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?

$5\sqrt{2} - 7$

$7 - 4\sqrt{3}$

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

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

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

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.

2011 AMC 10B Problem 22

Solution:

Problem 22 in Other Years