2016 AIME I Problems/Problem 4

Revision as of 18:06, 4 March 2016 by Gundraja (talk | contribs)

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 other vertex of the triangular pyramid. Then $\angle CAB = 120^\circ$ so $[ABC] = \frac{1}{2} \cdot AB \cdot AC \cdot \text{sin}(120^\circ) = 36\sqrt{2}$. Let $D$ be the foot of the altitude from $B$ to $AC$.