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?

$3\sqrt{2}$

$\dfrac{13}{3}$

$4\sqrt{2}$

$6$

$\dfrac{13}{2}$

Solution:

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.

2011 AMC 12A Problem 15

Solution:

Problem 15 in Other Years