2011 AMC 12B Problems/Problem 18
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?
Solution
We can use the Pythagorean Theorem to split one of the triangular faces into two 30-60-90 triangles with side lengths 
and 
.
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 .
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 
and .
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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 
, the pyramid has side length 
.
Thus, the height of the cube plus the height of the smaller pyramid equals the height of the larger pyramid.
.
$x = \frac{\sqrt{2}}{2+\sqrt{2}} \cdot \frac{2-\sqrt{2}}{2-\sqrt{2}} = \frac{2\sqrt{2}-2}{4-2} = \sqrt{2}-1} =$ (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 | |
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