2016 AIME I Problems/Problem 4

Problem

A right prism with height $h$ has bases that are regular hexagons with sides of length $12$ . A vertex $A$ of the prism and its three adjacent vertices are the vertices of a triangular pyramid. The dihedral angle (the angle between the two planes) formed by the face of the pyramid that lies in a base of the prism and the face of the pyramid that does not contain $A$ measures $60$ degrees. Find $h^2$ .

Solution

Let $B$ and $C$ be the vertices adjacent to $A$ on the same base as $A$ , and let $D$ be the last vertex of the triangular pyramid. Then $\angle CAB = 120^\circ$ . Let $X$ be the foot of the altitude from $A$ to $\overline{BC}$ . Then since $\triangle ABX$ is a $30-60-90$ triangle, $AX = 6$ . Since the dihedral angle between $\triangle ABC$ and $\triangle BCD$ is $60^\circ$ , $\triangle AXD$ is a $30-60-90$ triangle and $AD = 6\sqrt{3} = h$ . Thus $h^2 = \boxed{108}$ .

(Solution by gundraja)

2016 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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2016 AIME I Problems/Problem 4

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