Maths Olympiad Prep

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Geometry Difficulty 5.5 AIME, harder Prove it Romania

Let ABCABCABCA'B'C' be a triangular regular prism, with lateral edges AAAA', BBBB', CCCC'. Consider the midpoint DD of the edge BCBC and the parallelogram ADBEADB'E. Let FF be the orthogonal projection of the point AA' on the line AEAE, dd be the intersection of the planes (ADE)(ADE) and (ACF)(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=AA2AB = 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 PFDEPF \parallel DE, this is equivalent to AF=2FEAF = 2FE. Because AABBAA' \parallel BB', BDEABD \parallel EA' and BBBDBB' \perp BD, the triangle AEAAEA' has a right angle at AA'. Then AEA'E and AAA'A are legs of this triangle, therefore AF=2FE    AFAE=2EFAE    AA2=2AE2    AA2=12BC2    BC=AA2AF = 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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