IMO Math

Let $ABC$ be a triangle with circumcircle $\Omega$ and incentre $I$. Let the line passing through $I$ and perpendicular to $CI$ intersect the segment $BC$ and the arc $BC$ (not containing $A$) of $\Omega$ at points $U$ and $V$ , respectively. Let the line passing through $U$ and parallel to $AI$ intersect $AV$ at $X$, and let the line passing through $V$ and parallel to $AI$ intersect $AB$ at $Y$ . Let $W$ and $Z$ be the midpoints of $AX$ and $BC$, respectively. Prove that if the points $I, X,$ and $Y$ are collinear, then the points $I, W ,$ and $Z$ are also collinear.

A site is any point $(x, y)$ in the plane such that $x$ and $y$ are both positive integers less than or equal to 20. Initially, each of the 400 sites is unoccupied. Amy and Ben take turns placing stones with Amy going first. On her turn, Amy places a new red stone on an unoccupied site such that the distance between any two sites occupied by red stones is not equal to $\sqrt{5}$. On his turn, Ben places a new blue stone on any unoccupied site. (A site occupied by a blue stone is allowed to be at any distance from any other occupied site.) They stop as soon as a player cannot place a stone. Find the greatest $K$ such that Amy can ensure that she places at least $K$ red stones, no matter how Ben places his blue stones.

Two squirrels, Bushy and Jumpy, have collected 2021 walnuts for the winter. Jumpy numbers the walnuts from 1 through 2021, and digs 2021 little holes in a circular pattern in the ground around their favorite tree. The next morning Jumpy notices that Bushy had placed one walnut into each hole, but had paid no attention to the numbering. Unhappy, Jumpy decides to reorder the walnuts by performing a sequence of 2021 moves. In the $k$-th move, Jumpy swaps the positions of the two walnuts adjacent to walnut $k$. Prove that there exists a value of $k$ such that, on the $k$-th move, Jumpy swaps some walnuts $a$ and $b$ such that $a<k<b$.

 Hello,  I just noticed that the blog was private for a while. I fixed the problem now. Sorry for the inconvenience.

Four real numbers $ p$, $ q$, $ r$, $ s$ satisfy $ p+q+r+s = 9$ and $ p^{2}+q^{2}+r^{2}+s^{2}= 21$. Prove that there exists a permutation $ \left(a,b,c,d\right)$ of $ \left(p,q,r,s\right)$ such that $ ab-cd \geq 2$.

As I said in an earlier post, I will also post problems that are not from the IMO but I believe that they are of the same level. Let $ABC$ be a triangle with $AB<AC$ and $I$ be your incenter. Let $M$ and $N$ be the midpoints of the sides $BC$ and $AC$, respectively. If the lines $AI$ and $IN$ are perpendicular, prove that the line $AI$ is tangent to the circumcircle of $\triangle IMC$.

 From now on, I will only stream on Twitch if someone wants one-on-one detailed help on a solution they don't understand on my blog. 

This was the huge surprise I was talking about in the other post! Now, we have a site like this for Physics as well, which you can find at  thinkphysics1.blogspot.com .  I have combined these two to form my organization named Thonks, and its website can be found at  thonks123123.github.io/Thonks .  I hope you like it!

Let $n$ be a positive integer. Define a chameleon to be any sequence of $3n$ letters, with exactly $n$ occurrences of each of the letters $a, b,$ and $c$. Define a swap to be the transposition of two adjacent letters in a chameleon. Prove that for any chameleon $X$ , there exists a chameleon $Y$ such that $X$ cannot be changed to $Y$ using fewer than $3n^2/2$ swaps.

 In an acute-angled triangle $ABC$ on the sides $AB$, $BC$, $AC$ the points $H$, $L$, $K$ so that $CH \perp AB$, $HL \parallel AC$, $HK \parallel BC$. Let $P$ and $Q$ feet of altitudes of a triangle $HBL$, drawn from the vertices $H$ and $B$ respectively. Prove that the feet of the altitudes of the triangle $AKH$, drawn from the vertices $A$ and $H$ lie on the line $PQ$.

 Hi,  I'm sorry but I won't be able to stream today.  I have a little pain in neck and it hurts a bit when I talk. I will stream again next week (Friday, April 15th) at 7:00 PM

Now, we are officially spread all throughout the world! We have views from every continent (well other than Antarctica, of course).  We already had it spread in all continents other than Africa but now we have many views from Africa as well. Before, it was technically "spread" in Africa, but there were minimal views from there.  I'm really happy about this :) Thank you so much to everyone who views my blog :D I hope it continues to help you.

Hi,  I have decided that sometimes, I will pick problems that are not from the IMO shortlists but I believe that it is an IMO-level problem (in other words, I think that the problem could have appeared in the IMO as well). Because after all, there are a limited number of problems.  

 A permutation $ \{x_1, x_2, \ldots, x_{2n}\}$ of the set $ \{1,2, \ldots, 2n\}$ where $ n$ is a positive integer, is said to have property $ T$ if $ |x_i - x_{i + 1}| = n$ for at least one $ i$ in $ \{1,2, \ldots, 2n - 1\}.$ Show that, for each $ n$, there are more permutations with property $ T$ than without.

  Find all triplets of positive integers $ (a,m,n)$ such that $ a^m + 1 \mid (a + 1)^n$.