2019 AMC 12A Problem 18

Attempt Problem 18 of the 2019 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 2019 AMC 12A solutions, or check the answer key.

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18.

A sphere with center OO has radius 6.6. A triangle with sides of length 15,15,15, 15, and 2424 is situated in space so that each of its sides is tangent to the sphere. What is the distance between OO and the plane determined by the triangle?

232\sqrt{3}

44

323\sqrt{2}

252\sqrt{5}

55

Answer: D
Concepts:sphereincircle, incenter, and inradiusPythagorean Theorem
Difficulty rating: 1910
Solution:

The sphere intersects the triangle's plane in a circle of radius 36d2,\sqrt{36 - d^2}, where dd is the distance from OO to the plane. Since each side is tangent to the sphere, this circle is the triangle's incircle.

The triangle has area 12249=108\tfrac{1}{2} \cdot 24 \cdot 9 = 108 and semiperimeter 27,27, so its inradius is 10827=4.\dfrac{108}{27} = 4.

Thus 36d2=4,\sqrt{36 - d^2} = 4, giving d2=20d^2 = 20 and d=25.d = 2\sqrt{5}.

Thus, the correct answer is D.

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