2019 AMC 12A Problem 18

Question

A sphere with center

O

has radius

6.

A triangle with sides of length

15,

15,

and

24

is situated in space so that each of its sides is tangent to the sphere. What is the distance between

O

and the plane determined by the triangle?

2\sqrt{3}

4

3\sqrt{2}

2\sqrt{5}

5

The sphere meets the plane in a circle tangent to all three sides, so that circle is the incircle

If

d

is the distance, the cross-circle has radius

\sqrt{36 - d^2},

which equals the inradius

The sphere intersects the triangle’s plane in a circle of radius

\sqrt{36 - d^2},

where

d

is the distance from

O

to the plane. Since each side is tangent to the sphere, this circle is the triangle’s incircle.

The triangle has area

\tfrac{1}{2} \cdot 24 \cdot 9 = 108

and semiperimeter

27,

so its inradius is

\dfrac{108}{27} = 4.

Thus

\sqrt{36 - d^2} = 4,

giving

d^2 = 20

and

d = 2\sqrt{5}.

Thus, the correct answer is D.

Problem 18 in other years