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 3D geometry

Difficulty rating: 2030

Solution:

Let the apex be $A$ and the base be square $BCDE.$ Then $AB=AD=1$ and $BD=\sqrt2,$ so $\triangle BAD$ is an isosceles right triangle.

Let the cube have edge length $x.$ Its intersection with the plane of $\triangle BAD$ is a rectangle of height $x$ and width $\sqrt2\,x,$ whose top corners lie on $AB$ and $AD.$ Because the legs $AB$ and $AD$ meet the base at $45^\circ,$ each portion of $BD$ outside the rectangle has length $x,$ so $\sqrt2=BD=\sqrt2\,x+2x,$ which reduces to $x=\dfrac{\sqrt2}{2+\sqrt2}=\sqrt2-1.$

The volume is $(\sqrt2-1)^3=5\sqrt2-7.$

Thus, the correct answer is A.

2011 AMC 12B Problem 18

Answer: A

Difficulty rating: 2030

Solution:

Problem 18 in Other Years