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Showing posts from February, 2022

### 1977 IMO SL #8

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.

### 1975 IMO SL #4

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 $0 \leq (n + 1)(a_n - a_{n+1}) \leq 2 \qquad \text{ for } n = 1, 2, 3, \ldots$

### Closing QOTD

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.

### 2014 IMO SL #N3

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 $1$.

### 2013 IMO SL #G2

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 $KA=KT$.

### 2017 IMO SL #N2

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.

### 1 million views!

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

### 1992 IMO SL #6

Let $\,{\mathbb{R}}\,$ denote the set of all real numbers. Find all functions $\,f: {\mathbb{R}}\rightarrow {\mathbb{R}}\,$ such that$f\left( x^{2}+f(y)\right) =y+\left( f(x)\right) ^{2}\hspace{0.2in}\text{for all}\,x,y\in \mathbb{R}.$

### 2019 IMO SL #G2

Let $ABC$ be an acute-angled triangle and let $D, E$, and $F$ be the feet of altitudes from $A, B$, and $C$ to sides $BC, CA$, and $AB$, respectively. Denote by $\omega_B$ and $\omega_C$ the incircles of triangles $BDF$ and $CDE$, and let these circles be tangent to segments $DF$ and $DE$ at $M$ and $N$, respectively. Let line $MN$ meet circles $\omega_B$ and $\omega_C$ again at $P \ne M$ and $Q \ne N$, respectively. Prove that $MP = NQ$.