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Serbian Mathematical Olympiad

Geometry Difficulty 5.4 AIME, harder Prove it Serbia

Given is a cyclic quadrilateral $ABCD$ . Points $M, N, P$ and $Q$ are the midpoints of sides $DA, AB, BC$ and $CD$ , respectively, and point $E$ is the intersection of diagonals $AC$ and $BD$ . The circles circumscribed about $\triangle EMN$ and $\triangle EPQ$ intersect at a point $F \neq E$ . Prove that $EF \perp AC$ holds.

Solution

Solution:

Triangles $EAB$ and $EDC$ are similar, and so are triangles $EBN$ and $ECQ$ . Therefore, in directed angles it holds that $\text{MFE= MNE= BEN= QEC= EQM}$ . Analogously it holds that $\text{QFE= EMQ}$ , from which it follows that $F$ is the orthocenter of triangle $EMQ$ . Hence, $EF \perp QM \| AC$ .

Second solution. Consider the translation $\mathcal{T}$ by the vector $\frac{1}{2} \overrightarrow{AC}$ . It holds that $\mathcal{T}(M)=$

$Q$ and $\mathcal{T}(N)=P$ ; denote $\mathcal{T}(E)=E^{\prime}$ . From the similarity of triangles $AED$ and $BEC$ it follows that $\triangle AEM \sim \triangle BEP$ , whence $\text{QE E= EMQ= MEA= BEP= QPE}$ , so point $E^{\prime}$ lies on the circle $PEQ$ . Therefore, the translation $\mathcal{T}$ maps the circle $MEN$ into the circle $PEQ$ , so the line joining their centers is parallel to $AC$ , from which the claim follows.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from sr; metadata (topic, difficulty) added by this project.