Track / Stage 4 / 297 of 340 #1197 of 2604

Problem 1197

AMC 12 late, AIME early

Geometry Difficulty 4.9 Prove it Romanian Mathematical Olympiad · 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 $4AM = 3AC$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official 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.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.