Let $ a,b,c,x,y$ be five real numbers such that $ a^3 + ax + y = 0$, $ b^3 + bx + y = 0$ and $ c^3 + cx + y = 0$. If $ a,b,c$ are all distinct numbers prove that their sum is zero. Ciprus

Click for solution Since $a,b,c$ are distinct, they are the real roots of $t^3+xt+y=0$, so sum of roots equals $0$.

For each nonnegative integer $n$ we define $A_n = 2^{3n}+3^{6n+2}+5^{6n+2}$. Find the greatest common divisor of the numbers $A_0,A_1,\ldots, A_{1999}$. Romania

Click for solution $A_0=35$, but $A_1$ is not a multiple of $5$ so it is either $7$ or $1$. Since $5^6=3^6=2^3=1 \pmod 7$, we have that $7$ divides every number so the gcd is $7$.

Let $S$ be a square with the side length 20 and let $M$ be the set of points formed with the vertices of $S$ and another 1999 points lying inside $S$. Prove that there exists a triangle with vertices in $M$ and with area at most equal with $\frac 1{10}$. Yugoslavia

Click for solution Let's count the number of disjoint triangles made in this way. For 1 point we have 4 triangles, and for every point we add 2 triangles. (regardless wether the point is on a line or inside a triangle) So for $1999$ points, we have $4+2.1998=4000$ disjoint triangles, and a total area of $400$, so by pigeonhole there is at least one with area below $\frac1{10}$.

Let $ABC$ be a triangle with $AB=AC$. Also, let $D\in[BC]$ be a point such that $BC>BD>DC>0$, and let $\mathcal{C}_1,\mathcal{C}_2$ be the circumcircles of the triangles $ABD$ and $ADC$ respectively. Let $BB'$ and $CC'$ be diameters in the two circles, and let $M$ be the midpoint of $B'C'$. Prove that the area of the triangle $MBC$ is constant (i.e. it does not depend on the choice of the point $D$). Greece

Click for solution In the above post, Iris Aliaj asked whether somebody understood the following solution (in fact, taken from Kalva): Kalva's solution wrote: Solution We show that M must lie on the line through A parallel to BC. Let O be the center of the circumcircle of ABD and O' the center of the other circumcircle. Let X be the midpoint of AD. Take coordinates with origin the midpoint of BC, x-axis along BC and y-axis through A. It is sufficient to show that the y-coordinates of O and O' sum to that of A, or equivalently that their mean is the same as the y-coordinate of X. But it is easy to see that the x-coordinate of X less that of O equals the x-coordinate of O' less that of X. Hence the y-coordinate differences must also be equal Here is a simpler solution: We will prove that the point M lies on the parallel to BC through A. Once this will be proven, it will obviously follow that the distance from M to BC equals the distance from A to BC, hence the triangles BCM and BCA have an equal altitude, and since the bases are equal, too, the areas are equal; in other words, the area of triangle BCM is equal to the area of triangle BCA, and hence independent of the position of D. Hence it remains only to show that M lies on the parallel to BC through A. Since BB' and CC' are diameters of corresponding circles, we have < BDB' = 90° and < CDC' = 90°. Hence, both points B' and C' lie on the perpendicular to BC at D. As a consequence, the points D, B' and C' lie on one line. Now, in the following I will only consider the case when the point C' lies between the points D and B'. The other cases can be done similarly. Now, since the points B, A, D and B' lie on one circle, < AB'D = < ABD. On the other hand, since the points C, A, D and C' lie on one circle, < ACD = 180° - < AC'D. Finally, since triangle ABC is isosceles, < ABC = < ACB. Hence, we have < AB'C' = < AB'D = < ABD = < ABC = < ACB = < ACD = 180° - < AC'D = < AC'B'. Thus, the triangle AB'C' is isosceles, too, meaning that AB' = AC'. Therefore, the point A lies on the perpendicular bisector of the segment B'C'. Hence, since M is the midpoint of the segment B'C', the line AM is perpendicular to B'C'. But the line B'C' itself is perpendicular to BC (since both points B' and C' lie on the perpendicular to BC at D); hence, it follows that the line AM is parallel to BC. In other words, the point M lies on the parallel to BC through A. This is what we wanted to show. This may look like a long proof, but the one you cited and you couldn't understand was actually longer, just the writer had left out many steps. Darij