1997 Turkey MO (2nd round)

December 12th - Day 1

1

Find all pairs of integers $(x, y)$ such that $5x^{2}-6xy+7y^{2}=383$.

2

Let $F$ be a point inside a convex pentagon $ABCDE$, and let $a_{1}$, $a_{2}$, $a_{3}$, $a_{4}$, $a_{5}$ denote the distances from $F$ to the lines $AB$, $BC$, $CD$, $DE$, $EA$, respectively. The points $F_{1}$, $F_{2}$, $F_{3}$, $F_{4}$, $F_{5}$ are chosen on the inner bisectors of the angles $A$, $B$, $C$, $D$, $E$ of the pentagon respectively, so that $AF_{1} = AF$ , $BF_{2} = BF$ , $CF_{3} = CF$ , $DF_{4} = DF$ and $EF_{5} = EF$ . If the distances from $F_{1}$, $F_{2}$, $F_{3}$, $F_{4}$, $F_{5}$ to the lines $EA$, $AB$, $BC$, $CD$, $DE$ are $b_{1}$, $b_{2}$, $b_{3}$, $b_{4}$, $b_{5}$, respectively. Prove that $a_{1} + a_{2} + a_{3} + a_{4} + a_{5} \leq b_{1} + b_{2} + b_{3} + b_{4} + b_{5}$

3

Let $n$ and $k$ be positive integers, where $n > 1$ is odd. Suppose $n$ voters are to elect one of the $k$ cadidates from a set $A$ according to the rule of "majoritarian compromise" described below. After each voter ranks the candidates in a column according to his/her preferences, these columns are concatenated to form a $k$ x $n$ voting matrix. We denote the number of ccurences of $a \in A$ in the $i$-th row of the voting matrix by $a_{i}$ . Let $l_{a}$ stand for the minimum integer $l$ for which $\sum^{l}_{i=1}{a_{i}}> \frac{n}{2}$. Setting $l'= min \{l_{a} | a \in A\}$, we will regard the voting matrices which make the set $\{a \in A | l_{a} = l' \}$ as admissible. For each such matrix, the single candidate in this set will get elected according to majoritarian compromise. Moreover, if $w_{1} \geq w_{2} \geq ... \geq w_{k} \geq 0$ are given, for each admissible voting matrix, $\sum^{k}_{i=1}{w_{i}a_{i}}$ is called the total weighted score of $a \in A$. We will say that the system $(w_{1},w_{2}, . . . , w_{k})$ of weights represents majoritarian compromise if the total score of the elected candidate is maximum among the scores of all candidates. (a) Determine whether there is a system of weights representing majoritarian compromise if $k = 3$. (b) Show that such a system of weights does not exist for $k > 3$.

December 13th - Day 2

1

Let $e > 0$ be a given real number. Find the least value of $f(e)$ (in terms of $e$ only) such that the inequality $a^{3}+ b^{3}+ c^{3}+ d^{3} \leq e^{2}(a^{2}+b^{2}+c^{2}+d^{2}) + f(e)(a^{4}+b^{4}+c^{4}+d^{4})$ holds for all real numbers $a, b, c, d$.

2

In a triangle $ABC$, the inner and outer bisectors of the $\angle A$ meet the line $BC$ at $D$ and $E$, respectively. Let $d$ be a common tangent of the circumcircle $(O)$ of $\triangle ABC$ and the circle with diameter $DE$ and center $F$. The projections of the tangency points onto $FO$ are denoted by $P$ and $Q$, and the length of their common chord is denoted by $m$. Prove that $PQ = m$

3

Let $D_{1}, D_{2}, . . . , D_{n}$ be rectangular parallelepipeds in space, with edges parallel to the $x, y, z$ axes. For each $D_{i}$, let $x_{i} , y_{i} , z_{i}$ be the lengths of its projections onto the $x, y, z$ axes, respectively. Suppose that for all pairs $D_{i}$ , $D_{j}$, if at least one of $x_{i} < x_{j}$ , $y_{i} < y_{j}$, $z_{i} < z_{j}$ holds, then $x_{i} \leq x_{j}$ , $y_{i} \leq y_{j}$, and $z_{i} < z_{j}$ . If the volume of the region $\bigcup^{n}_{i=1}{D_{i}}$ equals 1997, prove that there is a subset $\{D_{i_{1}}, D_{i_{2}}, . . . , D_{i_{m}}\}$ of the set $\{D_{1}, . . . , D_{n}\}$ such that $(i)$ if $k \not= l $ then $D_{i_{k}} \cap D_{i_{l}} = \emptyset $, and $(ii)$ the volume of $\bigcup^{m}_{k=1}{D_{i_{k}}}$ is at least 73.