


20. C8 (TUN 3) Let $A B C D$ be a convex quadrilateral and draw regular triangles $A B M, C D P, B C N, A D Q$, the first two outward and the other two inward. Prove that $M N=A C$. What can be said about the quadrilateral $M N P Q$ ?

Solution: 20. Since $M N$ is the image of $A C$ under rotation about $B$ for $60^{\circ}$, we have $M N=A C$. Similarly, $P Q$ is the image of $A C$ under rotation about $D$ through $60^{\circ}$, from which it follows that $P Q \| M N$. Hence either $M, N, P, Q$ are collinear or $M N P Q$ is a parallelogram.


3. A3 (USS 4) ${ }^{\mathrm{IMO} 3}$ Consider the infinite sequences $\left\{x_{n}\right\}$ of positive real numbers with the following properties:  $$ x_{0}=1 \quad \text { and for all } i \geq 0, x_{i+1} \leq x_{i} $$  (a) Prove that for every such sequence there is an $n \geq 1$ such that $\frac{x_{0}^{2}}{x_{1}}+$ $\frac{x_{1}^{2}}{x_{2}}+\cdots+\frac{x_{n-1}^{2}}{x_{n}} \geq 3.999$. (b) Find such a sequence for which $\frac{x_{0}^{2}}{x_{1}}+\frac{x_{1}^{2}}{x_{2}}+\cdots+\frac{x_{n-1}^{2}}{x_{n}}<4$ for all $n$.

Solution: 3. (a) By the Cauchy-Schwarz inequality we have $\left(x_{0}^{2} / x_{1}+\cdots+x_{n-1}^{2} / x_{n}\right)$. $\left(x_{1}+\cdots+x_{n}\right) \geq\left(x_{0}+\cdots+x_{n-1}\right)^{2}$. Let us set $X_{n-1}=x_{1}+x_{2}+$ $\cdots+x_{n-1}$. Using $x_{0}=1$, the last inequality can be rewritten as  $$ \frac{x_{0}^{2}}{x_{1}}+\cdots+\frac{x_{n-1}^{2}}{x_{n}} \geq \frac{\left(1+X_{n-1}\right)^{2}}{X_{n-1}+x_{n}} \geq \frac{4 X_{n-1}}{X_{n-1}+x_{n}}=\frac{4}{1+x_{n} / X_{n-1}} $$  Since $x_{n} \leq x_{n-1} \leq \cdots \leq x_{1}$, it follows that $X_{n-1} \geq(n-1) x_{n}$. Now (1) yields $x_{0}^{2} / x_{1}+\cdots+x_{n-1}^{2} / x_{n} \geq 4(n-1) / n$, which exceeds 3.999 for $n>4000$. (b) The sequence $x_{n}=1 / 2^{n}$ obviously satisfies the required condition.  Second solution to part (a). For each $n \in \mathbb{N}$, let us find a constant $c_{n}$ such that the inequality $x_{0}^{2} / x_{1}+\cdots+x_{n-1}^{2} / x_{n} \geq c_{n} x_{0}$ holds for any sequence $x_{0} \geq x_{1} \geq \cdots \geq x_{n}>0$. For $n=1$ we can take $c_{1}=1$. Assuming that $c_{n}$ exists, we have  $$ \frac{x_{0}^{2}}{x_{1}}+\left(\frac{x_{1}^{2}}{x_{2}}+\cdots+\frac{x_{n}^{2}}{x_{n+1}}\right) \geq \frac{x_{0}^{2}}{x_{1}}+c_{n} x_{1} \geq 2 \sqrt{x_{0}^{2} c_{n}}=x_{0} \cdot 2 \sqrt{c_{n}} $$  Thus we can take $c_{n+1}=2 \sqrt{c_{n}}$. Then inductively $c_{n}=2^{2-1 / 2^{n-2}}$, and since $c_{n} \rightarrow 4$ as $n \rightarrow \infty$, the result follows. Third solution. Since $\left\{x_{n}\right\}$ is decreasing, there exists $\lim _{n \rightarrow \infty} x_{n}=x \geq 0$. If $x>0$, then $x_{n-1}^{2} / x_{n} \geq x_{n} \geq x$ holds for each $n$, and the result is trivial.  If otherwise $x=0$, then we note that $x_{n-1}^{2} / x_{n} \geq 4\left(x_{n-1}-x_{n}\right)$ for each $n$, with equality if and only if $x_{n-1}=2 x_{n}$. Hence  $$ \lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{x_{k-1}^{2}}{x_{k}} \geq \lim _{n \rightarrow \infty} \sum_{k=1}^{n} 4\left(x_{k-1}-x_{k}\right)=4 x_{0}=4 $$  Equality holds if and only if $x_{n-1}=2 x_{n}$ for all $n$, and consequently $x_{n}=1 / 2^{n}$.


4. A4 (BUL 2) Determine all real values of the parameter $a$ for which the equation  $$ 16 x^{4}-a x^{3}+(2 a+17) x^{2}-a x+16=0 $$  has exactly four distinct real roots that form a geometric progression.

Solution: 4. Suppose that $a$ satisfies the requirements of the problem and that $x, q x$, $q^{2} x, q^{3} x$ are the roots of the given equation. Then $x \neq 0$ and we may assume that $|q|>1$, so that $|x|<|q x|<\left|q^{2} x\right|<\left|q^{3} x\right|$. Since the equation is symmetric, $1 / x$ is also a root and therefore $1 / x=q^{3} x$, i.e., $q=x^{-2 / 3}$. It follows that the roots are $x, x^{1 / 3}, x^{-1 / 3}, x^{-1}$. Now by Vieta's formula we have $x+x^{1 / 3}+x^{-1 / 3}+x^{-1}=a / 16$ and $x^{4 / 3}+x^{2 / 3}+2+x^{-2 / 3}+x^{-4 / 3}=$ $(2 a+17) / 16$. On setting $z=x^{1 / 3}+x^{-1 / 3}$ these equations become  $$ \begin{aligned} z^{3}-2 z & =a / 16 \\ \left(z^{2}-2\right)^{2}+z^{2}-2 & =(2 a+17) / 16 \end{aligned} $$  Substituting $a=16\left(z^{3}-2 z\right)$ in the second equation leads to $z^{4}-2 z^{3}-$ $3 z^{2}+4 z+15 / 16=0$. We observe that this polynomial factors as $(z+$ $3 / 2)(z-5 / 2)\left(z^{2}-z-1 / 4\right)$. Since $|z|=\left|x^{1 / 3}+x^{-1 / 3}\right| \geq 2$, the only viable value is $z=5 / 2$. Consequently $a=170$ and the roots are $1 / 8,1 / 2,2,8$.


5. A5 (NET 2) ${ }^{\mathrm{IMO}}$ Let $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6}$ be a regular hexagon. Each of its diagonals $A_{i-1} A_{i+1}$ is divided into the same ratio $\frac{\lambda}{1-\lambda}$, where $0<\lambda<1$, by a point $B_{i}$ in such a way that $A_{i}, B_{i}$, and $B_{i+2}$ are collinear ( $i \equiv$ $1, \ldots, 6(\bmod 6))$. Compute $\lambda$.

Solution: 5. We first observe that $\triangle A_{5} B_{4} A_{4} \cong$ $\triangle A_{3} B_{2} A_{2}$. Since $\angle A_{5} A_{3} A_{2}=90^{\circ}$, we have $\angle A_{2} B_{4} A_{4}=\angle A_{2} B_{4} A_{3}+$ $\angle A_{3} B_{4} A_{4}=\left(90^{\circ}-\angle B_{2} A_{2} A_{3}\right)+$ $\left(\angle B_{4} A_{5} A_{4}+\angle A_{5} A_{4} B_{4}\right)=90^{\circ}+$ $\angle B_{4} A_{5} A_{4}=120^{\circ}$. Hence $B_{4}$ belongs to the circle with center $A_{3}$ and radius $A_{3} A_{4}$, so $A_{3} A_{4}=A_{3} B_{4}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-461.jpg?height=311&width=428&top_left_y=1245&top_left_x=866)  Thus $\lambda=A_{3} B_{4} / A_{3} A_{5}=A_{3} A_{4} / A_{3} A_{5}=1 / \sqrt{3}$.


6. A6 (VIE 1) ${ }^{\text {IMO6 }}$ Let $S$ be a square with sides of length 100 and let $L$ be a path within $S$ that does not meet itself and that is composed of linear segments $A_{0} A_{1}, A_{1} A_{2}, \ldots, A_{n-1} A_{n}$ with $A_{0} \neq A_{n}$. Suppose that for every point $P$ of the boundary of $S$ there is a point of $L$ at a distance from $P$ not greater than $\frac{1}{2}$. Prove that there are two points $X$ and $Y$ in $L$ such that the distance between $X$ and $Y$ is not greater than 1 and the length of that part of $L$ that lies between $X$ and $Y$ is not smaller than 198.

Solution: 6. Denote by $d(U, V)$ the distance between points or sets of points $U$ and $V$. For $P, Q \in L$ we shall denote by $L_{P Q}$ the part of $L$ between points $P$ and $Q$ and by $l_{P Q}$ the length of this part. Let us denote by $S_{i}(i=1,2,3,4)$ the vertices of $S$ and by $T_{i}$ points of $L$ such that $S_{i} T_{i} \leq 1 / 2$ in such a way that $l_{A_{0} T_{1}}$ is the least of the $l_{A_{0} T_{i}}$ 's, $S_{2}$ and $S_{4}$ are neighbors of $S_{1}$, and $l_{A_{0} T_{2}}<l_{A_{0} T_{4}}$. Now we shall consider the points of the segment $S_{1} S_{4}$. Let $D$ and $E$ be the sets of points defined as follows: $D=\left\{X \in\left[S_{1} S_{4}\right] \mid d\left(X, L_{A_{0} T_{2}}\right) \leq 1 / 2\right\}$ and $E=\left\{X \in\left[S_{1} S_{4}\right] \mid d\left(X, L_{T_{2} A_{n}}\right) \leq 1 / 2\right\}$. Clearly $D$ and $E$ are closed, nonempty (indeed, $S_{1} \in D$ and $S_{4} \in E$ ) subsets of [ $\left.S_{1} S_{4}\right]$. Since their union is a connected set $S_{1} S_{4}$, it follows that they must have a nonempty intersection. Let $P \in D \cap E$. Then there exist points $X \in L_{A_{0} T_{2}}$ and $Y \in L_{T_{2} A_{n}}$ such that $d(P, X) \leq 1 / 2, d(P, Y) \leq 1 / 2$, and consequently $d(X, Y) \leq 1$. On the other hand, $T_{2}$ lies between $X$ and $Y$ on $L$, and thus $L_{X Y}=L_{X T_{2}}+L_{T_{2} Y} \geq X T_{2}+T_{2} Y \geq\left(P S_{2}-X P-S_{2} T_{2}\right)+\left(P S_{2}-Y P-\right.$ $\left.S_{2} T_{2}\right) \geq 99+99=198$.


