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

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==Problem==
 
==Problem==
−
A right prism with height <math>h</math> has bases that are regular hexagons with sides of length 12. 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 ==
+
==Diagram==
−
[asy]
 
−
import three;
 
−
size(7cm);
 
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currentprojection = orthographic(5,-1,1.5);
 
  
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triple T(int i){ return 12*dir(90,60*i) + (0,0,6*sqrt(3)); }
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<asy>
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triple B(int i){ return 12*dir(90,60*i); }
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import bsp;
  
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for(int i = 0; i < 6; ++i){
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typedef path3[] shape;
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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));
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shape operator *(transform3 T, shape p){
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draw(T(i)--B(i),1 < i && i < 4 ? linetype("4 4") : i == 5 ? rgb(0,0.6,1) : black);
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shape os;
 +
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) {
 +
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)
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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);
 +
 
 +
path3 P=path(CBot)--cycle;
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Cyl.push(P);
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Cyl.push(shift(H*Z)*P);
 +
 
 +
return Cyl;
 
}
 
}
  
−
triple A = B(5), U = T(5), B = B(4), F = B(6);
 
  
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draw(B--U--F--cycle,rgb(0,0.6,1));
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size(10cm,0);
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draw(arc(A/2,2,30,300,90,300),rgb(1,0.4,0.1));
+
 
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draw(U--A/2--A,rgb(1,0.4,0.1));
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currentprojection=orthographic(1,1,1);
 +
 
 +
shape cyl1 = cylinder(R=1, H=2);
 +
 
 +
shape[] group={cyl1};
 +
 
 +
face[] hidden, visible;
 +
addshapes(hidden, group, drawpen=linewidth(bp));
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addshapes(visible, group, drawpen=dotted, fillpen=nullpen);
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add(hidden);
 +
add(visible);
 +
 
 +
//shipout(format="pdf");
 +
</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>.
  
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dot(A);
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~gundraja
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label(scale(0.8)*"<math>A</math>",A,dir(200));
 
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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>.
+
== 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.
  
−
Diagram Credits: chezbgone2
+
-A1001
  
 
== 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}}
 +
[[Category:Introductory Geometry Problems]]
 +
[[Category:3D Geometry Problems]]

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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