Problem 01

If $p$ and $q$ are natural numbers so that $\frac{p}{q}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+ \ldots -\frac{1}{1318}+\frac{1}{1319},$ Prove that $p$ is divisible with $1979$ .

Solution

We first write $\begin{align*} \frac{p}{q} &=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots-\frac{1}{1318}+\frac{1}{1319}\\ &=1+\frac{1}{2}+\cdots+\frac{1}{1319}-2\cdot\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{1318}\right)\\ &=1+\frac{1}{2}+\cdots+\frac{1}{1319}-\left(1+\frac{1}{2}+\cdots+\frac{1}{659}\right)\\ &=\frac{1}{660}+\frac{1}{661}+\cdots+\frac{1}{1319} \end{align*}$

Now, observe that $\begin{align*} \frac{1}{660}+\frac{1}{1319}=\frac{660+1319}{660\cdot 1319}=\frac{1979}{660\cdot 1319} \end{align*}$ and similarly $\frac{1}{661}+\frac{1}{1318}=\frac{1979}{661\cdot 1318}$ and $\frac{1}{662}+\frac{1}{1317}=\frac{1979}{662\cdot 1317}$ , and so on. We see that the original equation becomes $\begin{align*} \frac{p}{q} =\frac{1979}{660\cdot 1319}+\frac{1979}{661\cdot 1318}+\cdots+\frac{1979}{989\cdot 990}=1979\cdot\frac{r}{s} \end{align*}$ where $s=660\cdot 661\cdots 1319$ and $r=\frac{s}{660\cdot 1319}+\frac{s}{661\cdot 1318}+\cdots+\frac{s}{989\cdot 990}$ are two integers. Finally consider $p=1979\cdot\frac{qr}{s}$ , and observe that $s\nmid 1979$ because $1979$ is a prime, it follows that $\frac{qr}{s}\in\mathbb{Z}$ . Hence we deduce that $p$ is divisible with $1979$ .

Problem 02

We consider a prism which has the upper and inferior basis the pentagons: $A_{1}A_{2}A_{3}A_{4}A_{5}$ and $B_{1}B_{2}B_{3}B_{4}B_{5}$ . Each of the sides of the two pentagons and the segments $A_{i}B_{j}$ with $i,j=1,\ldots$ ,5 is colored in red or blue. In every triangle which has all sides colored there exists one red side and one blue side. Prove that all the 10 sides of the two basis are colored in the same color.

Solution

Let us prove first that the edges $A_1A_2,A_2A_3,\cdots ,A_5A_1$ are of the same color. Assume the contrary, and let w.l.o.g. $A_1A_2$ be red and $A_2A_3$ be green. Three of the segments $A_2B_l, (l = 1, 2, 3, 4, 5)$ , say $A_2B_i,A_2B_j,A_2B_k$ ,have to be of the same color, let it w.l.o.g. be red. Then $A_1B_i,A_1B_j,A_1B_k$ must be green. At least one of the sides of triangle $B_iB_jB_k$ , say $B_iB_j$ ,must be an edge of the prism. Then looking at the triangles $A_1B_iB_j$ and $A_2B_iB_j$ we deduce that $B_iB_j$ can be neither green nor red, which is acontradiction. Hence all five edges of the pentagon $A_1A_2A_3A_4A_5$ have thesame color. Similarly, all five edges of $B_1B_2B_3B_4B_5$ have the same color.We now show that the two colors are the same. Assume otherwise, i.e.,that w.l.o.g. the $A$ edges are painted red and the $B$ edges green. Let us call segments of the form $A_iB_j$ diagonal ( $i$ and $j$ may be equal). We now count the diagonal segments by grouping the red segments based on their $A$ point, and the green segments based on their $B$ point. As above, the assumption that three of $A_iB_j$ for fixed $i$ are red leads to a contradiction. Hence at most two diagonal segments out of each $A_i$ may be red, which counts up to at most $10$ red segments. Similarly, at most $10$ diagonal segments can be green. But then we can paint at most $20$ diagonal segments out of $25$ , which is a contradiction. Hence all edges in the pentagons $A_1A_2A_3A_4A_5$ and $B_1B_2B_3B_4B_5$ have the same color.

