2011 AMC 12A Problem 15

Attempt Problem 15 of the 2011 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 12A solutions, or check the answer key.

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15.

The circular base of a hemisphere of radius 22 rests on the base of a square pyramid of height 6.6. The hemisphere is tangent to the other four faces of the pyramid. What is the edge-length of the base of the pyramid?

323\sqrt{2}

133\dfrac{13}{3}

424\sqrt{2}

66

132\dfrac{13}{2}

Answer: A
Concepts:3D geometrypyramiddistance formula
Difficulty rating: 1870
Solution:

Let the base have side s,s, centered at the origin, with apex at height 6.6. Cut with the vertical plane through the apex and the midpoints of two opposite base edges. The slant face appears as the line from (s2,0)\left(\tfrac{s}{2}, 0\right) to (0,6).(0, 6).

This line is 2sx+16y=1.\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 2:2: 14s2+136=2. \dfrac{1}{\sqrt{\tfrac{4}{s^2} + \tfrac{1}{36}}} = 2.

Then 4s2+136=14,\tfrac{4}{s^2} + \tfrac{1}{36} = \tfrac14, so 4s2=29\tfrac{4}{s^2} = \tfrac{2}{9} and s2=18,s^2 = 18, giving s=32.s = 3\sqrt2.

Thus, the correct answer is A.

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