7. B1 (CAN 2) Let $p(x)$ be a cubic polynomial with integer coefficients with leading coefficient 1 and with one of its roots equal to the product of the other two. Show that $2 p(-1)$ is a multiple of $p(1)+p(-1)-2(1+p(0))$.

Solution: 7. Let $a, b, a b$ be the roots of the cubic polynomial $P(x)=(x-a)(x-b)(x-$ $a b)$. Observe that  $$ \begin{aligned} 2 p(-1) & =-2(1+a)(1+b)(1+a b) \\ p(1)+p(-1)-2(1+p(0)) & =-2(1+a)(1+b) \end{aligned} $$  The statement of the problem is trivial if both the expressions are equal to zero. Otherwise, the quotient $\frac{2 p(-1)}{p(1)+p(-1)-2(1+p(0))}=1+a b$ is rational and consequently $a b$ is rational. But since $(a b)^{2}=-P(0)$ is an integer, it follows that $a b$ is also an integer. This completes the proof.


8. B2 (POL 4) A convex, closed figure lies inside a given circle. The figure is seen from every point of the circumference at a right angle (that is, the two rays drawn from the point and supporting the convex figure are perpendicular). Prove that the center of the circle is a center of symmetry of the figure.

Solution: 8. Let $\mathcal{F}$ be the given figure. Consider any chord $A B$ of the circumcircle $\gamma$ that supports $\mathcal{F}$. The other supporting lines to $\mathcal{F}$ from $A$ and $B$ intersect $\gamma$ again at $D$ and $C$ respectively so that $\angle D A B=\angle A B C=90^{\circ}$. Then $A B C D$ is a rectangle, and hence $C D$ must support $\mathcal{F}$ as well, from which it follows that $\mathcal{F}$ is inscribed in the rectangle $A B C D$ touching each of its sides. We easily conclude that $\mathcal{F}$ is the intersection of all such rectangles. Now, since the center $O$ of $\gamma$ is the center of symmetry of all these rectangles, it must be so for their intersection $\mathcal{F}$ as well.


9. B3 (GBR 1) Let $A B C$ be a triangle, and let $P$ be a point inside it such that $\measuredangle P A C=\measuredangle P B C$. The perpendiculars from $P$ to $B C$ and $C A$ meet these lines at $L$ and $M$, respectively, and $D$ is the midpoint of $A B$. Prove that $D L=D M$.

Solution: 9. Let $X$ and $Y$ be the midpoints of the segments $A P$ and $B P$. Then $D Y P X$ is a parallelogram. Since $X$ and $Y$ are the circumcenters of $\triangle A P M$ and $\triangle B P L$, it follows that $X M=$ $X P=D Y$ and $Y L=Y P=D X$. Furthermore, $\angle D X M=\angle D X P+$ $\angle P X M=\angle D X P+2 \angle P A M=$ $\angle D Y P+2 \angle P B L=\angle D Y P+$ $\angle P Y L=\angle D Y L$. Therefore, the triangles $D X M$ and $L Y D$ are con- ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-462.jpg?height=379&width=451&top_left_y=1247&top_left_x=863) gruent, implying $D M=D L$.


1. (AUS 1) The localities $P_{1}, P_{2}, \ldots, P_{1983}$ are served by ten international airlines $A_{1}, A_{2}, \ldots, A_{10}$. It is noticed that there is direct service (without stops) between any two of these localities and that all airline schedules offer round-trip flights. Prove that at least one of the airlines can offer a round trip with an odd number of landings.

Solution: 1. Suppose that there are $n$ airlines $A_{1}, \ldots, A_{n}$ and $N>2^{n}$ cities. We shall prove that there is a round trip by at least one $A_{i}$ containing an odd number of stops. For $n=1$ the statement is trivial, since one airline serves at least 3 cities and hence $P_{1} P_{2} P_{3} P_{1}$ is a round trip with 3 landings. We use induction on $n$, and assume that $n>1$. Suppose the contrary, that all round trips by $A_{n}$ consist of an even number of stops. Then we can separate the cities into two nonempty classes $Q=\left\{Q_{1}, \ldots, Q_{r}\right\}$ and $R=\left\{R_{1}, \ldots, R_{s}\right\}$ (where $r+s=N$ ), so that each flight by $A_{n}$ runs between a $Q$-city and an $R$-city. (Indeed, take any city $Q_{1}$ served by $A_{n}$; include each city linked to $Q_{1}$ by $A_{n}$ in $R$, then include in $Q$ each city linked by $A_{n}$ to any $R$-city, etc. Since all round trips are even, no contradiction can arise.) At least one of $r, s$ is larger than $2^{n-1}$, say $r>2^{n-1}$. But, only $A_{1}, \ldots, A_{n-1}$ run between cities in $\left\{Q_{1}, \ldots, Q_{r}\right\}$; hence by the induction hypothesis at least one of them flies a round trip with an odd number of landings, a contradiction. It only remains to notice that for $n=10,2^{n}=1024<1983$. Remark. If there are $N=2^{n}$ cities, there is a schedule with $n$ airlines that contain no odd round trip by any of the airlines. Let the cities be $P_{k}$, $k=0, \ldots, 2^{n}-1$, and write $k$ in the binary system as an $n$-digit number $\overline{a_{1} \ldots a_{n}}$ (e.g., $1=(0 \ldots 001)_{2}$ ). Link $P_{k}$ and $P_{l}$ by $A_{i}$ if the $i$ th digits $k$ and $l$ are distinct but the first $i-1$ digits are the same. All round trips under $A_{i}$ are even, since the $i$ th digit alternates.


10. (FIN 1) Let $p$ and $q$ be integers. Show that there exists an interval $I$ of length $1 / q$ and a polynomial $P$ with integral coefficients such that  $$ \left|P(x)-\frac{p}{q}\right|<\frac{1}{q^{2}} $$  for all $x \in I$.

Solution: 10. Choose $P(x)=\frac{p}{q}\left((q x-1)^{2 n+1}+1\right), I=[1 / 2 q, 3 / 2 q]$. Then all the coefficients of $P$ are integers, and  $$ \left|P(x)-\frac{p}{q}\right|=\left|\frac{p}{q}(q x-1)^{2 n+1}\right| \leq\left|\frac{p}{q}\right| \frac{1}{2^{2 n+1}}, $$  for $x \in I$. The desired inequality follows if $n$ is chosen large enough.


11. (FIN $\mathbf{2}^{\prime}$ ) Let $f:[0,1] \rightarrow \mathbb{R}$ be continuous and satisfy:  $$ \begin{aligned} b f(2 x) & =f(x), & & 0 \leq x \leq 1 / 2 \\ f(x) & =b+(1-b) f(2 x-1), & & 1 / 2 \leq x \leq 1 \end{aligned} $$  where $b=\frac{1+c}{2+c}, c>0$. Show that $0<f(x)-x<c$ for every $x, 0<x<1$.

Solution: 11. First suppose that the binary representation of $x$ is finite: $x=0, a_{1} a_{2} \ldots a_{n}$ $=\sum_{j=1}^{n} a_{j} 2^{-j}, a_{i} \in\{0,1\}$. We shall prove by induction on $n$ that  $$ f(x)=\sum_{j=1}^{n} b_{0} \ldots b_{j-1} a_{j}, \quad \text { where } b_{k}=\left\{\begin{array}{lr} -b & \text { if } a_{k}=0 \\ 1-b & \text { if } a_{k}=1 \end{array}\right. $$  (Here $a_{0}=0$.) Indeed, by the recursion formula, $a_{1}=0 \Rightarrow f(x)=b f\left(\sum_{j=1}^{n-1} a_{j+1} 2^{-j}\right)=b \sum_{j=1}^{n-1} b_{1} \ldots b_{j} a_{j+1}$ hence $f(x)=$ $\sum_{j=0}^{n-1} b_{0} \ldots b_{j} a_{j+1}$ as $b_{0}=b_{1}=b ;$ $a_{1}=1 \Rightarrow f(x)=b+(1-b) f\left(\sum_{j=1}^{n-1} a_{j+1} 2^{-j}\right)=\sum_{j=0}^{n-1} b_{0} \ldots b_{j} a_{j+1}$, as $b_{0}=b, b_{1}=1-b$. Clearly, $f(0)=0, f(1)=1, f(1 / 2)=b>1 / 2$. Assume $x=\sum_{j=0}^{n} a_{j} 2^{-j}$, and for $k \geq 2, v=x+2^{-n-k+1}, u=x+2^{-n-k}=(v+x) / 2$. Then $f(v)=$ $f(x)+b_{0} \ldots b_{n} b^{k-2}$ and $f(u)=f(x)+b_{0} \ldots b_{n} b^{k-1}>(f(v)+f(x)) / 2$. This means that the point $(u, f(u))$ lies above the line joining $(x, f(x))$ and $(v, f(v))$. By induction, every $(x, f(x))$, where $x$ has a finite binary expansion, lies above the line joining $(0,0)$ and $(1 / 2, b)$ if $0<x<1 / 2$, or above the line joining $(1 / 2, b)$ and $(1,1)$ if $1 / 2<x<1$. It follows immediately that $f(x)>x$. For the second inequality, observe that  $$ \begin{aligned} f(x)-x & =\sum_{j=1}^{\infty}\left(b_{0} \ldots b_{j-1}-2^{-j}\right) a_{j} \\ & <\sum_{j=1}^{\infty}\left(b^{j}-2^{-j}\right) a_{j}<\sum_{j=1}^{\infty}\left(b^{j}-2^{-j}\right)=\frac{b}{1-b}-1=c . \end{aligned} $$  By continuity, these inequalities also hold for $x$ with infinite binary representations.


12. (GBR 4) $)^{\mathrm{IMO} 1}$ Find all functions $f$ defined on the positive real numbers and taking positive real values that satisfy the following conditions: (i) $f(x f(y))=y f(x)$ for all positive real $x, y$. (ii) $f(x) \rightarrow 0$ as $x \rightarrow+\infty$.

Solution: 12. Putting $y=x$ in (1) we see that there exist positive real numbers $z$ such that $f(z)=z$ (this is true for every $z=x f(x)$ ). Let $a$ be any of them.  Then $f\left(a^{2}\right)=f(a f(a))=a f(a)=a^{2}$, and by induction, $f\left(a^{n}\right)=a^{n}$. If $a>1$, then $a^{n} \rightarrow+\infty$ as $n \rightarrow \infty$, and we have a contradiction with (2). Again, $a=f(a)=f(1 \cdot a)=a f(1)$, so $f(1)=1$. Then, $a f\left(a^{-1}\right)=$ $f\left(a^{-1} f(a)\right)=f(1)=1$, and by induction, $f\left(a^{-n}\right)=a^{-n}$. This shows that $a \nless 1$. Hence, $a=1$. It follows that $x f(x)=1$, i.e., $f(x)=1 / x$ for all $x$. This function satisfies (1) and (2), so $f(x)=1 / x$ is the unique solution.


13. (LUX 2) Let $E$ be the set of $1983^{3}$ points of the space $\mathbb{R}^{3}$ all three of whose coordinates are integers between 0 and 1982 (including 0 and 1982). A coloring of $E$ is a map from $E$ to the set $\{$ red, blue $\}$. How many colorings of $E$ are there satisfying the following property: The number of red vertices among the 8 vertices of any right-angled parallelepiped is a multiple of 4 ?

