Question

A hemisphere with radius $200$ sits on top of a horizontal circular disk with radius $200,$ and the hemisphere and disk have the same center. Let $\mathcal{T}$ be the region of points $P$ in the disk such that a sphere of radius $42$ can be placed on top of the disk at $P$ and lie completely inside the hemisphere. The area of $\mathcal{T}$ divided by the area of the disk is $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p + q.$

Answer

Answer: 79

Concepts:sphere 3D geometry circle area difference of squares

Difficulty rating: 2180

Solution:

A sphere of radius $42$ resting on the disk at $P$ has its center $42$ directly above $P.$ It lies inside the hemisphere of radius $200$ exactly when its center is within $200 - 42 = 158$ of the common center $O.$ If $d$ is the distance from $O$ to $P,$ the center of the sphere is at distance $\sqrt{d^2 + 42^2}$ from $O,$ so the condition is $d^2 + 42^2 \le 158^2.$

By difference of squares, $d^2 \le 158^2 - 42^2$ $= 116 \cdot 200$ $= 23200.$ Thus $\mathcal{T}$ is a disk of radius $\sqrt{23200},$ and the ratio of areas is $\frac{23200}{200^2} = \frac{23200}{40000} = \frac{29}{50}.$ Therefore $p + q = 29 + 50 = 79.$

2026 AIME I Problem 3

Answer: 79

Difficulty rating: 2180

Solution:

Problem 3 in Other Years