Problem 03

Two circles in a plane intersect. $A$ is one of the points of intersection. Starting simultaneously from $A$ two points move with constant speed, each travelling along its own circle in the same sense. The two points return to $A$ simultaneously after one revolution. Prove that there is a fixed point $P$ in the plane such that the two points are always equidistant from $P.$

Solution 1

Let $O_1, O_2$ be the centers of the 2 circles and let the points B, C move on the circles $(O_1), (O_2)$ , recpectively. Let A' be the other intersection of the 2 circles different from the point A. Since the angles $\angle BO_1A = \angle CO_2A$ are equal, the isosceles triangles $\triangle AO_1B \sim \triangle AO_2C$ are similar. Consequently, the angles $\angle BAO_1 = \angle CAO_2$ are equal and the angles

$\angle BAC = BAO_1 + \angle O_1AO_2 + \angle O_2AC =$

$= BAO_1 + \angle O_1AO_2 - \angle CAO_2 = \angle O_1AO_2$

are also equal. The angles $\angle A'BA = \frac{\angle A'O_1A}{2} = \angle O_2O_1A$ are equal and the angles $\angle A'CA = \frac{\angle A'O_2A}{2} = \angle O_1O_2A$ are also equal. As a result, the point A' is on the side BC of the triangle $\triangle ABC$ , which is always similar to the fixed triangle $\triangle O_1AO_2$ .

Lemma (Yaglom's Geometric Transformations II, pp 68-69). Assume that a figure F' (the triangle $\triangle ABC$ ) moves in such a way that it remains at all times similar to a fixed figure F (the triangle $\triangle O_1AO_2$ ) and in addition, some point of the figure F' does not move at all (the point A). If a point B of the figure F' describes a curve $\Gamma$ (the circle $(O_1)$ ), then any other point M of the figure F' describes a curve $\Gamma'$ similar to the curve $\Gamma$ .

Consequently, the midpoint M of the segment BC decribes a circle (Q). When the points B, C, become identical with the point A, the midpoint M of BC also becomes identical with A. Hence, the circle (Q) passes through the point A. The line BC always passes through A'. Just after the start of motion counter-clockwise, the points B, C, A' follow on the line BC in this order. The points B, C both pass through the point A' one after another and just before the end of one period of motion, the points A', B, C follow on the line BC In this order. From the continuity principle, the midpoint M of BC also passes through the point A' and consequently, the circle (Q), the locus of the midpoints M, also passes through the point A'. Thus the circle (Q) belongs to the same pencil with the circles $(O_1), (O_2)$ and it is centered on their center line $O_1O_2$ . Let p be the perpendicular bisector of the segment BC intersectong the circle (Q) at a point P different from the point M. The point A' and the circle (Q) passing through this point are fixed and the angle $\angle A'MP = 90^\circ$ is right. Therefore, A'P is a fixed diameter of the circle (Q), which means that the point P is also fixed. Q.E.D.

When the points B, C are both diametrally opposite to the point A on their respective circles, the segment $O_1O_2$ is the midline of the triangle $\triangle ABC$ , i.e., $BC \parallel O_1O_2$ , $BC \perp AA'$ . The cyclic quadrilateral AA'MP inscribed in the circle (Q) is then a rectangle. Its circumcenter Q is the intersection of its diagonals AM, A'P intersecting at their midpoint and the midpoint of AM is identical with the midpoint of $O_1O_2$ . Thus the fixed point P is the reflection of the intersection A' of the circles $(O_1), (O_2)$ in the midpoint Q of the segment $O_1O_2$ .