Solution: 13. Given any coloring of the $3 \times 1983-2$ points of the axes, we prove that there is a unique coloring of $E$ having the given property and extending this coloring. The first thing to notice is that given any rectangle $R_{1}$ parallel to a coordinate plane and whose edges are parallel to the axes, there is an even number $r_{1}$ of red vertices on $R_{1}$. Indeed, let $R_{2}$ and $R_{3}$ be two other rectangles that are translated from $R_{1}$ orthogonally to $R_{1}$ and let $r_{2}, r_{3}$ be the numbers of red vertices on $R_{2}$ and $R_{3}$ respectively. Then $r_{1}+r_{2}$, $r_{1}+r_{3}$, and $r_{2}+r_{3}$ are multiples of 4 , so $r_{1}=\left(r_{1}+r_{2}+r_{1}+r_{3}-r_{2}-r_{3}\right) / 2$ is even. Since any point of a coordinate plane is a vertex of a rectangle whose remaining three vertices lie on the corresponding axes, this determines uniquely the coloring of the coordinate planes. Similarly, the coloring of the inner points of the parallelepiped is completely determined. The solution is hence $2^{3 \times 1983-2}=2^{5947}$.


14. (POL 2) ${ }^{\mathrm{IMO} 5}$ Prove or disprove: From the interval $[1, \ldots, 30000]$ one can select a set of 1000 integers containing no arithmetic triple (three consecutive numbers of an arithmetic progression).

Solution: 14. Let $T_{n}$ be the set of all nonnegative integers whose ternary representations consist of at most $n$ digits and do not contain a digit 2 . The cardinality of $T_{n}$ is $2^{n}$, and the greatest integer in $T_{n}$ is $11 \ldots 1=3^{0}+3^{1}+\cdots+3^{n-1}=$ $\left(3^{n}-1\right) / 2$. We claim that there is no arithmetic triple in $T_{n}$. To see this, suppose $x, y, z \in T_{n}$ and $2 y=x+z$. Then $2 y$ has only 0 's and 2's in its ternary representation, and a number of this form can be the sum of two integers $x, z \in T_{n}$ in only one way, namely $x=z=y$. But $\left|T_{10}\right|=2^{10}=1024$ and $\max T_{10}=\left(3^{10}-1\right) / 2=29524<30000$. Thus the answer is yes.


15. (POL 3) Decide whether there exists a set $M$ of natural numbers satisfying the following conditions: (i) For any natural number $m>1$ there are $a, b \in M$ such that $a+b=m$. (ii) If $a, b, c, d \in M, a, b, c, d>10$ and $a+b=c+d$, then $a=c$ or $a=d$.

Solution: 15. There is no such set. Suppose $M$ satisfies $(a)$ and $(b)$ and let $q_{n}=$ $|\{a \in M: a \leq n\}|$. Consider the differences $b-a$, where $a, b \in M$ and $10<a<b \leq k$. They are all positive and less than $k$, and (b) implies that they are $\left(\frac{q_{k}}{-} \frac{q_{10}}{2}\right)$ different integers. Hence $\binom{q_{k}-q_{10}}{2}<k$, so $q_{k} \leq \sqrt{2 k}+10$. It follows from (a) that among the numbers of the form $a+b$, where $a, b \in M, a \leq b \leq n$, or $a \leq n<b \leq 2 n$, there are all integers from the interval $[2,2 n+1]$. Thus $\binom{q_{n}+1}{2}+q_{n}\left(q_{2 n}-q_{n}\right) \geq 2 n$ for every $n \in \mathbb{N}$. Set $Q_{k}=\sqrt{2 k}+10$. We have  $$ \begin{aligned} \binom{q_{n}+1}{2}+q_{n}\left(q_{2 n}-q_{n}\right) & =\frac{1}{2} q_{n}+\frac{1}{2} q_{n}\left(2 q_{2 n}-q_{n}\right) \\ & \leq \frac{1}{2} q_{n}+\frac{1}{2} q_{n}\left(2 Q_{2 n}-q_{n}\right) \\ & \leq \frac{1}{2} Q_{n}+\frac{1}{2} Q_{n}\left(2 Q_{2 n}-Q_{n}\right) \\ & \leq 2(\sqrt{2}-1) n+(20+\sqrt{2} / 2) \sqrt{n}+55 \end{aligned} $$  which is less than $n$ for $n$ large enough, a contradiction.


16. (GDR 1) Let $F(n)$ be the set of polynomials $P(x)=a_{0}+a_{1} x+\cdots+a_{n} x^{n}$, with $a_{0}, a_{1}, \ldots, a_{n} \in \mathbb{R}$ and $0 \leq a_{0}=a_{n} \leq a_{1}=a_{n-1} \leq \cdots \leq a_{[n / 2]}=$ $a_{[(n+1) / 2]}$. Prove that if $f \in F(m)$ and $g \in F(n)$, then $f g \in F(m+n)$.

Solution: 16. Set $h_{n, i}(x)=x^{i}+\cdots+x^{n-i}, 2 i \leq n$. The set $F(n)$ is the set of linear combinations with nonnegative coefficients of the $h_{n, i}$ 's. This is a convex cone. Hence, it suffices to prove that $h_{n, i} h_{m, j} \in F(m+n)$. Indeed, setting $p=n-2 i$ and $q=m-2 j$ and assuming $p \leq q$ we obtain  $$ h_{n, i}(x) h_{m, j}(x)=\left(x^{i}+\cdots+x^{i+p}\right)\left(x^{j}+\cdots+x^{j+q}\right)=\sum_{k=i+j}^{n-i+j} h_{m+n, k} $$  which proves the claim.


17. (GDR 3) Let $P_{1}, P_{2}, \ldots, P_{n}$ be distinct points of the plane, $n \geq 2$. Prove that  $$ \max _{1 \leq i<j \leq n} P_{i} P_{j}>\frac{\sqrt{3}}{2}(\sqrt{n}-1) \min _{1 \leq i<j \leq n} P_{i} P_{j} $$

Solution: 17. Set $a=\min P_{i} P_{j}, b=\max P_{i} P_{j}$. We use the following lemma.  Lemma. There exists a disk of radius less than or equal to $b / \sqrt{3}$ containing all the $P_{i}$ 's. Assuming that this is proved, the disks with center $P_{i}$ and radius $a / 2$ are disjoint and included in a disk of radius $b / \sqrt{3}+a / 2$; hence comparing areas,  $$ n \pi \cdot \frac{a^{2}}{4}<\pi \cdot\left(\frac{b}{\sqrt{3}}+a / 2\right)^{2} \quad \text { and } \quad b>\sqrt{3} / 2 \cdot(\sqrt{n}-1) a $$  Proof of the lemma. If a nonobtuse triangle with sides $a \geq b \geq c$ has a circumscribed circle of radius $R$, we have $R=a /(2 \sin \alpha) \leq a / \sqrt{3}$. Now we show that there exists a disk $D$ of radius $R$ containing $A=$ $\left\{P_{1}, \ldots, P_{n}\right\}$ whose border $C$ is such that $C \cap A$ is not included in an open semicircle, and hence contains either two diametrically opposite points and $R \leq b / 2$, or an acute-angled triangle and $R \leq b / \sqrt{3}$. Among all disks whose borders pass through three points of $A$ and that contain all of $A$, let $D$ be the one of least radius. Suppose that $C \cap A$ is contained in an arc of central angle less than $180^{\circ}$, and that $P_{i}, P_{j}$ are its endpoints. Then there exists a circle through $P_{i}, P_{j}$ of smaller radius that contains $A$, a contradiction. Thus $D$ has the required property, and the assertion follows.


18. (FRG 3) ${ }^{\mathrm{IMO} 3}$ Let $a, b, c$ be positive integers satisfying $(a, b)=(b, c)=$ $(c, a)=1$. Show that $2 a b c-a b-b c-c a$ is the largest integer not representable as  $$ x b c+y c a+z a b $$  with nonnegative integers $x, y, z$.

Solution: 18. Let $\left(x_{0}, y_{0}, z_{0}\right)$ be one solution of $b c x+c a y+a b z=n$ (not necessarily nonnegative). By subtracting $b c x_{0}+c a y_{0}+a b z_{0}=n$ we get  $$ b c\left(x-x_{0}\right)+c a\left(y-y_{0}\right)+a b\left(z-z_{0}\right)=0 . $$  Since $(a, b)=(a, c)=1$, we must have $a \mid x-x_{0}$ or $x-x_{0}=a s$. Substituting this in the last equation gives  $$ b c s+c\left(y-y_{0}\right)+b\left(z-z_{0}\right)=0 $$  Since $(b, c)=1$, we have $b \mid y-y_{0}$ or $y-y_{0}=b t$. If we substitute this in the last equation we get $b c s+b c t+b\left(z-z_{0}\right)=0$, or $c s+c t+z-z_{0}=0$, or $z-z_{0}=-c(s+t)$. In $x=x_{0}+a s$ and $y=y_{0}+b t$, we can choose $s$ and $t$ such that $0 \leq x \leq a-1$ and $0 \leq y \leq b-1$. If $n>2 a b c-b c-c a-a b$, then $a b z=n-b c x-a c y>2 a b c-a b-b c-c a-b c(a-1)-c a(b-1)=-a b$ or $z>-1$, i.e., $z \geq 0$. Hence, it is representable as $b c x+c a y+a b z$ with $x, y, z \geq 0$. Now we prove that $2 a b c-b c-c a-a b$ is not representable as $b c x+c a y+a b z$ with $x, y, z \geq 0$. Suppose that $b c x+c a y+a b z=2 a b c-a b-b c-c a$ with $x, y, z \geq 0$. Then  $$ b c(x+1)+c a(y+1)+a b(z+1)=2 a b c $$  with $x+1, y+1, z+1 \geq 1$. Since $(a, b)=(a, c)=1$, we have $a \mid x+1$ and thus $a \leq x+1$. Similarly $b \leq y+1$ and $c \leq z+1$. Thus $b c a+c a b+a b c \leq 2 a b c$, a contradiction.


19. (ROM 1) Let $\left(F_{n}\right)_{n \geq 1}$ be the Fibonacci sequence $F_{1}=F_{2}=1, F_{n+2}=$ $F_{n+1}+F_{n}(n \geq 1)$, and $P(x)$ the polynomial of degree 990 satisfying  $$ P(k)=F_{k}, \quad \text { for } k=992, \ldots, 1982 $$  Prove that $P(1983)=F_{1983}-1$.

