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AMC 12 Mock Exam 5 Problem 14

Problem 14 of 25

A pyramid has a square base, and all eight of its edges have length 66. A sphere inside the pyramid is tangent to the base and to each of the four triangular faces. What is the radius of the sphere?

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Solution

By symmetry the center of the sphere lies on the pyramid's axis. The half-diagonal of the base is 323\sqrt2, so the height is 36−18=32\sqrt{36-18}=3\sqrt2. Each lateral face is equilateral with side 66, so its altitude (the slant height) is 333\sqrt3.

Cut the pyramid by the plane through the apex and the midpoints of two opposite base edges. The cross-section is an isosceles triangle with base 66, legs 333\sqrt3, and height 323\sqrt2, and the sphere meets this plane in a great circle tangent to all three sides of this triangle. That circle is the incircle of the cross-section, so r=areasemiperimeter=12⋅6⋅323+33=321+3 (=36−322≈1.55).r=\frac{\text{area}}{\text{semiperimeter}}=\frac{\tfrac12\cdot6\cdot3\sqrt2}{3+3\sqrt3}=\boxed{\frac{3\sqrt2}{1+\sqrt3}}\ \left(=\frac{3\sqrt6-3\sqrt2}{2}\approx1.55\right).

Tagged: solid geometry: sphere inscribed in a square pyramid