This solution was posted and copyrighted by yetti. The original thread for this problem can be found here:

Solution 2

Let $B$ and $C$ be the antipodes of $A$ on the two circles and let $D$ be the foot of the altitude from $A$ , which is the other intersection point of the two circles. Also, let $M$ be the midpoint of $\overline{BC}$ , and construct rectangle $ADMZ$ . Our claim is that $Z$ is the fixed point. We let $X$ and $Y$ be the two points; by the condition the angles are the same. So we have a spiral similarity $\triangle AXY \sim \triangle ABC.$ $[asy] size(10cm); pair A = dir(110); pair B = dir(210); pair C = dir(330); pair D = foot(A, B, C); pair M = midpoint(B--C); pair Z = A+M-D; pair X = midpoint(A--B)+dir(200)*abs(A-B)/2; pair Y = -D+2*foot(midpoint(A--C), X, D); draw(circumcircle(A, B, D), blue); draw(circumcircle(A, C, D), blue); draw(circumcircle(A, D, M), dashed+deepgreen); pair N = midpoint(X--Y); filldraw(A--X--Y--cycle, invisible, purple); filldraw(A--B--C--cycle, invisible, blue); draw(A--D, blue); draw(A--Z--M, deepgreen); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$M$", M, dir(M)); dot("$Z$", Z, dir(Z)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$N$", N, dir(N)); /* TSQ Source: A = dir 110 B = dir 210 C = dir 330 D = foot A B C M = midpoint B--C Z = A+M-D X = midpoint(A--B)+dir(200)*abs(A-B)/2 Y = -D+2*foot midpoint A--C X D circumcircle A B D blue circumcircle A C D blue circumcircle A D M dashed deepgreen N = midpoint X--Y A--X--Y--cycle 0.1 red / purple A--B--C--cycle 0.1 yellow / blue A--D blue A--Z--M deepgreen */ [/asy]$ Now let $N$ be the midpoint of $\overline{XY}$ . By spiral similarity, since $N$ maps to $M$ , it follows $M$ , $N$ , $A$ , and $D$ are cyclic too. So actually $N$ lies on the circumcircle of rectangle $ADMZ$ , meaning $\overline{ZN} \perp \overline{XY}$ , hence $ZX = ZY$ as needed.

Remark: The special point $Z$ can be identified by selecting the special case $X \to A$ , $Y \to A$ and $X=B$ , $Y=C$ .

Problem 04

We consider a point $P$ in a plane $p$ and a point $Q \not\in p$ . Determine all the points $R$ from $p$ for which $\frac{QP+PR}{QR}$

is maximum.

Solution

Let $T$ be the orthogonal projection of the point $Q$ on the plane $p$ . Then, the line $PT$ is the orthogonal projection of the line $PQ$ on the plane $p$ , and thus forms the least angle with the line $PQ$ among all lines through the point $P$ which lie in the plane $p$ ; hence, $\measuredangle RPQ\geq\measuredangle TPQ$ , and equality holds if and only if the point $R$ lies on the ray $PT$ (the only exception is when $PQ\perp p$ ; in this case, $P = T$ , so the ray $PT$ is undefined, and equality holds for all points $R$ in the plane $p$ , since we always have $\angle RPQ = \angle TPQ = 90^{\circ}$ ).

From $\measuredangle RPQ\geq\measuredangle TPQ$ , it follows that $\frac{\measuredangle RPQ}{2}\geq\frac{\measuredangle TPQ}{2}$ ; also, since the angles $\angle RPQ$ and $\angle TPQ$ are $180^{\circ}$ , their half-angles $\frac{\measuredangle RPQ}{2}$ and $\frac{\measuredangle TPQ}{2}$ are < 90°, so that from $\frac{\measuredangle RPQ}{2}\geq\frac{\measuredangle TPQ}{2}$ we can conclude that $\sin\frac{\measuredangle RPQ}{2}\geq\sin\frac{\measuredangle TPQ}{2}$ . Equality holds, as in the above, if and only if the point $R$ lies on the ray $PT$ (and, respectively, for all points $R$ in the plane $p$ if $PQ\perp p$ ).

