2011 AMC 12A Problem 15

Question

The circular base of a hemisphere of radius

2

rests on the base of a square pyramid of height

6.

The hemisphere is tangent to the other four faces of the pyramid. What is the edge-length of the base of the pyramid?

3\sqrt{2}

\dfrac{13}{3}

4\sqrt{2}

6

\dfrac{13}{2}

Take the cross-section through the apex and the midpoints of two opposite base edges

In that cross-section, the slant line runs from

\left(\tfrac{s}{2}, 0\right)

to

(0, 6),

and its distance from the center is

2

Let the base have side

s,

centered at the origin, with apex at height

6.

Cut with the vertical plane through the apex and the midpoints of two opposite base edges. The slant face appears as the line from

\left(\tfrac{s}{2}, 0\right)

to

(0, 6).

This line is

\tfrac{2}{s}x + \tfrac16 y = 1.

The hemisphere is tangent to the face, so the distance from the origin to this line is the radius

2:

\dfrac{1}{\sqrt{\tfrac{4}{s^2} + \tfrac{1}{36}}} = 2.

Then

\tfrac{4}{s^2} + \tfrac{1}{36} = \tfrac14,

so

\tfrac{4}{s^2} = \tfrac{2}{9}

and

s^2 = 18,

giving

s = 3\sqrt2.

Thus, the correct answer is A.

Problem 15 in other years