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Romanian Mathematical Olympiad - Final Round

Geometry Difficulty 5.8 AIME, harder Prove it Romania

The right prism $ABCA'B'C'$ , with $AB = AC = BC = a$ , has the property that there exists a unique point $M \in (BB')$ so that $AM \perp MC'$ . Find the measure of the angle of the straight line $AM$ and the plane $(ACC')$ .
Mircea Fianu, Cristian Mangra

Solution

We first prove that $M$ is the midpoint of the edge $[BB']$ . Indeed, if this is not the case, denote $M'$ the reflection of $M$ across the midpoint of $[BB']$ . Then $\Delta MAB \equiv \Delta M'C'B'$ and $\Delta M'AB \equiv \Delta MC'B'$ imply $[MA] \equiv [M'C']$ and $[M'A] \equiv [MC']$ . Therefore $\Delta MAC' \equiv \Delta M'C'A$ , hence $m(\angle AMC') = m(\angle AM'C') = 90^\circ$ , which contradicts the unicity of $M$ .

Denote $BB' = 2h$ . Then $AM = MC' = \sqrt{a^2 + h^2}$ , $AC' = \sqrt{a^2 + 4h^2}$ and Pythagoras' Theorem yields $a = h\sqrt{2}$ .

If $O$ is the center of the face $(ACC'A')$ , then $MO \perp (ACC')$ , hence the required angle is $MAO$ .
Since $MO = OA = a\sqrt{3}/2$ , the triangle $MOA$ is right and isosceles, so $m(\angle MAO) = 45^\circ$ .

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