Solution: 19. For all $n$, there exists a unique polynomial $P_{n}$ of degree $n$ such that $P_{n}(k)=F_{k}$ for $n+2 \leq k \leq 2 n+2$ and $P_{n}(2 n+3)=F_{2 n+3}-1$. For $n=0$, we have $F_{1}=F_{2}=1, F_{3}=2, P_{0}=1$. Now suppose that $P_{n-1}$ has been constructed and let $P_{n}$ be the polynomial of degree $n$ satisfying $P_{n}(X+2)-P_{n}(X+1)=P_{n-1}(X)$ and $P_{n}(n+2)=F_{n+2}$. (The mapping $\mathbb{R}_{n}[X] \rightarrow \mathbb{R}_{n-1}[X] \times \mathbb{R}, P \mapsto(Q, P(n+2)$ ), where $Q(X)=$ $P(X+2)-P(X+1)$, is bijective, since it is injective and those two spaces have the same dimension; clearly $\operatorname{deg} Q=\operatorname{deg} P-1$.) Thus for $n+2 \leq k \leq 2 n+2$ we have $P_{n}(k+1)=P_{n}(k)+F_{k-1}$ and $P_{n}(n+2)=F_{n+2}$; hence by induction on $k, P_{n}(k)=F_{k}$ for $n+2 \leq k \leq 2 n+2$ and  $$ P_{n}(2 n+3)=F_{2 n+2}+P_{n-1}(2 n+1)=F_{2 n+3}-1 $$  Finally, $P_{990}$ is exactly the polynomial $P$ of the terms of the problem, for $P_{990}-P$ has degree less than or equal to 990 and vanishes at the 991 points $k=992, \ldots, 1982$.


2. (BEL 1) Let $n$ be a positive integer. Let $\sigma(n)$ be the sum of the natural divisors $d$ of $n$ (including 1 and $n$ ). We say that an integer $m \geq 1$ is superabundant (P.Erdös, 1944) if $\forall k \in\{1,2, \ldots, m-1\}, \frac{\sigma(m)}{m}>\frac{\sigma(k)}{k}$. Prove that there exists an infinity of superabundant numbers.

Solution: 2. By definition, $\sigma(n)=\sum_{d \mid n} d=\sum_{d \mid n} n / d=n \sum_{d \mid n} 1 / d$, hence $\sigma(n) / n=$ $\sum_{d \mid n} 1 / d$. In particular, $\sigma(n!) / n!=\sum_{d \mid n!} 1 / d \geq \sum_{k=1}^{n} 1 / k$. It follows that the sequence $\sigma(n) / n$ is unbounded, and consequently there exist an infinite number of integers $n$ such that $\sigma(n) / n$ is strictly greater than $\sigma(k) / k$ for $k<n$.


20. (ROM 3) Solve the system of equations  $$ \begin{gathered} x_{1}\left|x_{1}\right|=x_{2}\left|x_{2}\right|+\left(x_{1}-a\right)\left|x_{1}-a\right| \\ x_{2}\left|x_{2}\right|=x_{3}\left|x_{3}\right|+\left(x_{2}-a\right)\left|x_{2}-a\right| \\ \quad \cdots \\ x_{n}\left|x_{n}\right|=x_{1}\left|x_{1}\right|+\left(x_{n}-a\right)\left|x_{n}-a\right| \end{gathered} $$  in the set of real numbers, where $a>0$.

Solution: 20. If $\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ satisfies the system with parameter $a$, then $\left(-x_{1},-x_{2}\right.$, $\ldots,-x_{n}$ ) satisfies the system with parameter $-a$. Hence it is sufficient to consider only $a \geq 0$. Let $\left(x_{1}, \ldots, x_{n}\right)$ be a solution. Suppose $x_{1} \leq a, x_{2} \leq a, \ldots, x_{n} \leq a$. Summing the equations we get  $$ \left(x_{1}-a\right)^{2}+\cdots+\left(x_{n}-a\right)^{2}=0 $$  and see that $(a, a, \ldots, a)$ is the only such solution. Now suppose that $x_{k} \geq a$ for some $k$. According to the $k$ th equation,  $$ x_{k+1}\left|x_{k+1}\right|=x_{k}^{2}-\left(x_{k}-a\right)^{2}=a\left(2 x_{k}-a\right) \geq a^{2} $$  which implies that $x_{k+1} \geq a$ as well (here $x_{n+1}=x_{1}$ ). Consequently, all $x_{1}, x_{2}, \ldots, x_{n}$ are greater than or equal to $a$, and as above $(a, a, \ldots, a)$ is the only solution.


21. (SWE 1) Find the greatest integer less than or equal to $\sum_{k=1}^{2^{1983}} k^{1 / 1983-1}$.

Solution: 21. Using the identity  $$ a^{n}-b^{n}=(a-b) \sum_{m=0}^{n-1} a^{n-m-1} b^{m} $$  with $a=k^{1 / n}$ and $b=(k-1)^{1 / n}$ one obtains  $$ 1<\left(k^{1 / n}-(k-1)^{1 / n}\right) n k^{1-1 / n} \text { for all integers } n>1 \text { and } k \geq 1 $$  This gives us the inequality $k^{1 / n-1}<n\left(k^{1 / n}-(k-1)^{1 / n}\right)$ if $n>1$ and $k \geq 1$. In a similar way one proves that $n\left((k+1)^{1 / n}-k^{1 / n}\right)<k^{1 / n-1}$ if $n>1$ and $k \geq 1$. Hence for $n>1$ and $m>1$ it holds that  $$ \begin{aligned} n \sum_{k=1}^{m}\left((k+1)^{1 / n}-k^{1 / n}\right) & <\sum_{k=1}^{m} k^{1 / n-1} \\ & <n \sum_{k=2}^{m}\left(k^{1 / n}-(k-1)^{1 / n}\right)+1, \end{aligned} $$  or equivalently,  $$ n\left((m+1)^{1 / n}-1\right)<\sum_{k=1}^{m} k^{1 / n-1}<n\left(m^{1 / n}-1\right)+1 $$  The choice $n=1983$ and $m=2^{1983}$ then gives  $$ 1983<\sum_{k=1}^{2^{1983}} k^{1 / 1983-1}<1984 $$  Therefore the greatest integer less than or equal to the given sum is 1983.


22. (SWE 4) Let $n$ be a positive integer having at least two different prime factors. Show that there exists a permutation $a_{1}, a_{2}, \ldots, a_{n}$ of the integers $1,2, \ldots, n$ such that  $$ \sum_{k=1}^{n} k \cdot \cos \frac{2 \pi a_{k}}{n}=0 $$

Solution: 22. Decompose $n$ into $n=s t$, where the greatest common divisor of $s$ and $t$ is 1 and where $s>1$ and $t>1$. For $1 \leq k \leq n$ put $k=v s+u$, where $0 \leq v \leq t-1$ and $1 \leq u \leq s$, and let $a_{k}=a_{v s+u}$ be the unique integer in the set $\{1,2,3, \ldots, n\}$ such that $v s+u t-a_{v s+u}$ is a multiple of $n$. To prove that this construction gives a permutation, assume that $a_{k_{1}}=a_{k_{2}}$, where $k_{i}=v_{i} s+u_{i}, i=1,2$. Then $\left(v_{1}-v_{2}\right) s+\left(u_{1}-u_{2}\right) t$ is a multiple of $n=s t$. It follows that $t$ divides $\left(v_{1}-v_{2}\right)$, while $\left|v_{1}-v_{2}\right| \leq t-1$, and that $s$ divides $\left(u_{1}-u_{2}\right)$, while $\left|u_{1}-u_{2}\right| \leq s-1$. Hence, $v_{1}=v_{2}, u_{1}=u_{2}$, and $k_{1}=k_{2}$. We have proved that $a_{1}, \ldots, a_{n}$ is a permutation of $\{1,2, \ldots, n\}$ and hence  $$ \sum_{k=1}^{n} k \cos \frac{2 \pi a_{k}}{n}=\sum_{v=0}^{t-1}\left(\sum_{u=1}^{s}(v s+u) \cos \left(\frac{2 \pi v}{t}+\frac{2 \pi u}{s}\right)\right) $$  Using $\sum_{u=1}^{s} \cos (2 \pi u / s)=\sum_{u=1}^{s} \sin (2 \pi u / s)=0$ and the additive formulas for cosine, one finds that  $$ \begin{aligned} \sum_{k=1}^{n} k \cos \frac{2 \pi a_{k}}{n}= & \sum_{v=0}^{t-1}\left(\cos \frac{2 \pi v}{t} \sum_{u=1}^{s} u \cos \frac{2 \pi u}{s}-\sin \frac{2 \pi v}{t} \sum_{u=1}^{s} u \sin \frac{2 \pi u}{s}\right) \\ = & \left(\sum_{u=1}^{s} u \cos \frac{2 \pi u}{s}\right)\left(\sum_{v=0}^{t-1} \cos \frac{2 \pi v}{t}\right) \\ & -\left(\sum_{u=1}^{s} u \sin \frac{2 \pi u}{s}\right)\left(\sum_{v=0}^{t-1} \sin \frac{2 \pi v}{t}\right)=0 \end{aligned} $$


23. (USS 1) ${ }^{\mathrm{IMO} 2}$ Let $K$ be one of the two intersection points of the circles $W_{1}$ and $W_{2}$. Let $O_{1}$ and $O_{2}$ be the centers of $W_{1}$ and $W_{2}$. The two common tangents to the circles meet $W_{1}$ and $W_{2}$ respectively in $P_{1}$ and $P_{2}$, the first tangent, and $Q_{1}$ and $Q_{2}$, the second tangent. Let $M_{1}$ and $M_{2}$ be the midpoints of $P_{1} Q_{1}$ and $P_{2} Q_{2}$, respectively. Prove that  $$ \angle O_{1} K O_{2}=\angle M_{1} K M_{2} $$

Solution: 23. We note that $\angle O_{1} K O_{2}=\angle M_{1} K M_{2}$ is equivalent to $\angle O_{1} K M_{1}=$ $\angle O_{2} K M_{2}$. Let $S$ be the intersection point of the common tangents, and let $L$ be the second point of intersection of $S K$ and $W_{1}$. Since $\triangle S O_{1} P_{1} \sim \triangle S P_{1} M_{1}$, we have $S K$. $S L=S P_{1}^{2}=S O_{1} \cdot S M_{1}$ which implies that points $O_{1}, L, K, M_{1}$ lie on a circle. Hence $\angle O_{1} K M_{1}=$ $\angle O_{1} L M_{1}=\angle O_{2} K M_{2}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-474.jpg?height=295&width=551&top_left_y=959&top_left_x=809)


24. (USS 2) Let $d_{n}$ be the last nonzero digit of the decimal representation of $n$ !. Prove that $d_{n}$ is aperiodic; that is, there do not exist $T$ and $n_{0}$ such that for all $n \geq n_{0}, d_{n+T}=d_{n}$.

Solution: 24. See the solution of (SL91-15).


25. (USS 3) Prove that every partition of 3-dimensional space into three disjoint subsets has the following property: One of these subsets contains all possible distances; i.e., for every $a \in \mathbb{R}_{+}$, there are points $M$ and $N$ inside that subset such that distance between $M$ and $N$ is exactly $a$.

