IMO Math

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

Let $\mathcal{A}$ denote the set of all polynomials in three variables $x, y, z$ with integer coefficients. Let $\mathcal{B}$ denote the subset of $\mathcal{A}$ formed by all polynomials which can be expressed as \begin{align*} (x + y + z)P(x, y, z) + (xy + yz + zx)Q(x, y, z) + xyzR(x, y, z) \end{align*}with $P, Q, R \in \mathcal{A}$. Find the smallest non-negative integer $n$ such that $x^i y^j z^k \in \mathcal{B}$ for all non-negative integers $i, j, k$ satisfying $i + j + k \geq n$.

In  this  post, Anonymous suggested that I should start a Twitch stream. Thanks for the suggestion! I will stream every Friday at 7:00 PM EST, starting March 25th. You can find my Twitch account here .  You can submit problems that you want me to solve during the stream at my email, imomath@imomath.xyz, or at my discord handle, math4l#4750.  If I get no problem requests, then I will pick a random problem to solve and explain. 

Today (March 18th) is my birthday! I was born on March 18th, 2009.  Thanks to my parents for supporting me in my journey so much :) I probably would've given up a long time back if it wasn't for them.  Now it's time for 2009 IMO SL #18, which is 2009 IMO SL #G3.  Let $ABC$ be a triangle. The incircle of $ABC$ touches the sides $AB$ and $AC$ at the points $Z$ and $Y$, respectively. Let $G$ be the point where the lines $BY$ and $CZ$ meet, and let $R$ and $S$ be points such that the two quadrilaterals $BCYR$ and $BCSZ$ are parallelogram. Prove that $GR=GS$. Let the incircle touch $\overline{BC}$ at $X$ and let the $A$-excircle $\omega_A$ touch $\overline{BC}$ at $X'$ and $\overline{AC}$ at $Y'$. Denote $\omega_R$ and $\omega_S$ as the circles centered at $R$ and $S$ respectively, both with radius $0$. Notice that $BR=CY=CX=BX'$ and $YR=BC=AY'-AY=YY'$, so we have that $\overline{BY}$ is the radical axis of $\omega_A$ and $\omega_R$. Similarly, $\overline{CZ}

We consider two sequences of real numbers $x_{1} \geq x_{2} \geq \ldots \geq x_{n}$ and $\ y_{1} \geq y_{2} \geq \ldots \geq y_{n}.$ Let $z_{1}, z_{2}, .\ldots, z_{n}$ be a permutation of the numbers $y_{1}, y_{2}, \ldots, y_{n}.$ Prove that $\sum \limits_{i=1}^{n} ( x_{i} -\ y_{i} )^{2} \leq \sum \limits_{i=1}^{n}$ $( x_{i} - z_{i})^{2}.$

I got several emails about how I haven't posted in a week, while I usually post during the first few days of the week. So don't worry, I'm still alive :D  Now let's get to the math.

Suppose that $ a_1$, $ a_2$, $ \ldots$, $ a_n$ are integers such that $ n\mid a_1 + a_2 + \ldots + a_n$. Prove that there exist two permutations $ \left(b_1,b_2,\ldots,b_n\right)$ and $ \left(c_1,c_2,\ldots,c_n\right)$ of $ \left(1,2,\ldots,n\right)$ such that for each integer $ i$ with $ 1\leq i\leq n$, we have \[ n\mid a_i - b_i - c_i \]

Let $ a$, $ b$, $ c$, $ d$ be positive real numbers such that $ abcd = 1$ and $ a + b + c + d > \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{d} + \dfrac{d}{a}$. Prove that \[ a + b + c + d < \dfrac{b}{a} + \dfrac{c}{b} + \dfrac{d}{c} + \dfrac{a}{d}\]

Let $S$ be a convex quadrilateral $ABCD$ and $O$ a point inside it. The feet of the perpendiculars from $O$ to $AB, BC, CD, DA$ are $A_1, B_1, C_1, D_1$ respectively. The feet of the perpendiculars from $O$ to the sides of $S_i$, the quadrilateral $A_iB_iC_iD_i$, are $A_{i+1}B_{i+1}C_{i+1}D_{i+1}$, where $i = 1, 2, 3.$ Prove that $S_4$ is similar to S.

Let $a_1, a_2, \ldots , a_n, \ldots  $ be a sequence of real numbers such that $0 \leq a_n \leq 1$ and $a_n - 2a_{n+1} + a_{n+2} \geq  0$ for $n = 1, 2, 3, \ldots$. Prove that \[0 \leq (n + 1)(a_n - a_{n+1}) \leq 2 \qquad \text{ for } n = 1, 2, 3, \ldots\]

Due to several emails requesting to close QOTD (people found them distracting to the main purpose of the blog), I will not be posting any more quotes. I have deleted the two quotes I posted as well. By the way, if you didn't know, you can contact me at imomath@imomath.xyz or at math4l#4750 on Discord for any questions or suggestions. 

For each positive integer $n$, the Bank of Cape Town issues coins of denomination $\frac1n$. Given a finite collection of such coins (of not necessarily different denominations) with total value at most most $99+\frac12$, prove that it is possible to split this collection into $100$ or fewer groups, such that each group has total value at most $1$.

Let $\omega$ be the circumcircle of a triangle $ABC$. Denote by $M$ and $N$ the midpoints of the sides $AB$ and $AC$, respectively, and denote by $T$ the midpoint of the arc $BC$ of $\omega$ not containing $A$. The circumcircles of the triangles $AMT$ and $ANT$ intersect the perpendicular bisectors of $AC$ and $AB$ at points $X$ and $Y$, respectively; assume that $X$ and $Y$ lie inside the triangle $ABC$. The lines $MN$ and $XY$ intersect at $K$. Prove that $KA=KT$.

Let $ p \geq 2$ be a prime number. Eduardo and Fernando play the following game making moves alternately: in each move, the current player chooses an index $i$ in the set $\{0,1,2,\ldots, p-1 \}$ that was not chosen before by either of the two players and then chooses an element $a_i$ from the set $\{0,1,2,3,4,5,6,7,8,9\}$. Eduardo has the first move. The game ends after all the indices have been chosen .Then the following number is computed: $$M=a_0+a_110+a_210^2+\cdots+a_{p-1}10^{p-1}= \sum_{i=0}^{p-1}a_i \cdot 10^i$$. The goal of Eduardo is to make $M$ divisible by $p$, and the goal of Fernando is to prevent this. Prove that Eduardo has a winning strategy. 

This blog just reached 1 million views... I'm glad to know that my solutions are helping people! Keep visiting this blog, more interesting solutions are to come.  As of now, I have no plans of closing this blog, and there will usually be 2-3 posts a week.   Also, if you have any questions or suggestions for this blog, please contact me at imomath@imomath.xyz