Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2011 AMC 12B Problems/Problem 18"
(text change; major formatting fixes)
Line 1: Line 1:
 +
==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?
  
−
(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>
+
<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>
  
−
 
+
==Solution==
−
----
 
−
 
 
−
 
 
−
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>.
Line 23: Line 20:
 
Thus, the height of the cube plus the height of the smaller pyramid equals the height of the larger pyramid.
 
Thus, the height of the cube plus the height of the smaller pyramid equals the height of the larger pyramid.
  
−
<math>x</math> + <math>\frac{x\sqrt{2}}{2}</math> = <math>\frac{\sqrt2}{2}</math>.
+
<math>x +\frac{x\sqrt{2}}{2} = \frac{\sqrt2}{2}</math>.
  
−
<math>x\left(1+\frac{\sqrt{2}}{2} \right)</math> = <math>\frac{\sqrt{2}}{2}</math>
+
<math>x\left(1+\frac{\sqrt{2}}{2} \right) =\frac{\sqrt{2}}{2}</math>
  
−
<math>x\left(2+\sqrt{2}\right)</math> = <math>\sqrt{2}</math>
+
<math>x\left(2+\sqrt{2}\right) = \sqrt{2}</math>
  
−
<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.
+
<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.
  
−
<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>
+
<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>
 +
==See Also==
 +
{{AMC12 box|year=2011|num-b=17|num-a=19|ab=B}}

Revision as of 18:01, 29 May 2011

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

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 triangle 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 $\frac{\sqrt{3}}{2}$.

<geogebra>b1c2e62d3f53df283ef11bf819f89f07e650057e</geogebra>

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 $x$, the pyramid has side length $\frac{x\sqrt{2}}{2}$.

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

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

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

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

$x = \frac{\sqrt{2}}{2+\sqrt{2}} \cdot \frac{2-\sqrt{2}}{2-\sqrt{2}} = \frac{2\sqrt{2}-2}{4-2} = \sqrt{2}-1} =$ (Error compiling LaTeX. Unknown error_msg)side length of cube.

$\left(\sqrt{2}-1}\right)^3 = (\sqrt{2})^3 + 3(\sqrt{2})^2(-1) + 3(\sqrt{2})(-1)^2 + (-1)^3 = 2\sqrt{2} - 6 +3\sqrt{2} - 1 =\textbf{(A)}\ 5\sqrt{2} - 7$ (Error compiling LaTeX. Unknown error_msg)

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems/Problem_18&oldid=38779"