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Difference between revisions of "2016 AIME I Problems/Problem 4"
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== Solution ==   
 
== Solution ==   
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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 <math>\angle CAB = 120^\circ</math> so <math>[ABC] = \frac{1}{2} \cdot AB \cdot AC \cdot \text{sin}(120^\circ) = 36\sqrt{2}</math>.  Let <math>D</math> be the foot of the altitude from <math>B</math> to <math>AC</math>.
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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 <math>\angle CAB = 120^\circ</math>. Let <math>X</math> be the foot of the altitude from <math>B</math> to <math>AC</math>. Then since <math>\triangle ABX</math> is a <math>30-60-90</math> triangle, <math>AX = 6</math>.  Since the dihedral angle between <math>\triangle ABC</math> and <math>\triangle BCD</math> is <math>60^\circ</math>, <math>\triangle AXD</math> is a <math>30-60-90</math> triangle and <math>AD = 6\sqrt{3} = h</math>. Thus <math>h^2 = \boxed{108}</math>.

Revision as of 17:10, 4 March 2016

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 $\angle CAB = 120^\circ$. Let $X$ be the foot of the altitude from $B$ to $AC$. 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 $h^2 = \boxed{108}$.

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