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

[asy] import three; size(7cm); currentprojection = orthographic(5,-1,1.5);

triple T(int i){ return 12*dir(90,60*i) + (0,0,6*sqrt(3)); } triple B(int i){ return 12*dir(90,60*i); }

for(int i = 0; i < 6; ++i){ draw(B(i)--B(i + 1),0 < i && i < 6 ? i == 4 || i == 5 ? rgb(0,0.6,1) : linetype("4 4") : black); draw(T(i)--T(i + 1)); draw(T(i)--B(i),1 < i && i < 4 ? linetype("4 4") : i == 5 ? rgb(0,0.6,1) : black); }

triple A = B(5), U = T(5), B = B(4), F = B(6);

draw(B--U--F--cycle,rgb(0,0.6,1)); draw(arc(A/2,2,30,300,90,300),rgb(1,0.4,0.1)); draw(U--A/2--A,rgb(1,0.4,0.1));

dot(A); label(scale(0.8)*" $A$ ",A,dir(200)); label(scale(0.8)*" $60^\circ$ ",A/2,dir(50),rgb(1,0.4,0.1)); label(scale(0.8)*" $h$ ",A--U,dir(0),rgb(0,0.6,1)); [/asy]

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 $\angle CAB = 120^\circ$ . 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 $h^2 = \boxed{108}$ .

Diagram Credits: chezbgone2

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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