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Difference between revisions of "2011 AMC 12B Problems/Problem 18"
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==Problem==
 
A pyramid has a square base with side of length 1 and has lateral faces that are equilateral triangles. A cube is placed within the pyramid so that one face is on the base of the pyramid and its opposite face has all its edges on the lateral faces of the pyramid. What is the volume of this cube?
 
A pyramid has a square base with side of length 1 and has lateral faces that are equilateral triangles. A cube is placed within the pyramid so that one face is on the base of the pyramid and its opposite face has all its edges on the lateral faces of the pyramid. What is the volume of this cube?
  
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(A) <math>5\sqrt{2} - 7</math>  (B) <math>7 - 4\sqrt{3}</math>  (C) <math>\frac{2\sqrt{2}}{27}</math>  (D) <math>\frac{\sqrt{2}}{9}</math>  (E) <math>\frac{\sqrt{3}}{9}</math>
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<math>\textbf{(A)}\ 5\sqrt{2} - 7 \qquad \textbf{(B)}\ 7 - 4\sqrt{3} \qquad \textbf{(C)}\ \frac{2\sqrt{2}}{27} \qquad \textbf{(D)}\ \frac{\sqrt{2}}{9} \qquad \textbf{(E)}\ \frac{\sqrt{3}}{9}</math>
  
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==Solution 1==
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Forgive me if this isn't a great solution, it's my first one.
 
  
 
We can use the Pythagorean Theorem to split one of the triangular faces into two 30-60-90 triangles with side lengths <math>\frac{1}{2}, 1</math> and <math>\frac{\sqrt{3}}{2}</math>.
 
We can use the Pythagorean Theorem to split one of the triangular faces into two 30-60-90 triangles with side lengths <math>\frac{1}{2}, 1</math> and <math>\frac{\sqrt{3}}{2}</math>.
  
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Next, take a cross-section of the pyramid, forming a triangle with the top of the triangle and the midpoints of two opposite sides of the square base.
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Next, take a cross-section of the pyramid, forming a triangle with the top of the pyramid and the midpoints of two opposite sides of the square base.
  
 
This triangle is isosceles with a base of 1 and two sides of length <math>\frac{\sqrt{3}}{2}</math>.
 
This triangle is isosceles with a base of 1 and two sides of length <math>\frac{\sqrt{3}}{2}</math>.
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The height of this triangle will equal the height of the pyramid. To find this height, split the triangle into two right triangles, with sides <math>\frac{1}{2}, \frac{\sqrt2}{2}</math> and <math>\frac{\sqrt{3}}{2}</math>.
 
The height of this triangle will equal the height of the pyramid. To find this height, split the triangle into two right triangles, with sides <math>\frac{1}{2}, \frac{\sqrt2}{2}</math> and <math>\frac{\sqrt{3}}{2}</math>.
  
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<geogebra>b1c2e62d3f53df283ef11bf819f89f07e650057e</geogebra>
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The cube, touching all four triangular faces, will form a similar pyramid that sits on top of the cube. If the cube has side length <math>x</math>, the small pyramid has height <math>\frac{x\sqrt{2}}{2}</math> (because the pyramids are similar).
  
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The cube, touching all four triangular faces, will form a similar pyramid which sits on top of the cube. If the cube has side length <math>x</math>, the pyramid has side length <math>\frac{x\sqrt{2}}{2}</math>.
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Thus, the height of the cube plus the height of the smaller pyramid equals the height of the larger pyramid.
  
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Thus, the height of the cube plus the height of the smaller pyramid equals the height of the larger pyramid.
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<math>x +\frac{x\sqrt{2}}{2} = \frac{\sqrt2}{2}</math>.
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<math>x\left(1+\frac{\sqrt{2}}{2} \right) =\frac{\sqrt{2}}{2}</math>
  
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<math>x</math> + <math>\frac{x\sqrt{2}}{2}</math> = <math>\frac{\sqrt2}{2}</math>.
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<math>x\left(2+\sqrt{2}\right) = \sqrt{2}</math>
  
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<math>x\left(1+\frac{\sqrt{2}}{2} \right)</math> = <math>\frac{\sqrt{2}}{2}</math>
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<math>x = \frac{\sqrt{2}}{2+\sqrt{2}}  \cdot \frac{2-\sqrt{2}}{2-\sqrt{2}} = \frac{2\sqrt{2}-2}{4-2} = \sqrt{2}-1 = </math>side length of cube.
  
