Difference between revisions of "2016 AIME I Problems/Problem 4"

Revision as of 17:33, 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 $h^2 = \boxed{108}$ .

Solution by gundraja

@@ Line 6: / Line 6: @@
 Solution by gundraja
+== See also ==
+{{AIME box|year=2016|n=I|num-b=3|num-a=5}}
+{{MAA Notice}}

2016 AIME I (Problems • Answer Key • Resources)
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Difference between revisions of "2016 AIME I Problems/Problem 4"

Revision as of 17:33, 4 March 2016

Problem

Solution

See also