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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 <math>B</math> and <math>C</math> be the vertices adjacent to <math>A</math> on the same base as <math>A</math>, and let <math>D</math> 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>A</math> to <math>\overline{BC}</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>.
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Let <math>B</math> and <math>C</math> be the vertices adjacent to <math>A</math> on the same base as <math>A</math>, and let <math>D</math> be the last vertex of the triangular pyramid.  Then <math>\angle CAB = 120^\circ</math>. Let <math>X</math> be the foot of the altitude from <math>A</math> to <math>\overline{BC}</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>.
  
 
(Solution by gundraja)
 
(Solution by gundraja)

Revision as of 18:28, 1 March 2017

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)

See also

2016 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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