Show that given any 9 points inside a square of side 1 we can always find 3 which form a triangle with area less than $\frac 18$. Bulgaria

Click for solution pbornsztein wrote: The problem asked for a strict inequality, thus there is some more work to do.... There are lots of cases!! Well, let's call A, B, C, D the four small squares, in clockwise order, and suppose all the triangles with vertices in 3 of our 9 points have area >= 1/8. If 3 points lie inside a square, then two of them must lie on the vertices of one of its sides, while the other one must lie on the opposite side. This is easy to prove by DusT's outlined argument. If 4 points lie inside a square, then they must coincide with its vertices. More than 4 points cannot lie inside the same square, or they'll generate a too small triangle. Now, suppose that exactly 4 points lie inside A. Then, the remaining 5 points can lie on just two sides of the big square: the one touching B and C, and the one touching C and D. By pigeonhole principle, 3 points must lie on the same side, generating a triangle with area 0. So, not more than 3 points can lie inside the same square. Suppose that exaclty 3 points lie inside A. Then, by the above argument both B and D have at least one point in common with A (meaning that the points lie on the common sides). Then, of the remaining 6 points, not more than 2 can lie inside B, and the same for D. Since not more than 3 of them can lie inside C, then exactly two must lie inside B or D. Suppose by symmetry it's B: then they must lie on the distant vertices wrt A, and one of them must be in common with C. A similar result holds for D, and so 4 points must lie in C: contradiction.

Let $\frac{x^2+y^2}{x^2-y^2} + \frac{x^2-y^2}{x^2+y^2} = k$. Compute the following expression in terms of $k$: \[ E(x,y) = \frac{x^8 + y^8}{x^8-y^8} - \frac{ x^8-y^8}{x^8+y^8}. \] Ciprus

Click for solution $k=\frac{x^2+y^2}{x^2-y^2}+\frac{x^2-y^2}{x^2+y^2}\\ k=\frac{x^4+2x^2y^2+y^4+x^4-2x^2y^2+y^4}{x^4-y^4}\\ k=\frac{2x^4+2y^4}{x^4-y^4}\\ \frac{k}{2}=\frac{x^4+y^4}{x^4-y^4}\\ \frac{k}{2}+\frac{2}{k}=\frac{x^4+y^4}{x^4-y^4}+\frac{x^4-y^4}{x^4+y^4}\\ \frac{k}{2}+\frac{2}{k}= \frac{x^8+2x^4y^4+y^8+x^8-2x^4y^4+y^8}{x^8-y^8}\\ \frac{k}{2}+\frac{2}{k}=\frac{2x^8+2y^8}{x^8-y^8}\\ \frac{\frac{k}{2}+\frac{2}{k}}{2}=\frac{x^8+y^8}{x^8-y^8}\\ \frac{\frac{k}{2}+\frac{2}{k}}{2}-\frac{2}{\frac{k}{2}+\frac{2}{k}}=\frac{x^8+y^8}{x^8-y^8}-\frac{x^8-y^8}{x^8+y^8}\\ \frac{\frac{k}{2}+\frac{2}{k}}{2}-\frac{2}{\frac{k}{2}+\frac{2}{k}}=\frac{k^2+4}{4k}-\frac{4k}{k^2+4}\\ \frac{\frac{k}{2}+\frac{2}{k}}{2}-\frac{2}{\frac{k}{2}+\frac{2}{k}}=\frac{k^4+8k^2+16-16k^2}{4k(k^2+4)}\\ \frac{\frac{k}{2}+\frac{2}{k}}{2}-\frac{2}{\frac{k}{2}+\frac{2}{k}}=\frac{(k^2-4)^2}{4k(k^2+4)}\\ \frac{x^8+y^8}{x^8-y^8}-\frac{x^8-y^8}{x^8+y^8}=\frac{(k^2-4)^2}{4k(k^2+4)}$

Let $ABC$ be a triangle and let $I$ be the incenter. Let $N$, $M$ be the midpoints of the sides $AB$ and $CA$ respectively. The lines $BI$ and $CI$ meet $MN$ at $K$ and $L$ respectively. Prove that $AI+BI+CI>BC+KL$. Greece

Click for solution The triangle inequality gives $BI+CI>BC$, so the result will follow from the stronger inequality $AI>KL$. A bit of angle chasing reveals $MK=MB$ and $NL=NC$. So we can express $KL$ in terms of these lengths with \[KL=MK+NL-MN=MB+NC-MC=\frac{AB+AC-BC}{2}\] which happens to be the length of the tangent line from $A$ to the incircle. Let $r$ be the inradius, then $AI=\sqrt{KL^{2}+r^{2}}>KL$, as desired.

Determine the triangle with sides $a,b,c$ and circumradius $R$ for which $R(b+c) = a\sqrt{bc}$. Romania

Click for solution Use Law of Sines: $a=2R\sin A$ and similar for other variables. So $\sin B+\sin C=2\sin A\sqrt{\sin B \sin C}\Rightarrow \sqrt{\sin B \sin C}=\frac{\sin B+\sin C}{2\sin A}$. But $\sqrt{\sin B \sin C}\leq \frac{\sin B+\sin C}{2}$ (AM-GM) and $\frac{\sin B+\sin C}{2}\leq \frac{\sin B+\sin C}{2\sin A}$ since $0<\sin A\leq 1$. Both equality conditions must be met, so $\sin A=1$, making $\angle A=90^\circ$, and $\sin B=\sin C$; both are acute, so both are equal. So the equation works if and only if $\Delta ABC$ is an isosceles right triangle.

Let $n_1$, $n_2$, $\ldots$, $n_{1998}$ be positive integers such that \[ n_1^2 + n_2^2 + \cdots + n_{1997}^2 = n_{1998}^2. \] Show that at least two of the numbers are even.

Click for solution sorry about that, here is the solution in full 1) if at least two of the $n_1, n_2, ..., n_{1997}$ are even we would be done. 2) if one of the $n_1, n_2, ..., n_{1997}$ is even, then $n_1^2+n_2^2+...+n_{1997}^2$ must be even since we have an even number of odd terms in that sum, so $n_{1998}$ must be even and we have our two even numbers. for the sake of contradiction, assume that $n_1, n_2, ..., n_{1997}$ are odd. note that if x is odd, then $x^2 \equiv 1 \mod 8$, so $n_1^2+...+n_{1997}^2 \equiv 1997 \equiv 5 \equiv n_{1998}^2 \mod 8$ this is a contradiction because 5 is not a square mod 8. Therefore at least one of $n_1, n_2, ..., n_{1997}$ is even and from 1) and 2) we see that there are at least two even numbers.