Let $S$ be a set of $9$ distinct integers all of whose prime factors are at most $3.$ Prove that $S$ contains $3$ distinct integers such that their product is a perfect cube.

Let $ABC$ be an acute angled triangle with $\angle{BAC}=60^\circ$ and $AB > AC$. Let $I$ be the incenter, and $H$ the orthocenter of the triangle $ABC$ . Prove that $2\angle{AHI}= 3\angle{ABC}$.

Consider $n$ disks $C_{1}; C_{2}; ... ; C_{n}$ in a plane such that for each $1 \leq i < n$, the center of $C_{i}$ is on the circumference of $C_{i+1}$, and the center of $C_{n}$ is on the circumference of $C_{1}$. Define the score of such an arrangement of $n$ disks to be the number of pairs $(i; j )$ for which $C_{i}$ properly contains $C_{j}$ . Determine the maximum possible score.

Let $x; y$ and $z$ be positive real numbers such that $\sqrt{x}+\sqrt{y}+\sqrt{z}= 1$. Prove that $\frac{x^{2}+yz}{\sqrt{2x^{2}(y+z)}}+\frac{y^{2}+zx}{\sqrt{2y^{2}(z+x)}}+\frac{z^{2}+xy}{\sqrt{2z^{2}(x+y)}}\geq 1.$

A regular $ (5 \times 5)$-array of lights is defective, so that toggling the switch for one light causes each adjacent light in the same row and in the same column as well as the light itself to change state, from on to off, or from off to on. Initially all the lights are switched off. After a certain number of toggles, exactly one light is switched on. Find all the possible positions of this light.

Click for solution label the light by its position (i;j) $1\leq\i\leq\ 5;1\leq\j\leq\ 5$ call (i;j) good if there is an algorithm make all lights but light(i;j) off notice that ;if (i;j) is good then the light symmetric to it through 2 diagonal is also good now we define an invariant consider set S (1;1);(1;2);(1;4);(1;5);(3;1);(3;2);(3;4);(3;5);(5;1);(5;2);(5;1);(5;3) the parity of M,the number of light on in S is invariant.A light not in S has even adjacent light in S(0 or 2);so toggle it will make M unchanged,or reduce 2 or increase 2,so parity is unchanged.A light in S has one adjacent light in S,so toggle it also make the parity of M unchanged Initially,M is 0,so a good position (i;j) cannot in S also (1;3);(2;1);(4;1);(2;5);(4;5) is not good since it is symmetric to a position in S so there are 5 possible good position and we can check that all of it works easy problem,right?