Let $\omega_1$ and $\omega_2$ with center $O_1$ and $O_2$ respectively, meet at points $A$ and $B$. Let $X$ and $Y$ be points on $\omega_1$. Lines $XA$ and $Y A$ meet $\omega_2$ at $Z$ and $W$, respectively, such that $A$ lies between $X$ and $Z$ and between $Y$ and $W$. Let $M$ be the midpoint of $O_1O_2$, $S$ be the midpoint of $XA$ and $T$ be the midpoint of $W A$. Prove that $MS = MT$ if and only if $X,~ Y ,~ Z$ and $W$ are concyclic.
Problem
Source: Saudi Arabia IMO TST Day III Problem 4
Tags: geometry unsolved, geometry
Sardor
22.07.2014 23:37
It's easy problem.Here my solution: Let $ P $ and $ Q $ are midpoints of $ YA $ and $ ZA $ ,respectively.We know that ( by projection ) $ MP =MS $ and $ MQ=MT $ . $ (i) $ Let $ X,Y,Z,W $ are concyclic , then $ P,S,T,Q $ are also concyclic $ \implies MP=MS=MT=MQ $ . $ (ii) $ .Let $ MS=MT \implies MP=MS=MT=MQ \implies P,S,T,Q $ are concyclic $ \implies X,Y,Z,W $ are concyclic.DONE!
junioragd
23.07.2014 17:52
A very easy problem: Take midpoints of AO1 and AO2 be P and Q. 1)If they are concyclic then <XYA=<WZA,from which we have <XO1A=AO2W,and from this we have <SPM=<MQA,and from we have PS=MQ and PM=QT and the triangles SPM and MQT are congruent by SAS,and we are finished 2)We have that the triangles are congruent by SSS,and the rest is reversed angle chasing
infiniteturtle
24.07.2014 00:58
What a strange problem; certainly the other geos are harder than this but it's the last on Day 3... Let the midpoint of $YA$ be $E$, midpoint of $ZA$ is $D$. Now note $XYZW$ cyclic $\iff STDE$ cyclic by PoP. So we want $MS=MT\implies MS=MT=MD=ME$. (The other direction is obvious.) But obviously $MS=MD,MT=ME$ always, so the problem is solved.
Zezohabibullah
23.02.2015 21:25
By Apollonius': $4MS^{2}=2SO_{1}^{2}+2SO_{2}^{2}-O_{1}O_{2}^{2}$ also since $O_{1}S$ is perpendicular to $AX$, hence $SO_{1}^{2}=O_{1}A^{2}-AS^{2}$ and by power of point: $SO_{2}^{2}=SA\cdot SZ+O_{2}A^{2}=SA^{2}+SA\cdot AZ+O_{2}A^{2}$ thus, $4MS^{2}=2O_{1}A^{2}+2O_{2}A^{2}-O_{1}O_{2}^{2}+XA\cdot AZ$ Similarly: $4MT^{2}=2O_{1}A^{2}+2O_{2}A^{2}-O_{1}O_{2}^{2}+YA\cdot AW$ $\Rightarrow 4(MS^{2}-MT^{2})=XA\cdot AZ-YA\cdot AW$ and the problem solved