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<math>x\left(2+\sqrt{2}\right)</math> = <math>\sqrt{2}</math>
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<math>\left(\sqrt{2}-1\right)^3 = (\sqrt{2})^3 + 3(\sqrt{2})^2(-1) + 3(\sqrt{2})(-1)^2 + (-1)^3 = 2\sqrt{2} - 6 +3\sqrt{2} - 1 =\textbf{(A)} 5\sqrt{2} - 7</math>
  
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<math>x</math> = <math>\frac{\sqrt{2}}{2+\sqrt{2}}</math> <math>\cdot</math> <math>\frac{2-\sqrt{2}}{2-\sqrt{2}}</math> = <math>\frac{2\sqrt{2}-2}{4-2}</math> = <math>\sqrt{2}-1}</math> = side length of cube.
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== Solution 2 ==
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Let the side length of the square be <math>s</math>. Let the top of the pyramid be <math>A</math> and the square base be <math>BCDE</math>. Then, let the smaller cube meet edge <math>AB</math> at <math>F</math> and edge <math>AC</math> at <math>G</math>. Let the closest vertice of the cube to <math>C</math> on the base of the pyramid be <math>H</math>.
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Consider diagonal <math>CE</math>. It has length <math>\sqrt{2}</math>. Since the diagonal of the smaller cube's base is <math>s\sqrt{2}</math>, note that the distance from <math>C</math> to <math>H</math> is <math>\dfrac{\sqrt{2} - s\sqrt{2}}{2}.</math> Also, note that <math>AG = s</math> and <math>CG = 1-s</math>. We can now use the Pythagorean Theorem on triangle <math>CHG</math>, the right angle at <math>G</math>, to solve for <math>s</math>.
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<cmath>s^2 + (\dfrac{\sqrt{2} - s\sqrt{2}}{2})^2 = (1-s)^2</cmath>
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<cmath>\rightarrow s = \sqrt{2} - 1 \rightarrow \boxed{\textbf{(A)}}.</cmath>
  
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<math>\left(\sqrt{2}-1}\right)^3</math> = <math>(\sqrt{2})^3 + 3(\sqrt{2})^2(-1) + 3(\sqrt{2})(-1)^2 + (-1)^3</math> = <math>2\sqrt{2} - 6 +3\sqrt{2} - 1</math> = <math>5\sqrt{2} - 7</math>
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==See Also==
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{{AMC12 box|year=2011|num-b=17|num-a=19|ab=B}}
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{{MAA Notice}}

Latest revision as of 18:26, 7 August 2023

Problem

A pyramid has a square base with side of length 1 and has lateral faces that are equilateral triangles. A cube is placed within the pyramid so that one face is on the base of the pyramid and its opposite face has all its edges on the lateral faces of the pyramid. What is the volume of this cube?

$\textbf{(A)}\ 5\sqrt{2} - 7 \qquad \textbf{(B)}\ 7 - 4\sqrt{3} \qquad \textbf{(C)}\ \frac{2\sqrt{2}}{27} \qquad \textbf{(D)}\ \frac{\sqrt{2}}{9} \qquad \textbf{(E)}\ \frac{\sqrt{3}}{9}$

Solution 1

We can use the Pythagorean Theorem to split one of the triangular faces into two 30-60-90 triangles with side lengths $\frac{1}{2}, 1$ and $\frac{\sqrt{3}}{2}$.

Next, take a cross-section of the pyramid, forming a triangle with the top of the pyramid and the midpoints of two opposite sides of the square base.

This triangle is isosceles with a base of 1 and two sides of length $\frac{\sqrt{3}}{2}$.

The height of this triangle will equal the height of the pyramid. To find this height, split the triangle into two right triangles, with sides $\frac{1}{2}, \frac{\sqrt2}{2}$ and $\frac{\sqrt{3}}{2}$.

The cube, touching all four triangular faces, will form a similar pyramid that sits on top of the cube. If the cube has side length $x$, the small pyramid has height $\frac{x\sqrt{2}}{2}$ (because the pyramids are similar).

Thus, the height of the cube plus the height of the smaller pyramid equals the height of the larger pyramid.

$x +\frac{x\sqrt{2}}{2} = \frac{\sqrt2}{2}$.

$x\left(1+\frac{\sqrt{2}}{2} \right) =\frac{\sqrt{2}}{2}$

$x\left(2+\sqrt{2}\right) = \sqrt{2}$

$x = \frac{\sqrt{2}}{2+\sqrt{2}} \cdot \frac{2-\sqrt{2}}{2-\sqrt{2}} = \frac{2\sqrt{2}-2}{4-2} = \sqrt{2}-1 =$side length of cube.

$\left(\sqrt{2}-1\right)^3 = (\sqrt{2})^3 + 3(\sqrt{2})^2(-1) + 3(\sqrt{2})(-1)^2 + (-1)^3 = 2\sqrt{2} - 6 +3\sqrt{2} - 1 =\textbf{(A)} 5\sqrt{2} - 7$

Solution 2

Let the side length of the square be $s$. Let the top of the pyramid be $A$ and the square base be $BCDE$. Then, let the smaller cube meet edge $AB$ at $F$ and edge $AC$ at $G$. Let the closest vertice of the cube to $C$ on the base of the pyramid be $H$. Consider diagonal $CE$. It has length $\sqrt{2}$. Since the diagonal of the smaller cube's base is $s\sqrt{2}$, note that the distance from $C$ to $H$ is $\dfrac{\sqrt{2} - s\sqrt{2}}{2}.$ Also, note that $AG = s$ and $CG = 1-s$. We can now use the Pythagorean Theorem on triangle $CHG$, the right angle at $G$, to solve for $s$. \[s^2 + (\dfrac{\sqrt{2} - s\sqrt{2}}{2})^2 = (1-s)^2\] \[\rightarrow s = \sqrt{2} - 1 \rightarrow \boxed{\textbf{(A)}}.\]

See Also

2011 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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