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

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A right prism with height <math>h</math> has bases that are regular hexagons with sides of length <math>12</math>. A vertex <math>A</math> 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 <math>A</math> measures <math>60</math> degrees. Find <math>h^2</math>.
 
A right prism with height <math>h</math> has bases that are regular hexagons with sides of length <math>12</math>. A vertex <math>A</math> 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 <math>A</math> measures <math>60</math> degrees. Find <math>h^2</math>.
  
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== Solution ==
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==Diagram==
−
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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(Solution by gundraja)
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<asy>
 +
import bsp;
 +
 
 +
typedef path3[] shape;
 +
 
 +
shape operator *(transform3 T, shape p){
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shape os;
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for(path3 g:p) os.push(T*g);
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return os;
 +
}
 +
 
 +
 
 +
path3 path(triple[] T){
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path3 P;
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for(triple i:T) P=P--i;
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return P;
 +
}
 +
 
 +
void addshapes(face[] F, shape[] shp, pen drawpen=currentpen, pen fillpen=white)
 +
{
 +
for(int i=0; i < shp.length; ++i)
 +
for(int j=0; j < shp[i].length; ++j) {
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path3 g=shp[i][j];
 +
picture pic=F.push(g);
 +
if(fillpen != nullpen) filldraw(pic,project(g),fillpen, drawpen);
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else draw(pic,project(g),drawpen);
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// filldraw(pic,g,currentlight.intensity(F[F.length-1].point)*fillpen, drawpen);
 +
}
 +
}
 +
 
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shape cylinder(real R=1, real H=1, int n=6){
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shape Cyl;
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triple[] CTop;
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triple[] CBot;
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for(int i=0; i <= n; ++i)
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CBot.push((R*cos(2pi*i/n), R*sin(2pi*i/n),0));
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CTop = CBot+(0,0,H);
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for(int i=0; i < n; ++i)
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Cyl.push(CBot[i]--CBot[i+1]--CTop[i+1]--CTop[i]--cycle);
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path3 P=path(CBot)--cycle;
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Cyl.push(P);
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Cyl.push(shift(H*Z)*P);
 +
 
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return Cyl;
 +
}
 +
 
 +
 
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size(10cm,0);
 +
 
 +
currentprojection=orthographic(1,1,1);
 +
 
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shape cyl1 = cylinder(R=1, H=2);
 +
 
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shape[] group={cyl1};
 +
 
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face[] hidden, visible;
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addshapes(hidden, group, drawpen=linewidth(bp));
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addshapes(visible, group, drawpen=dotted, fillpen=nullpen);
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add(hidden);
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add(visible);
 +
 
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//shipout(format="pdf");
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</asy>
 +
 
 +
~gundraja
 +
 
 +
== Solution 1 == 
 +
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>.
 +
 
 +
~gundraja
 +
 
 +
== Solution 2 ==
 +
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. Notice that we can already find some lengths. We have <math>AB=AC=12</math> (given) and <math>BC=BD=\sqrt{144+h^2}</math> by the Pythagorean Theorem. Let <math>M</math> be the midpoint of <math>BC</math>. Then, we have <math>AM=6</math> (<math>30-60-90</math>) triangles and <math>DM=\sqrt{36+h^2}</math> by the Pythagorean Theorem. Applying the Law of Cosines, since <math>\angle AMD=60^{\circ}</math>, we get <cmath>h^2=36+h^2+36-\frac12 \cdot 12 \sqrt{36+h^2} \implies h^2=\boxed{108},</cmath> as desired.
 +
 
 +
-A1001
  
 
== See also ==
 
== See also ==

Latest revision as of 21:00, 9 October 2024

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 $h^2$.

Diagram

[asy] import bsp; typedef path3[] shape; shape operator *(transform3 T, shape p){ shape os; for(path3 g:p) os.push(T*g); return os; } path3 path(triple[] T){ path3 P; for(triple i:T) P=P--i; return P; } void addshapes(face[] F, shape[] shp, pen drawpen=currentpen, pen fillpen=white) { for(int i=0; i < shp.length; ++i) for(int j=0; j < shp[i].length; ++j) { path3 g=shp[i][j]; picture pic=F.push(g); if(fillpen != nullpen) filldraw(pic,project(g),fillpen, drawpen); else draw(pic,project(g),drawpen); // filldraw(pic,g,currentlight.intensity(F[F.length-1].point)*fillpen, drawpen); } } shape cylinder(real R=1, real H=1, int n=6){ shape Cyl; triple[] CTop; triple[] CBot; for(int i=0; i <= n; ++i) CBot.push((R*cos(2pi*i/n), R*sin(2pi*i/n),0)); CTop = CBot+(0,0,H); for(int i=0; i < n; ++i) Cyl.push(CBot[i]--CBot[i+1]--CTop[i+1]--CTop[i]--cycle); path3 P=path(CBot)--cycle; Cyl.push(P); Cyl.push(shift(H*Z)*P); return Cyl; } size(10cm,0); currentprojection=orthographic(1,1,1); shape cyl1 = cylinder(R=1, H=2); shape[] group={cyl1}; face[] hidden, visible; addshapes(hidden, group, drawpen=linewidth(bp)); addshapes(visible, group, drawpen=dotted, fillpen=nullpen); add(hidden); add(visible); //shipout(format="pdf"); [/asy]

~gundraja

Solution 1

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 $h^2 = \boxed{108}$.

~gundraja

Solution 2

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. Notice that we can already find some lengths. We have $AB=AC=12$ (given) and $BC=BD=\sqrt{144+h^2}$ by the Pythagorean Theorem. Let $M$ be the midpoint of $BC$. Then, we have $AM=6$ ($30-60-90$) triangles and $DM=\sqrt{36+h^2}$ by the Pythagorean Theorem. Applying the Law of Cosines, since $\angle AMD=60^{\circ}$, we get \[h^2=36+h^2+36-\frac12 \cdot 12 \sqrt{36+h^2} \implies h^2=\boxed{108},\] as desired.

-A1001

See also

2016 AIME I (Problems • Answer Key • Resources)
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Problem 3 		Followed by
Problem 5
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All AIME Problems and Solutions

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