Let $M$ be the midpoint of the base $AC$ of an isosceles $\triangle ABC$. Points $E$ and $F$ on the sides $AB$ and $BC$ respectively are chosen so that $AE \neq CF$ and $\angle FMC = \angle MEF = \alpha$. Determine $\angle AEM$. (6 points) Maxim Prasolov
Problem
Source: Tournament of Towns Spring 2016
Tags: geometry
ak12sr99
22.02.2017 22:17
The answer is $\alpha$. From the given condition $AC$ is tangent to the circumcircle of $\Delta MEF$ (at $M$), so its center $O'$ lies on $BM$. Suppose this circle cuts $AB$ and $AC$ again at $K$ and $L$ respectively. In the figure let $AE=x, CL=y, BK=z, BF=w$. Now, $AE.AK=AM^2=CM^2=CL.CF$ and $BK.BE=BF.BL$ imply $x(a-z)=y(a-w)$ and $z(a-x)=w(a-y)$ which implies $\frac{x}{y}=\frac{a-w}{a-z}=\frac{x+w-a}{y+z-a}$ and $\frac{w}{z}=\frac{a-x}{a-y}=\frac{x+w-a}{y+z-a}$ whereupon $\frac{x}{y}=\frac{z}{w}=\frac{a-x-z}{a-y-w} \implies BK:KE:EA = BF:FL:LC$ so that $KFLE$ is an isoceles trapezium with parallel sides parallel to $AC$. $\angle FMC = \angle AMK = \alpha \implies \angle KEF = \angle KMF = 180^o - 2\alpha \implies \angle AEM = \alpha$ (on straight line $AEK)$. EDIT: The previous solution was incomplete.
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Severus
26.10.2018 17:37
Sorry for the bump, but there is a 2 liner solution to this one.
Construct $E'$ on $AB$ such that $AE'=CF$. Then $\angle ME'F=\alpha$. So points $E,F,E',M$ are concyclic. Since $E'M=MF$, so $\angle AEM=\angle FEM=\alpha.$ Done.
Tsikaloudakis
19.10.2021 20:49
$\begin{gathered} \left. \begin{gathered} EZ//AC \hfill \\ KZ// = AC \hfill \\ \end{gathered} \right\}\mathop {}\limits_{}^{} \Rightarrow \mathop {}\limits_{}^{} \left\{ \begin{gathered} ME = MZ \hfill \\ MK = MF \hfill \\ \end{gathered} \right.\mathop {}\limits_{}^{} \Rightarrow \mathop {}\limits_{}^{} A\hat EM = C\hat ZM\mathop {}\limits_{}^{} ,\mathop {}\limits_{}^{} Z\hat EF = Z\hat MF\mathop {}\limits_{}^{} \Rightarrow \hfill \\ E,Z,F,M\mathop {}\limits_{}^{} ,\mathop {}\limits_{}^{\mathop {}\limits_{}^{} } concyclic\mathop {}\limits_{}^{} \Rightarrow \mathop {}\limits_{}^{} M\hat ZF = M\hat EF = A\hat EM = \alpha \hfill \\ \end{gathered} $
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