2026 AIME I Problem 3

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

Small hint:

The small sphere’s center sits

42

directly above

P,

and the sphere fits inside the hemisphere exactly when its center is within

200 - 42

of the common center

Big hint:

With

d

the distance from the center of the disk to

P,

the condition is

d^2 + 42^2 \le 158^2,

and

158^2 - 42^2

factors as a difference of squares

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.