Maths Olympiad Prep

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Geometry Difficulty 5.5 AIME, harder Prove it Romania

Let ABCA′B′C′ABCA'B'C' be a triangular regular prism, with lateral edges AA′AA', BB′BB', CC′CC'. Consider the midpoint DD of the edge BCBC and the parallelogram ADB′EADB'E. Let FF be the orthogonal projection of the point A′A' on the line AEAE, dd be the intersection of the planes (ADE)(ADE) and (A′CF)(A'CF) and PP be the intersection of the line dd with the plane (ABC)(ABC). Prove that PP is the baricentre of the triangle ABCABC if and only if AB=AA′2AB = AA'\sqrt{2}.
Valeriu Bărbieru
Figure 1

Solution

AD is a median in △ABC\triangle ABC, so the point PP is the baricenter of the triangle ABCABC if and only if AP=2PDAP = 2PD. Since PF∥DEPF \parallel DE, this is equivalent to AF=2FEAF = 2FE. Because AA′∥BB′AA' \parallel BB', BD∥EA′BD \parallel EA' and BB′⊥BDBB' \perp BD, the triangle AEA′AEA' has a right angle at A′A'. Then A′EA'E and A′AA'A are legs of this triangle, therefore AF=2FE  ⟺  AF⋅AE=2EF⋅AE  ⟺  A′A2=2A′E2  ⟺  A′A2=12BC2  ⟺  BC=A′A2AF = 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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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.