Solution: 25. Suppose the contrary, that $\mathbb{R}^{3}=P_{1} \cup P_{2} \cup P_{3}$ is a partition such that $a_{1} \in \mathbb{R}^{+}$is not realized by $P_{1}, a_{2} \in \mathbb{R}^{+}$is not realized by $P_{2}$ and $a_{3} \in \mathbb{R}^{+}$ not realized by $P_{3}$, where w.l.o.g. $a_{1} \geq a_{2} \geq a_{3}$. If $P_{1}=\emptyset=P_{2}$, then $P_{3}=\mathbb{R}^{3}$, which is impossible. If $P_{1}=\emptyset$, and $X \in P_{2}$, the sphere centered at $X$ with radius $a_{2}$ is included in $P_{3}$ and $a_{3} \leq a_{2}$ is realized, which is impossible. If $P_{1} \neq \emptyset$, let $X_{1} \in P_{1}$. The sphere $S$ centered in $X_{1}$, of radius $a_{1}$ is included in $P_{2} \cap P_{3}$. Since $a_{1} \geq a_{3}, S \not \subset P_{3}$. Let $X_{2} \in P_{2} \cap S$. The circle $\left\{Y \in S \mid d\left(X_{2}, Y\right)=a_{2}\right\}$ is included in $P_{3}$, but $a_{2} \leq a_{1}$; hence it has radius $r=a_{2} \sqrt{1-a_{2}^{2} /\left(4 a_{1}^{2}\right)} \geq a_{2} \sqrt{3} / 2$ and $a_{3} \leq a_{2} \leq a_{2} \sqrt{3}<2 r$; hence $a_{3}$ is realized by $P_{3}$.


3. (BEL 3) $)^{\mathrm{IMO} 4}$ We say that a set $E$ of points of the Euclidian plane is "Pythagorean" if for any partition of $E$ into two sets $A$ and $B$, at least one of the sets contains the vertices of a right-angled triangle. Decide whether the following sets are Pythagorean: (a) a circle; (b) an equilateral triangle (that is, the set of three vertices and the points of the three edges).

Solution: 3. (a) A circle is not Pythagorean. Indeed, consider the partition into two semicircles each closed at one and open at the other end. (b) An equilateral triangle, call it $P Q R$, is Pythagorean. Let $P^{\prime}, Q^{\prime}$, and $R^{\prime}$ be the points on $Q R, R P$, and $P Q$ such that $P R^{\prime}: R^{\prime} Q=Q P^{\prime}$ : $P^{\prime} R=R Q^{\prime}: Q^{\prime} P=1: 2$. Then $Q^{\prime} R^{\prime} \perp P Q$, etc. Suppose that $P Q R$ is not Pythagorean, and consider a partition into $A, B$, neither of which contains the vertices of a right-angled triangle. At least two of $P^{\prime}, Q^{\prime}$, and $R^{\prime}$ belong to the same class, say $P^{\prime}, Q^{\prime} \in A$. Then $[P R] \backslash\left\{Q^{\prime}\right\} \subset B$ and hence $R^{\prime} \in A$ (otherwise, if $R^{\prime \prime}$ is the foot of the perpendicular from $R^{\prime}$ to $P R, \triangle R R^{\prime} R^{\prime \prime}$ is right-angled with all vertices in $B$ ). But this implies again that $[P Q] \backslash\left\{R^{\prime}\right\} \subset B$, and thus $B$ contains vertices of a rectangular triangle. This is a contradiction.


4. (BEL 5) On the sides of the triangle $A B C$, three similar isosceles triangles $A B P(A P=P B), A Q C(A Q=Q C)$, and $B R C(B R=R C)$ are constructed. The first two are constructed externally to the triangle $A B C$, but the third is placed in the same half-plane determined by the line $B C$ as the triangle $A B C$. Prove that $A P R Q$ is a parallelogram.

Solution: 4. The rotational homothety centered at $C$ that sends $B$ to $R$ also sends $A$ to $Q$; hence the triangles $A B C$ and $Q R C$ are similar. For the same reason, $\triangle A B C$ and $\triangle P B R$ are similar. Moreover, $B R=C R$; hence $\triangle C R Q \cong$ $\triangle R B P$. Thus $P R=Q C=A Q$ and $Q R=P B=P A$, so $A P Q R$ is a parallelogram.


5. (BRA 1) Consider the set of all strictly decreasing sequences of $n$ natural numbers having the property that in each sequence no term divides any other term of the sequence. Let $A=\left(a_{j}\right)$ and $B=\left(b_{j}\right)$ be any two such sequences. We say that $A$ precedes $B$ if for some $k, a_{k}<b_{k}$ and $a_{i}=b_{i}$ for $i<k$. Find the terms of the first sequence of the set under this ordering.

Solution: 5. Each natural number $p$ can be written uniquely in the form $p=2^{q}(2 r-1)$. We call $2 r-1$ the odd part of $p$. Let $A_{n}=\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ be the first sequence. Clearly the terms of $A_{n}$ must have different odd parts, so those parts must be at least $1,3, \ldots, 2 n-1$. Being the first sequence, $A_{n}$ must have the numbers $2 n-1,2 n-3, \ldots, 2 k+1$ as terms, where $k=[n+1 / 3]$ (then $3(2 k-1)<2 n-1<3(2 k+1)$ ). Smaller odd numbers $2 s+1$ (with $s<k$ ) obviously cannot be terms of $A_{n}$. In this way we have obtained the $n-k$ odd numbers of $A_{n}$. The other $k$ terms must be even, and by the same reasoning as above they must be precisely the terms of $2 A_{k}$ (twice the terms of $A_{k}$ ). Therefore $A_{n}$ is defined recursively as  $$ \begin{gathered} A_{0}=\emptyset, \quad A_{1}=\{1\}, \quad A_{2}=\{3,2\} \\ A_{n}=\{2 n-1,2 n-3, \ldots, 2 k+1\} \cup 2 A_{k} . \end{gathered} $$


6. (CAN 2) Suppose that $\left\{x_{1}, x_{2}, \ldots, x_{n}\right\}$ are positive integers for which $x_{1}+x_{2}+\cdots+x_{n}=2(n+1)$. Show that there exists an integer $r$ with $0 \leq r \leq n-1$ for which the following $n-1$ inequalities hold:  $$ \begin{aligned} x_{r+1}+\cdots+x_{r+i} & \leq 2 i+1 & & \forall i, 1 \leq i \leq n-r ; \\ x_{r+1}+\cdots+x_{n}+x_{1}+\cdots+x_{i} & \leq 2(n-r+i)+1 & & \forall i, 1 \leq i \leq r-1 . \end{aligned} $$  Prove that if all the inequalities are strict, then $r$ is unique and that otherwise there are exactly two such $r$.

Solution: 6. The existence of $r$ : Let $S=\left\{x_{1}+x_{2}+\cdots+x_{i}-2 i \mid i=1,2, \ldots, n\right\}$. Let $\max S$ be attained for the first time at $r^{\prime}$. If $r^{\prime}=n$, then $x_{1}+x_{2}+\cdots+x_{i}-2 i<2$ for $1 \leq i \leq n-1$, so one can take $r=r^{\prime}$. Suppose that $r^{\prime}<n$. Then for $l<n-r^{\prime}$ we have $x_{r^{\prime}+1}+x_{r^{\prime}+2}+\cdots+x_{r^{\prime}+l}=$ $\left(x_{1}+\cdots+x_{r^{\prime}+l}-2\left(r^{\prime}+l\right)\right)-\left(x_{1}+\cdots+x_{r^{\prime}}-2 r^{\prime}\right)+2 l \leq 2 l$; also, for $i<r^{\prime}$ we have $\left(x_{r^{\prime}+1}+\cdots+x_{n}\right)+\left(x_{1}+\cdots+x_{i}-2 i\right)<\left(x_{r^{\prime}+1}+\cdots+\right.$ $\left.x_{n}\right)+\left(x_{1}+\cdots+x_{r^{\prime}}-2 r^{\prime}\right)=\left(x_{1}+\cdots+x_{n}\right)-2 r^{\prime}=2\left(n-r^{\prime}\right)+2 \Rightarrow$ $x_{r^{\prime}+1}+\cdots+x_{n}+x_{1}+\cdots+x_{i} \leq 2\left(n+i-r^{\prime}\right)+1$, so we can again take $r=r^{\prime}$ 。 For the second part of the problem, we relabel the sequence so that $r=0$ works. Suppose that the inequalities are strict. We have $x_{1}+x_{2}+\cdots+x_{k} \leq 2 k$, $k=1, \ldots, n-1$. Now, $2 n+2=\left(x_{1}+\cdots+x_{k}\right)+\left(x_{k+1}+\cdots+x_{n}\right) \leq$ $2 k+x_{k+1}+\cdots+x_{n} \Rightarrow x_{k+1}+\cdots+x_{n} \geq 2(n-k)+2>2(n-k)+1$. So we cannot begin with $x_{k+1}$ for any $k>0$. Now assume that there is an equality for some $k$. There are two cases: (i) Suppose $x_{1}+x_{2}+\cdots+x_{i} \leq 2 i(i=1, \ldots, k)$ and $x_{1}+\cdots+x_{k}=2 k+1$, $x_{1}+\cdots+x_{k+l} \leq 2(k+l)+1(1 \leq l \leq n-1-k)$. For $i \leq k-1$ we have $x_{i+1}+\cdots+x_{n}=2(n+1)-\left(x_{1}+\cdots+x_{i}\right)>2(n-i)+1$, so we cannot take $r=i$. If there is a $j \geq 1$ such that $x_{1}+x_{2}+\cdots+x_{k+j} \leq 2(k+j)$, then also $x_{k+j+1}+\cdots+x_{n}>2(n-k-j)+1$. If $(\forall j \geq 1) x_{1}+\cdots+x_{k+j}=$ $2(k+j)+1$, then $x_{n}=3$ and $x_{k+1}=\cdots=x_{n-1}=2$. In this case we directly verify that we cannot take $r=k+j$. However, we can also take $r=k$ : for $k+l \leq n-1, x_{k+1}+\cdots+x_{k+l} \leq 2(k+l)+1-(2 k+1)=2 l$, also $x_{k+1}+\cdots+x_{n}=2(n-k)+1$, and moreover $x_{1} \leq 2, x_{1}+x_{2} \leq 4, \ldots$. (ii) Suppose $x_{1}+\cdots+x_{i} \leq 2 i(1 \leq i \leq n-2)$ and $x_{1}+\cdots+x_{n-1}=2 n-1$. Then we can obviously take $r=n-1$. On the other hand, for any $1 \leq i \leq n-2, x_{i+1}+\cdots+x_{n-1}+x_{n}=\left(x_{1}+\cdots+x_{n-1}\right)-\left(x_{1}+\cdots+\right.$ $\left.x_{i}\right)+3>2(n-i)+1$, so we cannot take another $r \neq 0$.


7. (CAN 5) Let $a$ be a positive integer and let $\left\{a_{n}\right\}$ be defined by $a_{0}=0$ and  $$ a_{n+1}=\left(a_{n}+1\right) a+(a+1) a_{n}+2 \sqrt{a(a+1) a_{n}\left(a_{n}+1\right)} \quad(n=1,2 \ldots) $$  Show that for each positive integer $n, a_{n}$ is a positive integer.

