Let $M$ be the intersection point of the medians $AD$ and $BE$ of a right triangle $ABC$ ($\angle C=90^\circ$). It is known that the circumcircles of triangles $AEM$ and $CDM$ are tangent. Find the angle $\angle BMC.$
Problem
Source: Kyiv mathematical festival 2018
Tags: Kyiv mathematical festival, geometry, tangent circles
sunken rock
07.05.2018 08:50
Let $O', O"$ be the circumcenters of the 2 triangles respectively. We have $\angle EMO'+\angle CMO"=\angle EAM+\angle MDC=90^\circ$. Best regards, sunken rock
mathafou
07.05.2018 12:21
I don't understand how you get that
the two circles being tangent at M, M is an homothetic center of these circles,
hence AC' // DC, etc
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Gluncho
07.05.2018 12:50
mathafou wrote: I don't understand how you get that
the two circles being tangent at M, M is an homothetic center of these circles,
hence AC' // DC, etc
his solution is correct