(a) Let $p$ be a fixed prime number. Find all positive integers $n \ge 2$ such that there exist $n + 1$ (not necessarily positive) integers satisfying the sum of any $n$ of them is a power of $p$. (b) Let $p$ be a fixed prime number. Find all positive integers $n \ge 2$ such that there exist $n + 1$ positive integers satisfying the sum of any $n$ of them is a power of $p$. Note: For both parts, the $n + 1$ numbers do not have to be distinct. (by JasonM#8428)

Mathinity and Golden Math are playing a game. The two players take turns to fill in the blanks with real numbers in the following polynomial. Each player fill exactly one blank on each turn, and they can pick any unfilled blank to fill (not necessarily fill them from left to right). Mathinity goes first. $x^{69}+$______ $x^{68} +$ _______ $x^{67} + ...+$_____ $x+$_________ After all the blanks are filled, if the resulted polynomial has at least $3$ distinct real roots, then Mathinity wins. Else, Golden Math wins. Who has a winning strategy? (by XDitto#0165)

Let $\omega$ denote the incircle of an acute, non-isosceles triangle $ABC$, which touches sides $BC$, $CA$, $AB$ at $A_1$, $B_1$, $C_1$ respectively. Let $I$ denote the center of $\omega$. On rays $IA_1$, $IB_1$, $IC_1$ take points $A_2$, $B_2$, $C_2$ such that $$IA_2 = AB_1 = AC_1,IB_2 = BA_1 = BC_1, IC_2 = CB_1 = CA_1$$Prove that there exist two distinct points lying on the circumcircle of triangles $AA_1A_2$, $BB_1B_2$ and $CC_1C_2$. (by JasonM#8428)

Creative Math and Pepemath are playing a game: Initially, Creative Math puts $2022$ frogs on the circumference of the unit circle. On each turn, Pepemath takes away one frog, then uses telepathy to make another frog jump $60$ degree clockwise along the circumference. After that, Creative Math will create a new frog and put it somewhere on the circumference. Then the turn ends. Pepemath wins if at the end of a turn, the center of mass of all $2022$ frogs is less than $\frac{1}{1000}$ unit distance from the origin. If he cannot win within $69420$ turns, then Creative Math wins. Who has a winning strategy? Note 1: All frogs are assumed to be equal point mass. Note 2: Given n equal point mass $P_1, P_2, ..., P_n$, their center of mass is the point represented by the vector $\frac{1}{n} \left( \overrightarrow{OP_1} +\overrightarrow{OP_2} +...+\overrightarrow{OP_n} \right)$. (by XDitto#0165)

Let $a, b, n$ be positive integers in which $a, b \le n$. You are locked in a room, with $n$ distinguishable keys and $n$ distinguishable locks in it. You know that each lock can be unlocked by a unique key, but you don’t know which key unlocks which lock. You also know each key initially has a durablility of $a$ and each lock initially has a durability of $b$. The only way out of the room is to unlock at least one of the locks, but each time you try to unlock a lock using a key, the durablility of both that lock and key will decrease by $1$. A key or lock cannot be used for any further unlocking attempts if its durability is $0$. For each $n$, what pairs of $a, b$ guarentee a strategy which you can certainly leave the room? (by ? ? ?#3107)

For $k \ge 3$, $P(k, n)$ denotes the number of dots that create a regular $k$-gon with side length $n$. For example, $P(3, 2) = 3$, $P(4, 3) = 9$, $P(5, 4) = 22$. (See the following figure shamelessly stolen from IGMO2022 R2 P1.) Is it possible to find some integers $k \ge 3$, $n_3, n_4, ..., n_k$ all greater than or equal to $2$, such that the following equation hold? $$\frac{1}{P(3, n_3)}-\frac{1}{P(4, n_4)}+\frac{1}{P(5, n_5)}- ... + (-1)^{k+1} \frac{1}{P(k, n_k)}=\frac{1}{2022}$$ (by XDitto#0165)