Solution: 7. Clearly, each $a_{n}$ is positive and $\sqrt{a_{n+1}}=\sqrt{a_{n}} \sqrt{a+1}+\sqrt{a_{n}+1} \sqrt{a}$. Notice that $\sqrt{a_{n+1}+1}=\sqrt{a+1} \sqrt{a_{n}+1}+\sqrt{a} \sqrt{a_{n}}$. Therefore  $$ \begin{aligned} & (\sqrt{a+1}-\sqrt{a})\left(\sqrt{a_{n}+1}-\sqrt{a_{n}}\right) \\ & \quad=\left(\sqrt{a+1} \sqrt{a_{n}+1}+\sqrt{a} \sqrt{a_{n}}\right)-\left(\sqrt{a_{n}} \sqrt{a+1}+\sqrt{a_{n}+1} \sqrt{a}\right) \\ & \quad=\sqrt{a_{n+1}+1}-\sqrt{a_{n+1}} \end{aligned} $$  By induction, $\sqrt{a_{n+1}}-\sqrt{a_{n}}=(\sqrt{a+1}-\sqrt{a})^{n}$. Similarly, $\sqrt{a_{n+1}}+\sqrt{a_{n}}=$ $(\sqrt{a+1}+\sqrt{a})^{n}$. Hence,  $$ \sqrt{a_{n}}=\frac{1}{2}\left[(\sqrt{a+1}+\sqrt{a})^{n}-(\sqrt{a+1}-\sqrt{a})^{n}\right] $$  from which the result follows.


8. (SPA 2) In a test, $3 n$ students participate, who are located in three rows of $n$ students in each. The students leave the test room one by one. If $N_{1}(t), N_{2}(t), N_{3}(t)$ denote the numbers of students in the first, second, and third row respectively at time $t$, find the probability that for each $t$ during the test,  $$ \left|N_{i}(t)-N_{j}(t)\right|<2, \quad i \neq j, \quad i, j=1,2, \ldots $$

Solution: 8. Situations in which the condition of the statement is fulfilled are the following: $S_{1}: N_{1}(t)=N_{2}(t)=N_{3}(t)$ $S_{2}: N_{i}(t)=N_{j}(t)=h, N_{k}(t)=h+1$, where $(i, j, k)$ is a permutation of the set $\{1,2,3\}$. In this case the first student to leave must be from row $k$. This leads to the situation $S_{1}$. $S_{3}: N_{i}(t)=h, N_{j}(t)=N_{k}(t)=h+1,((i, j, k)$ is a permutation of the set $\{1,2,3\})$. In this situation the first student leaving the room belongs to row $j$ (or $k$ ) and the second to row $k$ (or $j$ ). After this we arrive at the situation $S_{1}$. Hence, the initial situation is $S_{1}$ and after each triple of students leaving the room the situation $S_{1}$ must recur. We shall compute the probability $P_{h}$ that from a situation $S_{1}$ with $3 h$ students in the room $(h \leq n)$ one arrives at a situation $S_{1}$ with $3(h-1)$ students in the room:  $$ P_{h}=\frac{(3 h) \cdot(2 h) \cdot h}{(3 h) \cdot(3 h-1) \cdot(3 h-2)}=\frac{3!h^{3}}{3 h(3 h-1)(3 h-2)} $$  Since the room becomes empty after the repetition of $n$ such processes, which are independent, we obtain for the probability sought  $$ P=\prod_{h=1}^{n} P_{h}=\frac{(3!)^{n}(n!)^{3}}{(3 n)!} $$


9. (USA 3) ${ }^{\mathrm{IMO} 6}$ If $a, b$, and $c$ are sides of a triangle, prove that  $$ a^{2} b(a-b)+b^{2} c(b-c)+c^{2} a(c-a) \geq 0 $$  Determine when there is equality.

Solution: 9. For any triangle of sides $a, b, c$ there exist 3 nonnegative numbers $x, y, z$ such that $a=y+z, b=z+x, c=x+y$ (these numbers correspond to the division of the sides of a triangle by the point of contact of the incircle). The inequality becomes  $$ (y+z)^{2}(z+x)(y-x)+(z+x)^{2}(x+y)(z-y)+(x+y)^{2}(y+z)(x-z) \geq 0 $$  Expanding, we get $x y^{3}+y z^{3}+z x^{3} \geq x y z(x+y+z)$. This follows from Cauchy's inequality $\left(x y^{3}+y z^{3}+z x^{3}\right)(z+x+y) \geq(\sqrt{x y z}(x+y+z))^{2}$ with equality if and only if $x y^{3} / z=y z^{3} / x=z x^{3} / y$, or equivalently $x=y=z$, i.e., $a=b=c$.


1. (FRA 1) Find all solutions of the following system of $n$ equations in $n$ variables:  $$ \begin{aligned} & x_{1}\left|x_{1}\right|-\left(x_{1}-a\right)\left|x_{1}-a\right|=x_{2}\left|x_{2}\right| \\ & x_{2}\left|x_{2}\right|-\left(x_{2}-a\right)\left|x_{2}-a\right|=x_{3}\left|x_{3}\right| \\ & \cdots \\ & x_{n}\left|x_{n}\right|-\left(x_{n}-a\right)\left|x_{n}-a\right|=x_{1}\left|x_{1}\right| \end{aligned} $$  where $a$ is a given number.

Solution: 1. This is the same problem as (SL83-20).


10. (GBR 1) Prove that the product of five consecutive positive integers cannot be the square of an integer.

Solution: 10. Suppose that the product of some five consecutive numbers is a square. It is easily seen that among them at least one, say $n$, is divisible neither by 2 nor 3 . Since $n$ is coprime to the remaining four numbers, it is itself a square of a number $m$ of the form $6 k \pm 1$. Thus $n=(6 k \pm 1)^{2}=24 r+1$, where $r=k(3 k \pm 1) / 2$. Note that neither of the numbers $24 r-1,24 r+5$ is one of our five consecutive numbers because it is not a square. Hence the five numbers must be $24 r, 24 r+1, \ldots, 24 r+4$. However, the number $24 r+4=(6 k \pm 1)^{2}+3$ is divisible by $6 r+1$, which implies that it is a square as well. It follows that these two squares are 1 and 4 , which is impossible.


11. (CAN 1) Let $n$ be a natural number and $a_{1}, a_{2}, \ldots, a_{2 n}$ mutually distinct integers. Find all integers $x$ satisfying  $$ \left(x-a_{1}\right) \cdot\left(x-a_{2}\right) \cdots\left(x-a_{2 n}\right)=(-1)^{n}(n!)^{2} $$

Solution: 11. Suppose that an integer $x$ satisfies the equation. Then the numbers $x-$ $a_{1}, x-a_{2}, \ldots, x-a_{2 n}$ are $2 n$ distinct integers whose product is $1 \cdot(-1)$. $2 \cdot(-2) \cdots n \cdot(-n)$. From here it is obvious that the numbers $x-a_{1}, x-a_{2}, \ldots, x-a_{2 n}$ are some reordering of the numbers $-n,-n+1, \ldots,-1,1, \ldots, n-1, n$. It follows that their sum is 0 , and therefore $x=\left(a_{1}+a_{2}+\cdots+a_{2 n}\right) / 2 n$. This is the only solution if $\left\{a_{1}, a_{2}, \ldots, a_{2 n}\right\}=\{x-n, \ldots, x-1, x+1, \ldots, x+n\}$ for some $x \in \mathbb{N}$. Otherwise there is no solution.


12. (NET 1) ${ }^{\mathrm{IMO} 2}$ Find two positive integers $a, b$ such that none of the numbers $a, b, a+b$ is divisible by 7 and $(a+b)^{7}-a^{7}-b^{7}$ is divisible by $7^{7}$.

Solution: 12. By the binomial formula we have  $$ \begin{aligned} (a+b)^{7}-a^{7}-b^{7} & =7 a b\left[\left(a^{5}+b^{5}\right)+3 a b\left(a^{3}+b^{3}\right)+5 a^{2} b^{2}(a+b)\right] \\ & =7 a b(a+b)\left(a^{2}+a b+b^{2}\right)^{2} \end{aligned} $$  Thus it will be enough to find $a$ and $b$ such that $7 \nmid a, b$ and $7^{3} \mid a^{2}+a b+b^{2}$. Such numbers must satisfy $(a+b)^{2}>a^{2}+a b+b^{2} \geq 7^{3}=343$, implying $a+b \geq 19$. Trying $a=1$ we easily find the example $(a, b)=(1,18)$.


13. (BUL 5) Prove that the volume of a tetrahedron inscribed in a right circular cylinder of volume 1 does not exceed $\frac{2}{3 \pi}$.

Solution: 13. Let $Z$ be the given cylinder of radius $r$, altitude $h$, and volume $\pi r^{2} h=1, k_{1}$ and $k_{2}$ the circles surrounding its bases, and $V$ the volume of an inscribed tetrahedron $A B C D$. We claim that there is no loss of generality in assuming that $A, B, C, D$ all lie on $k_{1} \cup k_{2}$. Indeed, if the vertices $A, B, C$ are fixed and $D$ moves along a segment $E F$ parallel to the axis of the cylinder $\left(E \in k_{1}, F \in k_{2}\right)$, the maximum distance of $D$ from the plane $A B C$ (and consequently the maximum value of $V$ ) is achieved either at $E$ or at $F$. Hence we shall consider only the following two cases: (i) $A, B \in k_{1}$ and $C, D \in k_{2}$. Let $P, Q$ be the projections of $A, B$ on the plane of $k_{2}$, and $R, S$ the projections of $C, D$ on the plane of $k_{1}$, respectively. Then $V$ is one-third of the volume $V^{\prime}$ of the prism $A R B S C P D Q$ with bases $A R B S$ and $C P D Q$. The area of the quadrilateral $A R B S$ inscribed in $k_{1}$ does not exceed the area of the square inscribed therein, which is $2 r^{2}$. Hence $3 V=V^{\prime} \leq 2 r^{2} h=2 / \pi$. (ii) $A, B, C \in k_{1}$ and $D \in k_{2}$. The area of the triangle $A B C$ does not exceed the area of an equilateral triangle inscribed in $k_{1}$, which is $3 \sqrt{3} r^{2} / 4$. Consequently, $V \leq \frac{\sqrt{3}}{4} r^{2} h=\frac{\sqrt{3}}{4 \pi}<\frac{2}{3 \pi}$.


14. (ROM 5) ${ }^{\mathrm{IMO} 4}$ Let $A B C D$ be a convex quadrilateral for which the circle with diameter $A B$ is tangent to the line $C D$. Show that the circle with diameter $C D$ is tangent to the line $A B$ if and only if the lines $B C$ and $A D$ are parallel.

