2011 AMC 10B Problem 22

Question

A pyramid has a square base with sides 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?

5\sqrt{2} - 7

7 - 4\sqrt{3}

\dfrac{2\sqrt{2}}{27}

\dfrac{\sqrt{2}}{9}

\dfrac{\sqrt{3}}{9}

Answer: A

Concepts:pyramid cube geometry volume

Difficulty rating: 2150

Small hint:

Use a diagonal cross-section through the pyramid and cube

Big hint:

The cube contributes a rectangle of height

x

and width

\sqrt2x

Solution:

Let the cube have side length

x

. Take a vertical diagonal cross-section of the pyramid through opposite vertices of the square base.

This cross-section is an isosceles right triangle with hypotenuse

\sqrt2

. The cube appears as a rectangle of height

x

and width

\sqrt2x

, leaving two congruent right isosceles triangles of leg

x

.

Thus

\sqrt2=\sqrt2x+2x

, so

x=\sqrt2-1

. The cube volume is

x^3=(\sqrt2-1)^3=5\sqrt2-7

.

Thus, A is the correct answer.