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Romanian Mathematical Olympiad

Geometry Difficulty 5.5 AIME, harder Prove it Romania

Let $ABCA'B'C'$ be a triangular regular prism, with lateral edges $AA'$ , $BB'$ , $CC'$ . Consider the midpoint $D$ of the edge $BC$ and the parallelogram $ADB'E$ . Let $F$ be the orthogonal projection of the point $A'$ on the line $AE$ , $d$ be the intersection of the planes $(ADE)$ and $(A'CF)$ and $P$ be the intersection of the line $d$ with the plane $(ABC)$ . Prove that $P$ is the baricentre of the triangle $ABC$ if and only if $AB = AA'\sqrt{2}$ .
Valeriu Bărbieru

Solution

AD is a median in $\triangle ABC$ , so the point $P$ is the baricenter of the triangle $ABC$ if and only if $AP = 2PD$ . Since $PF \parallel DE$ , this is equivalent to $AF = 2FE$ . Because $AA' \parallel BB'$ , $BD \parallel EA'$ and $BB' \perp BD$ , the triangle $AEA'$ has a right angle at $A'$ . Then $A'E$ and $A'A$ are legs of this triangle, therefore $AF = 2FE \iff AF \cdot AE = 2EF \cdot AE \iff A'A^2 = 2A'E^2 \iff A'A^2 = \frac{1}{2}BC^2 \iff BC = A'A\sqrt{2}$ .

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