Solution: 14. Let $M$ and $N$ be the midpoints of $A B$ and $C D$, and let $M^{\prime}, N^{\prime}$ be their projections on $C D$ and $A B$, respectively. We know that $M M^{\prime}=A B /$, and hence  $$ S_{A B C D}=S_{A M D}+S_{B M C}+S_{C M D}=\frac{1}{2}\left(S_{A B D}+S_{A B C}\right)+\frac{1}{4} A B \cdot C D $$  The line $A B$ is tangent to the circle with diameter $C D$ if and only if $N N^{\prime}=C D / 2$, or equivalently,  $$ S_{A B C D}=S_{A N D}+S_{B N C}+S_{A N B}=\frac{1}{2}\left(S_{B C D}+S_{A C D}\right)+\frac{1}{4} A B \cdot C D $$  By (1), this is further equivalent to $S_{A B C}+S_{A B D}=S_{B C D}+S_{A C D}$. But since $S_{A B C}+S_{A C D}=S_{A B D}+S_{B C D}=S_{A B C D}$, this reduces to $S_{A B C}=S_{B C D}$, i.e., to $B C \| A D$.


15. (LUX 2) Angles of a given triangle $A B C$ are all smaller than $120^{\circ}$. Equilateral triangles $A F B, B D C$ and $C E A$ are constructed in the exterior of $\triangle A B C$. (a) Prove that the lines $A D, B E$, and $C F$ pass through one point $S$. (b) Prove that $S D+S E+S F=2(S A+S B+S C)$.

Solution: 15. (a) Since rotation by $60^{\circ}$ around $A$ transforms the triangle $C A F$ into $\triangle E A B$, it follows that $\measuredangle(C F, E B)=60^{\circ}$. We similarly deduce that $\measuredangle(E B, A D)=\measuredangle(A D, F C)=60^{\circ}$. Let $S$ be the intersection point of $B E$ and $A D$. Since $\measuredangle C S E=\measuredangle C A E=60^{\circ}$, it follows that $E A S C$ is cyclic. Therefore $\measuredangle(A S, S C)=60^{\circ}=\measuredangle(A D, F C)$, which implies that $S$ lies on $C F$ as well. (b) A rotation of $E A S C$ around $E$ by $60^{\circ}$ transforms $A$ into $C$ and $S$ into a point $T$ for which $S E=S T=S C+C T=S C+S A$. Summing the equality $S E=S C+S A$ and the analogous equalities $S D=S B+S C$ and $S F=S A+S B$ yields the result.


16. (POL 1) ${ }^{\mathrm{IMO} 6}$ Let $a, b, c, d$ be odd positive integers such that $a<b<c<$ $d$, $a d=b c$, and $a+d=2^{k}, b+c=2^{m}$ for some integers $k$ and $m$. Prove that $a=1$.

Solution: 16. From the first two conditions we can easily conclude that $a+d>b+c$ (indeed, $(d+a)^{2}-(d-a)^{2}=(c+b)^{2}-(c-b)^{2}=4 a d=4 b c$ and $d-a>c-b>0)$. Thus $k>m$. From $d=2^{k}-a$ and $c=2^{m}-b$ we get $a\left(2^{k}-a\right)=b\left(2^{m}-b\right)$, or equivalently,  $$ (b+a)(b-a)=2^{m}\left(b-2^{k-m} a\right) $$  Since $2^{k-m} a$ is even and $b$ is odd, the highest power of 2 that divides the right-hand side of $(1)$ is $m$. Hence $(b+a)(b-a)$ is divisible by $2^{m}$ but not by $2^{m+1}$, which implies $b+a=2^{m_{1}} p$ and $b-a=2^{m_{2}} q$, where $m_{1}, m_{2} \geq 1$, $m_{1}+m_{2}=m$, and $p, q$ are odd. Furthermore, $b=\left(2^{m_{1}} p+2^{m_{2}} q\right) / 2$ and $a=\left(2^{m_{1}} p-2^{m_{2}} q\right) / 2$ are odd, so either $m_{1}=1$ or $m_{2}=1$. Note that $m_{1}=1$ is not possible, since it would imply that $b-a=2^{m-1} q \geq 2^{m-1}$, although $b+c=2^{m}$ and $b<c$ imply that $b<2^{m-1}$. Hence $m_{2}=1$ and $m_{1}=m-1$. Now since $a+b<b+c=2^{m}$, we obtain $a+b=2^{m-1}$ and $b-a=2 q$, where $q$ is an odd integer. Substituting these into (1) and dividing both sides by $2^{m}$ we get  $$ q=2^{m-2}+q-2^{k-m} a \quad \Longrightarrow \quad 2^{k-m} a=2^{m-2} . $$  Since $a$ is odd and $k>m$, it follows that $a=1$. Remark. Now it is not difficult to prove that all fourtuplets $(a, b, c, d)$ that satisfy the given conditions are of the form $\left(1,2^{m-1}-1,2^{m-1}+1,2^{2 m-2}-\right.$ 1 ), where $m \in \mathbb{N}, m \geq 3$.


17. (FRG 3) In a permutation $\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ of the set $1,2, \ldots, n$ we call a pair $\left(x_{i}, x_{j}\right)$ discordant if $i<j$ and $x_{i}>x_{j}$. Let $d(n, k)$ be the number of such permutations with exactly $k$ discordant pairs. Find $d(n, 2)$ and $d(n, 3)$.

Solution: 17. For any $m=0,1, \ldots, n-1$, we shall find the number of permutations $\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ with exactly $k$ discordant pairs such that $x_{n}=n-m$. This $x_{n}$ is a member of exactly $m$ discordant pairs, and hence the permutation $\left(x_{1}, \ldots, x_{n-1}\right.$ of the set $\{1,2, \ldots, n\} \backslash\{m\}$ must have exactly $k-m$ discordant pairs: there are $d(n-1, k-m)$ such permutations. Therefore  $$ \begin{aligned} d(n, k) & =d(n-1, k)+d(n-1, k-1) \cdots+d(n-1, k-n+1) \\ & =d(n-1, k)+d(n, k-1) \end{aligned} $$  (note that $d(n, k)$ is 0 if $k<0$ or $k>\binom{n}{2}$ ). We now proceed to calculate $d(n, 2)$ and $d(n, 3)$. Trivially, $d(n, 0)=1$. It follows that $d(n, 1)=d(n-1,1)+d(n, 0)=d(n-1,1)+1$, which yields $d(n, 1)=d(1,1)+n-1=n-1$.  Further, $d(n, 2)=d(n-1,2)+d(n, 1)=d(n-1,2)+n-1=d(2,2)+$ $2+3+\cdots+n-1=\left(n^{2}-n-2\right) / 2$. Finally, using the known formula $1^{2}+2^{2}+\cdots+k^{2}=k(k+1)(2 k+1) / 6$, we have $d(n, 3)=d(n-1,3)+d(n, 2)=d(n-1,3)+\left(n^{2}-n-2\right) / 2=$ $d(2,3)+\sum_{i=3}^{n}\left(n^{2}-n-2\right) / 2=\left(n^{3}-7 n+6\right) / 6$.


18. (USA 5) Inside triangle $A B C$ there are three circles $k_{1}, k_{2}, k_{3}$ each of which is tangent to two sides of the triangle and to its incircle $k$. The radii of $k_{1}, k_{2}, k_{3}$ are 1, 4, and 9 . Determine the radius of $k$.

Solution: 18. Suppose that circles $k_{1}\left(O_{1}, r_{1}\right), k_{2}\left(O_{2}, r_{2}\right)$, and $k_{3}\left(O_{3}, r_{3}\right)$ touch the edges of the angles $\angle B A C, \angle A B C$, and $\angle A C B$, respectively. Denote also by $O$ and $r$ the center and radius of the incircle. Let $P$ be the point of tangency of the incircle with $A B$ and let $F$ be the foot of the perpendicular from $O_{1}$ to $O P$. From $\triangle O_{1} F O$ we obtain $\cot (\alpha / 2)=2 \sqrt{r r_{1}} /\left(r-r_{1}\right)$ and analogously $\cot (\beta / 2)=2 \sqrt{r r_{2}} /\left(r-r_{2}\right), \cot (\gamma / 2)=2 \sqrt{r r_{3}} /\left(r-r_{3}\right)$. We will now use a well-known trigonometric identity for the angles of a triangle:  $$ \cot \frac{\alpha}{2}+\cot \frac{\beta}{2}+\cot \frac{\gamma}{2}=\cot \frac{\alpha}{2} \cdot \cot \frac{\beta}{2} \cdot \cot \frac{\gamma}{2} . $$  (This identity follows from $\tan (\gamma / 2)=\cot (\alpha / 2+\beta / 2)$ and the formula for the cotangent of a sum.) Plugging in the obtained cotangents, we get  $$ \begin{aligned} \frac{2 \sqrt{r r_{1}}}{r-r_{1}}+\frac{2 \sqrt{r r_{2}}}{r-r_{2}}+\frac{2 \sqrt{r r_{3}}}{r-r_{3}}= & \frac{2 \sqrt{r r_{1}}}{r-r_{1}} \cdot \frac{2 \sqrt{r r_{2}}}{r-r_{2}} \cdot \frac{2 \sqrt{r r_{3}}}{r-r_{3}} \Rightarrow \\ & \sqrt{r_{1}}\left(r-r_{2}\right)\left(r-r_{3}\right)+\sqrt{r_{2}}\left(r-r_{1}\right)\left(r-r_{3}\right) \\ & +\sqrt{r_{3}}\left(r-r_{1}\right)\left(r-r_{2}\right)=4 r \sqrt{r_{1} r_{2} r_{3}} . \end{aligned} $$  For $r_{1}=1, r_{2}=4$, and $r_{3}=9$ we get $(r-4)(r-9)+2(r-1)(r-9)+3(r-1)(r-4)=24 r \Rightarrow 6(r-1)(r-11)=0$. Clearly, $r=11$ is the only viable value for $r$.


19. (CAN 5) The triangular array $\left(a_{n, k}\right)$ of numbers is given by $a_{n, 1}=1 / n$, for $n=1,2, \ldots, a_{n, k+1}=a_{n-1, k}-a_{n, k}$, for $1 \leq k \leq n-1$. Find the harmonic mean of the 1985th row.

Solution: 19. First, we shall prove that the numbers in the $n$th row are exactly the numbers  $$ \frac{1}{n\binom{n-1}{0}}, \frac{1}{n\binom{n-1}{1}}, \frac{1}{n\binom{n-1}{2}}, \ldots, \frac{1}{n\binom{n-1}{n-1}} $$  The proof of this fact can be done by induction. For small $n$, the statement can be easily verified. Assuming that the statement is true for some $n$, we have that the $k$ th element in the $(n+1)$ st row is, as is directly verified,  $$ \frac{1}{n\binom{n-1}{k-1}}-\frac{1}{(n+1)\binom{n}{k-1}}=\frac{1}{(n+1)\binom{n}{k}} $$  Thus (1) is proved. Now the geometric mean of the elements of the $n$th row becomes:  $$ \frac{1}{n \sqrt[n]{\binom{n-1}{0} \cdot\binom{n-1}{1} \cdots\binom{n-1}{n-1}}} \geq \frac{1}{n\left(\frac{\binom{n-1}{0}+\binom{n-1}{1}+\cdots+\binom{n-1}{n-1}}{n}\right)}=\frac{1}{2^{n-1}} $$  The desired result follows directly from substituting $n=1984$.