On the other hand, we obviously have $\cos\frac{\measuredangle QRP-\measuredangle PQR}{2}\leq 1$ with equality if and only if $\frac{\measuredangle QRP-\measuredangle PQR}{2}=0^{\circ}$ , i. e. if and only if < QRP = < PQR, i. e. if and only if triangle PQR is isosceles with base $QR$ , i. e. if and only if $PR = PQ$ , i. e. if and only if the point $R$ lies on the sphere with center $P$ and radius $PQ$ .

Now, applying the Mollweide theorem in triangle $QPR$ , we get

$\frac{QP+PR}{QR}=\frac{\cos\frac{\measuredangle QRP-\measuredangle PQR}{2}}{\sin\frac{\measuredangle RPQ}{2}}$ $\leq\frac{1}{\sin\frac{\measuredangle RPQ}{2}}$ (since $\cos\frac{\measuredangle QRP-\measuredangle PQR}{2}\leq 1$ ) $\leq\frac{1}{\sin\frac{\measuredangle TPQ}{2}}$ (since $\sin\frac{\measuredangle RPQ}{2}\geq\sin\frac{\measuredangle TPQ}{2}$ ),

and equality holds here if and only if equality holds in both of the inequalities $\sin\frac{\measuredangle RPQ}{2}\geq\sin\frac{\measuredangle TPQ}{2}$ and $\cos\frac{\measuredangle QRP-\measuredangle PQR}{2}\leq 1$ that we have used, i. e. if and only if the point $R$ lies both on the ray $PT$ (this condition should be ignored if $PQ\perp p$ ) and on the sphere with center $P$ and radius $PQ$ .

Hence, the point $R$ for which the ratio $\frac{QP+PR}{QR}$ is maximum is the point of intersection of the ray $PT$ with the sphere with center $P$ and radius $PQ$ (or, respectively, it can be any arbitrary point on the intersection of the plane $p$ with the sphere with center $P$ and radius $PQ$ if $PQ\perp p$ ).

Problem 05

Determine all real numbers a for which there exists positive reals $x_{1}, \ldots, x_{5}$ which satisfy the relations $\sum_{k=1}^{5} kx_{k}=a,$ $\sum_{k=1}^{5} k^{3}x_{k}=a^{2},$ $\sum_{k=1}^{5} k^{5}x_{k}=a^{3}.$

Problem 06

Let $A$ and $E$ be opposite vertices of an octagon. A frog starts at vertex $A.$ From any vertex except $E$ it jumps to one of the two adjacent vertices. When it reaches $E$ it stops. Let $a_n$ be the number of distinct paths of exactly $n$ jumps ending at $E$ . Prove that: $a_{2n-1}=0, \quad a_{2n}={(2+\sqrt2)^{n-1} - (2-\sqrt2)^{n-1} \over\sqrt2}.$

Solution

First the part $a_{2n-1}=0$ It's pretty obvious. Let's make a sequence $b$ of 1 and -1, setting 1 when the frog jumps left and -1 when it jumps right. If the frog would want to come to vertex E from vertex A, then $\sum b_{i}$ from $i =1$ to $i =2n-1$ should be equal to either 4 or -4. But this sum is odd, so we have $a_{2n-1}= 0$

Now the less obvious part. Let's name $f(X,y)$ the number of ways, in which we can come to vertex X in y moves.

Then $f(E,2n) = f(F,2n-1)+f(D, 2n-1) = f(C, 2n-2)+f(G, 2n-2) =$ $f(D, 2n-3)+f(B, 2n-3)+f(F, 2n-3)+f(H, 2n-3) =$ $2f(D, 2n-5)+2f(F, 2n-5)+4f(H,2n-5)+4f(B, 2n-5) =$ $4[ f(B,2n-5)+f(D, 2n-5)+f(F,2n-5)+f(H, 2n-5) ]-2 [ f(F,2n-5)+f(D,2n-5)] =$ $4f(E,2n-2)-2f(E,2n-4)$