aime26-02

Proposed solution

To solve this problem, we start by understanding the geometry of the hemisphere and the disk. The hemisphere has a radius of 200, and it sits perfectly on top of a circular disk of the same radius. We need to find the region \( \mathcal{T} \) on the disk where a smaller sphere of radius 42 can fit entirely within the hemisphere when placed on top of the disk.

First, consider the hemisphere's equation in 3D coordinates. The hemisphere's equation is given by:
\[
x^2 + y^2 + z^2 = 200^2 \quad \text{for} \quad z \geq 0
\]
where the center of the hemisphere is at the origin \((0,0,0)\).

The smaller sphere of radius 42 must fit entirely within this hemisphere. Therefore, the center of this smaller sphere must be at most \(200 - 42 = 158\) units away from the origin in the vertical direction (z-axis). This is because the maximum height from the disk to the top of the hemisphere is 200, and the smaller sphere's radius is 42.

Next, we consider the projection of this region onto the disk. The disk lies in the \(xy\)-plane at \(z = 0\). The region \( \mathcal{T} \) is the set of points \((x, y)\) in the disk such that the distance from the origin to the center of the smaller sphere is at most 158 units. This describes a circle of radius 158 in the \(xy\)-plane.

The area of the disk with radius 200 is:
\[
\pi \times 200^2 = 40000\pi
\]

The area of the circle with radius 158 is:
\[
\pi \times 158^2 = 24964\pi
\]

The ratio of the area of region \( \mathcal{T} \) to the area of the disk is:
\[
\frac{24964\pi}{40000\pi} = \frac{24964}{40000}
\]

We simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 4:
\[
\frac{24964 \div 4}{40000 \div 4} = \frac{6241}{10000}
\]

Thus, the fraction \(\frac{6241}{10000}\) is already in its simplest form. The sum of the numerator and the denominator is:
\[
6241 + 10000 = 16241
\]

Therefore, the final answer is:
\[
\boxed{16241}
\]