2. (CAN 2) Prove: (a) There are infinitely many triples of positive integers $m, n, p$ such that $4 m n-m-n=p^{2}-1$. (b) There are no positive integers $m, n, p$ such that $4 m n-m-n=p^{2}$.

Solution: 2. (a) For $m=t(t-1) / 2$ and $n=t(t+1) / 2$ we have $4 m n-m-n=$ $\left(t^{2}-1\right)^{2}-1$ (b) Suppose that $4 m n-m-n=p^{2}$, or equivalently, $(4 m-1)(4 n-1)=$ $4 p^{2}+1$. The number $4 m-1$ has at least one prime divisor, say $q$, that is of the form $4 k+3$. Then $4 p^{2} \equiv-1(\bmod q)$. However, by Fermat's theorem we have  $$ 1 \equiv(2 p)^{q-1}=\left(4 p^{2}\right)^{\frac{q-1}{2}} \equiv(-1)^{\frac{q-1}{2}}(\bmod q) $$  which is impossible since $(q-1) / 2=2 k+1$ is odd.


20. (USA 2) Determine all pairs $(a, b)$ of positive real numbers with $a \neq 1$ such that  $$ \log _{a} b<\log _{a+1}(b+1) $$  ### 3.26 The Twenty-Sixth IMO

Solution: 20. Define the set $S=\mathbb{R}^{+} \backslash\{1\}$. The given inequality is equivalent to $\ln b / \ln a<\ln (b+1) / \ln (a+1)$. If $b=1$, it is obvious that each $a \in S$ satisfies this inequality. Suppose now that $b$ is also in $S$. Let us define on $S$ a function $f(x)=\ln (x+1) / \ln x$. Since $\ln (x+1)>\ln x$ and $1 / x>1 / x+1>0$, we have  $$ f^{\prime}(x)=\frac{\frac{\ln x}{x+1}-\frac{\ln (x+1)}{x}}{\ln ^{2} x}<0 \quad \text { for all } x $$  Hence $f$ is always decreasing. We also note that $f(x)<0$ for $x<1$ and that $f(x)>0$ for $x>1$ (at $x=1$ there is a discontinuity). Let us assume $b>1$. From $\ln b / \ln a<\ln (b+1) / \ln (a+1)$ we get $f(b)>$ $f(a)$. This holds for $b>a$ or for $a<1$. Now let us assume $b<1$. This time we get $f(b)<f(a)$. This holds for $a<b$ or for $a>1$. Hence all the solutions to $\log _{a} b<\log _{a+1}(b+1)$ are $\{b=1, a \in S\}$, $\{a>b>1\},\{b>1>a\},\{a<b<1\}$, and $\{b<1<a\}$.


3. (USS 3) Find all positive integers $n$ such that  $$ n=d_{6}^{2}+d_{7}^{2}-1 $$  where $1=d_{1}<d_{2}<\cdots<d_{k}=n$ are all positive divisors of the number $n$.

Solution: 3. From the equality $n=d_{6}^{2}+d_{7}^{2}-1$ we see that $d_{6}$ and $d_{7}$ are relatively prime and $d_{7}\left|d_{6}^{2}-1=\left(d_{6}-1\right)\left(d_{6}+1\right), d_{6}\right| d_{7}^{2}-1=\left(d_{7}-1\right)\left(d_{7}+1\right)$. Suppose that $d_{6}=a b, d_{7}=c d$ with $1<a<b, 1<c<d$. Then $n$ has 7 divisors smaller than $d_{7}$, namely $1, a, b, c, d, a b, a c$, which is impossible. Hence, one of the two numbers $d_{6}$ and $d_{7}$ is either a prime $p$ or the square of a prime $p^{2}$, where $p$ is not 2 . Let it be $d_{i}, i \in\{6,7\}$; then $d_{i} \mid\left(d_{j}-1\right)\left(d_{j}+1\right)$ implies that $d_{j} \equiv \pm 1\left(\bmod d_{i}\right)$, and consequently $\left(d_{i}^{2}-1\right) / d_{j} \equiv \pm 1$ as well. But either $d_{j}$ or $\left(d_{i}^{2}-1\right) / d_{j}$ is less than $d_{i}$, and therefore equals $d_{i}-1$. We thus conclude that $d_{7}=d_{6}+1$. Setting $d_{6}=x, d_{7}=x+1$ we obtain that $n=x^{2}+(x+1)^{2}-1=2 x(x+1)$ is even. (i) Assume that one of $x, x+1$ is a prime $p$. The other one has at most 6 divisors and hence must be of the form $2^{3}, 2^{4}, 2^{5}, 2 q, 2 q^{2}, 4 q$, where $q$ is an odd prime. The numbers $2^{3}$ and $2^{4}$ are easily eliminated, while $2^{5}$ yields the solution $x=31, x+1=32, n=1984$. Also, $2 q$ is eliminated because $n=4 p q$ then has only 4 divisors less than $x ; 2 q^{2}$ is eliminated because $n=4 p q^{2}$ has at least 6 divisors less than $x ; 4 q$ is also eliminated because $n=8 p q$ has 6 divisors less than $x$. (ii) Assume that one of $x, x+1$ is $p^{2}$. The other one has at most 5 divisors ( $p$ excluded), and hence is of the form $2^{3}, 2^{4}, 2 q$, where $q$ is an odd prime. The number $2^{3}$ yields the solution $x=8, x+1=9, n=144$, while $2^{4}$ is easily eliminated. Also, $2 q$ is eliminated because $n=4 p^{2} q$ has 6 divisors less than $x$. Thus there are two solutions in total: 144 and 1984.


4. (MON 1) ${ }^{\mathrm{IMO} 5}$ Let $d$ be the sum of the lengths of all diagonals of a convex polygon of $n(n>3)$ vertices and let $p$ be its perimeter. Prove that  $$ \frac{n-3}{2}<\frac{d}{p}<\frac{1}{2}\left(\left[\frac{n}{2}\right]\left[\frac{n+1}{2}\right]-2\right) . $$

Solution: 4. Consider the convex $n$-gon $A_{1} A_{2} \ldots A_{n}$ (the indices are considered modulo $n)$. For any diagonal $A_{i} A_{j}$ we have $A_{i} A_{j}+A_{i+1} A_{j+1}>A_{i} A_{i+1}+A_{j} A_{j+1}$. Summing all such $n(n-3) / 2$ inequalities, we obtain $2 d>(n-3) p$, proving the first inequality. Let us now prove the second inequality. We notice that for each diagonal $A_{i} A_{i+j}$ (we may assume w.l.o.g. that $j \leq[n / 2]$ ) the following relation holds:  $$ A_{i} A_{i+j}<A_{i} A_{i+1}+\cdots+A_{i+j-1} A_{i+j} . $$  If $n=2 k+1$, then summing the inequalities (1) for $j=2,3, \ldots, k$ and $i=1,2, \ldots, n$ yields $d<(2+3+\cdots+k) p=([n / 2][n+1 / 2]-2) p / 2$. If $n=2 k$, then summing the inequalities (1) for $j=2,3, \ldots, k-1$, $i=1,2, \ldots, n$ and for $j=k, i=1,2, \ldots, k$ again yields $d<(2+3+\cdots+$ $(k-1)+k / 2) p=\frac{1}{2}([n / 2][n+1 / 2]-2) p$.


5. (FRG 1) ${ }^{\mathrm{IMO} 1}$ Let $x, y, z$ be nonnegative real numbers with $x+y+z=1$. Show that  $$ 0 \leq x y+y z+z x-2 x y z \leq \frac{7}{27} $$

Solution: 5. Let $f(x, y, z)=x y+y z+z x-2 x y z$. The first inequality follows immediately by adding $x y \geq x y z, y z \geq x y z$, and $z x \geq x y z$ (in fact, a stronger inequality $x y+y z+z x-9 x y z \geq 0$ holds). Assume w.l.o.g. that $z$ is the smallest of $x, y, z$. Since $x y \leq(x+y)^{2} / 4=$ $(1-z)^{2} / 4$ and $z \leq 1 / 2$, we have  $$ \begin{aligned} x y+y z+z x-2 x y z & =(x+y) z+x y(1-2 z) \\ & \leq(1-z) z+\frac{(1-z)^{2}(1-2 z)}{4} \\ & =\frac{7}{27}-\frac{(1-2 z)(1-3 z)^{2}}{108} \leq \frac{7}{27} \end{aligned} $$


6. (CAN 3) Let $c$ be a positive integer. The sequence $\left\{f_{n}\right\}$ is defined as follows:  $$ f_{1}=1, \quad f_{2}=c, \quad f_{n+1}=2 f_{n}-f_{n-1}+2 \quad(n \geq 2) $$  Show that for each $k \in \mathbb{N}$ there exists $r \in \mathbb{N}$ such that $f_{k} f_{k+1}=f_{r}$.

Solution: 6. From the given recurrence we infer $f_{n+1}-f_{n}=f_{n}-f_{n-1}+2$. Consequently, $f_{n+1}-f_{n}=\left(f_{2}-f_{1}\right)+2(n-1)=c-1+2(n-1)$. Summing up for $n=1,2, \ldots, k-1$ yields the explicit formula  $$ f_{k}=f_{1}+(k-1)(c-1)+(k-1)(k-2)=k^{2}+b k-b, $$  where $b=c-4$. Now we easily obtain $f_{k} f_{k+1}=k^{4}+2(b+1) k^{3}+\left(b^{2}+b+\right.$ $1) k^{2}-\left(b^{2}+b\right) k-b$. We are looking for an $r$ for which the last expression equals $f_{r}$. Setting $r=k^{2}+p k+q$ we get by a straightforward calculation that $p=b+1, q=-b$, and $r=k^{2}+(b+1) k-b=f_{k}+k$. Hence $f_{k} f_{k+1}=f_{f_{k}+k}$ for all $k$.


7. (FRG 5) (a) Decide whether the fields of the $8 \times 8$ chessboard can be numbered by the numbers $1,2, \ldots, 64$ in such a way that the sum of the four numbers in each of its parts of one of the forms ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-186.jpg?height=171&width=587&top_left_y=1708&top_left_x=531) is divisible by four. (b) Solve the analogous problem for ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-186.jpg?height=159&width=575&top_left_y=1974&top_left_x=539)

Solution: 7. It clearly suffices to solve the problem for the remainders modulo 4 (16 of each kind). (a) The remainders can be placed as shown in Figure 1, so that they satisfy the conditions.  | 10321032 |  | | :---: | :---: | | 23012301 |  | | 32103210 |  | | 01230123 | p | | 10321032 | q r s | | 23012301 | t | | 32103210 |  | | 01230123 |  |  Fig. 1 Fig. 2 (b) Suppose that the required numbering exists. Consider a part of the chessboard as in Figure 2. By the stated condition, all the numbers $p+q+r+s, q+r+s+t, p+q+r+t, p+r+s+t$ give the same remainder modulo 4 , and so do $p, q, r, s$. We deduce that all numbers on black cells of the board, except possibly the two corner cells, give the same remainder, which is impossible.