2019 AMC 10A Problem 21

The plane of the triangle cuts the sphere in a circle. Each side of the triangle is tangent to the sphere and lies in the plane, so each side is tangent to this circle. A circle in the triangle's plane tangent to all three sides is the incircle, so the cross-section is the incircle and its center is the foot of the perpendicular from $O$ to the plane.

Compute the inradius. The isosceles triangle with sides $15, 15, 24$ has altitude to the base $\sqrt{15^2 - 12^2} = 9$, so its area is $\frac12 \cdot 24 \cdot 9 = 108$. The semiperimeter is $\frac{15 + 15 + 24}{2} = 27$, so
$$r = \frac{\text{Area}}{s} = \frac{108}{27} = 4.$$

Now let $I$ be the incenter and $P$ a point of tangency on a side. Then $OP = 6$ (radius of the sphere), $IP = 4$ (radius of the cross-section), and $OI \perp IP$. By the Pythagorean theorem,
$$OI = \sqrt{6^2 - 4^2} = \sqrt{20} = 2\sqrt5.$$

The answer is $\boxed{\textbf{(D)}\ 2\sqrt5}$.

Reduce 3D to 2D by slicing: a sphere meets a plane in a circle whose radius, the sphere's radius, and the distance to the plane form a right triangle. "Tangent to all three sides" identifies that circle as the incircle, and $r = \text{Area}/s$ finds its radius quickly for a triangle whose area is easy. Recognizing the $9$-$12$-$15$ right triangle inside the isosceles triangle saves Heron's formula.

• Using the circumradius or the altitude 9 instead of the inradius, or forgetting that the cross-section circle is smaller than the sphere.
• Answering the inradius $4$ (choice B) instead of finishing with the Pythagorean step, or computing $\sqrt{6^2 + 4^2}$.