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

m (→‎Solution)
(→‎Solution)
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== Solution ==   
 
== Solution ==   
 +
[asy]
 +
import three;
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size(7cm);
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currentprojection = orthographic(5,-1,1.5);
 +
 +
triple T(int i){ return 12*dir(90,60*i) + (0,0,6*sqrt(3)); }
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triple B(int i){ return 12*dir(90,60*i); }
 +
 +
for(int i = 0; i < 6; ++i){
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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)*"<math>A</math>",A,dir(200));
 +
label(scale(0.8)*"<math>60^\circ</math>",A/2,dir(50),rgb(1,0.4,0.1));
 +
label(scale(0.8)*"<math>h</math>",A--U,dir(0),rgb(0,0.6,1));
 +
[/asy]
 +
 
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>.
 
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>.
 +
 +
Diagram Credits: chezbgone2
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2016|n=I|num-b=3|num-a=5}}
 
{{AIME box|year=2016|n=I|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:12, 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

[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

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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