Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.9 AIME Prove it Romania

Let $VABCD$ be a regular pyramid, having the square $ABCD$ as basis. Suppose that on the line $AC$ lies a point $M$ such that $VM = MB$ and $(VMB) \perp (VAB)$ . Prove that $4AM = 3AC$ .

Solution

Since $MV = MB = MD$ and $MO \perp (VBD)$ , it follows that $O$ is the circumcenter of the triangle $VBD$ . Furthermore, triangle $VBD$ is isosceles and right-angled, implying that the lateral faces of the pyramid are equilateral triangles.
Let $P$ be the midpoint of the edge $VB$ . The angle of the planes $(VAB)$ and $(VBM)$ is $\angle APM$ , hence $\angle APM = 90^\circ$ .
Triangles $MPA$ and $POA$ are similar, yielding $\frac{MA}{PA} = \frac{PA}{OA}$ .
Therefore $AM = \frac{PA^2}{OA} = \frac{3}{4}AC$ , as claimed.

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