AIME 2026 I · 第 3 题

AIME 2026 I — Problem 3

专题: Contest Math
难度: L4
来源: AIME

题目详情

Problem

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 $p+q$ .

解析

Solution 1

$\mathcal T$ will be the shape of a circle by symmetry. To find the radius, imagine the case in which the sphere is in contact with the very edge of $\mathcal T$ . Let the center of the sphere be $S$ , and the center of the hemisphere be $H$ . Imagine the vertical plane containing $S$ and $H$ .

On the plane, a right triangle exists with vertices $S$ , $H$ , and $P$ (the point of tangency between sphere and disk). We know $SH$ is $200 - 42 = 158$ , and $SP$ is $42$ , the radius of the sphere.

Now we can use the Pythagorean theorem to get the radius of $\mathcal T$ as $2\sqrt{5800}$ . Finding the area of $\mathcal T$ and dividing by the area of the disk will give you a final answer of $\frac{29}{50}$ , or $\boxed{079}$ .

Note: Using the Pythagorean theorem on big numbers can be tricky, so we can scale down the triangle by a factor of two and rescale it when we find the third side.

~ Logibyte

Solution 2

By symmetry, $\mathcal T$ is a disk. Consider the boundary case where the sphere of radius $42$ is internally tangent to the hemisphere. Let $H$ be the center of the hemisphere, $A$ the point of tangency, and let line $HA$ intersect the small sphere again at $M$ . Let $P$ be the point on the disk directly below the center of the small sphere.

Since the hemisphere has radius $200$ , $HA=200$ . The diameter of the small sphere is $84$ , so $HM=200-84=116$ . By Power of a Point at $H$ , $HP^2=HA\cdot HM=200\cdot116.$

Thus the radius $r$ of $\mathcal T$ satisfies $r^2=200\cdot116$ .

Our desired ratio is

$\frac{\pi r^2}{\pi(200)^2}=\frac{200\cdot116}{200^2}=\frac{29}{50}$ . Therefore, $p+q=\boxed{79}$ .

~ MathKing555

Video Solution (Fast and Easy 🔥🚀)

2026 AIME I #3

By piacademyus.org

Video Solution

https://www.youtube.com/watch?v=YtrhwNWhFic - Continuum Math