2019 AMC 10A Problem 21

Step 1: turn the 3D picture into a flat one.

1. A flat plane cuts a sphere in a circle. Look at the circle where the triangle's plane cuts the sphere.
2. Each side of the triangle lies in that plane and touches the sphere at one point. So each side touches the circle at one point.
3. The only circle inside a triangle that touches all three sides is the inscribed circle. So the cut circle is the inscribed circle of the triangle.
4. The center of that circle, call it $I$, is the point of the plane directly below $O$. So the distance we want is $OI$.

Step 2: find the radius $r$ of the inscribed circle.

5. The triangle has sides $15, 15, 24$. Drop the height from the top corner to the side of length $24$; it splits that side into $12$ and $12$. The height is $\sqrt{15^2 - 12^2} = \sqrt{81} = 9$.
6. Area: $\tfrac12 \cdot 24 \cdot 9 = 108$.
7. There is a handy formula: area $=$ $r$ $\times$ (half the perimeter). It holds because joining $I$ to the three corners cuts the triangle into three pieces, each with height $r$. Half the perimeter is $\frac{15 + 15 + 24}{2} = 27$, so
$$r = \frac{108}{27} = 4.$$

Step 3: use a right triangle.

8. Pick a point $P$ where a side touches the sphere. Then $OP = 6$ (the sphere's radius) and $IP = 4$ (the circle's radius).
9. $OI$ goes straight down to the plane, so the angle at $I$ is a right angle. By the Pythagorean theorem,
$$OI = \sqrt{6^2 - 4^2} = \sqrt{20} = 2\sqrt5.$$

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

To handle a sphere and a plane, slice: the sphere's radius, the radius of the slice circle, and the distance to the plane always make a right triangle. "Touches all three sides" tells you the slice circle is the inscribed circle, and area $=$ $r \times$ half-perimeter finds its radius fast. Spotting the $9$-$12$-$15$ right triangle inside gives the area with no heavy formula.

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