Find the largest $K$ satisfying the following: Given any closed intervals $A_1,\ldots, A_N$ of length $1$ where $N$ is an arbitrary positive integer. If their union is $[0,2021]$, then we can always find $K$ intervals from $A_1,\ldots, A_N$ such that the intersection of any two of them is empty.

Find all integers $n=2k+1>1$ so that there exists a permutation $a_0, a_1,\ldots,a_{k}$ of $0, 1, \ldots, k$ such that \[a_1^2-a_0^2\equiv a_2^2-a_1^2\equiv \cdots\equiv a_{k}^2-a_{k-1}^2\pmod n.\] Proposed by usjl

Let $n$ be a positive odd integer. $C$ is a set consists of integral points on a plane, which is defined by \[ C = \{(i, j): i, j = 0, 1, \dots, 2n-1\} \]and forms a $2n \times 2n$ array. On every point there is a Guinea pig, which is facing toward one of the following directions: positive/negative $x$-axis, or positive/negative $y$-axis. Jeff wants to keep $n^2+1$ of the Guinea pigs on the plane and remove all the others. After that, the Guinea pigs on the plane will move as the following: 1. In every round, the Guinea pigs move toward by an unit, and keep facing the same direction. 2. If a Guinea pig move to a point $(i, j)$ which is not in $C$, it will further move to another point $(p, q)$ in $C$, such that $p \equiv i \pmod {2n}$ and $q \equiv j \pmod {2n}$. (For example, if a Guinea pig move from $(2, 0)$ to $(2, -1)$, it will then further move to $(2, 2n-1)$.) The next round begins after all the Guinea pigs settle up. Jeff's goal is to keep the appropriate Guinea pigs on the plane, so that in every single round, any two Guinea pigs will never move to the same endpoint, and will never move to the startpoints(in that round) of each other simultaneously. Prove that Jeff can always succeed wherever the Guinea pigs initially face. Proposed by Weijiun Kao Edit: By the way, it can be proven that the number $n^2+1$ is optimal, i.e. if the Guinea pigs face appropriately, Jeff can only keep at most $n^2+1$ of them on the plane to avoid any collision.

Let $I$ be the incenter of triangle $ABC$ and let $D$ the foot of altitude from $I$ to $BC$. Suppose the reflection point $D’$ of $D$ with respect to $I$ satisfying $\overline{AD’} = \overline{ID’}$. Let $\Gamma$ be the circle centered at $D’$ that passing through $A$ and $I$, and let $X$, $Y\neq A$ be the intersection of $\Gamma$ and $AB$, $AC$, respectively. Suppose $Z$ is a point on $\Gamma$ so that $AZ$ is perpendicular to $BC$. Prove that $AD$, $D’Z$, $XY$ concurrent at a point.

Let $n$ be a given positive integer. Alice and Bob play a game. In the beginning, Alice determines an integer polynomial $P(x)$ with degree no more than $n$. Bob doesn’t know $P(x)$, and his goal is to determine whether there exists an integer $k$ such that no integer roots of $P(x) = k$ exist. In each round, Bob can choose a constant $c$. Alice will tell Bob an integer $k$, representing the number of integer $t$ such that $P(t) = c$. Bob needs to pay one dollar for each round. Find the minimum cost such that Bob can guarantee to reach his goal. Proposed by ltf0501