2021 AIME II Problems/Problem 10

Revision as of 16:23, 28 March 2021 by Arcticturn (talk | contribs) (Solution)

Problem

Two spheres with radii $36$ and one sphere with radius $13$ are each externally tangent to the other two spheres and to two different planes $\mathcal{P}$ and $\mathcal{Q}$. The intersection of planes $\mathcal{P}$ and $\mathcal{Q}$ is the line $\ell$. The distance from line $\ell$ to the point where the sphere with radius $13$ is tangent to plane $\mathcal{P}$ is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Solution 1

The centers of the three spheres form a 49-49-72 triangle. Consider the points at which the plane is tangent to the two bigger spheres; the line segment connecting these two points should be parallel to the 72 side of this triangle. Take its midpoint $M$, which is 36 away from the midpoint of the 72 side $A$, and connect these two midpoints.

Now consider the point at which the plane is tangent to the small sphere, and connect $M$ with the small sphere's tangent point $B$. Extend $MB$ through B until it hits the ray from $A$ through the center of the small sphere (convince yourself that these two intersect). Call this intersection $D$, the center of the small sphere $C$, we want to find $BD$.

By Pythagorus AC= $\sqrt{49^2-36^2}=\sqrt{1105}$, and we know $MB=36,BC=13$. We know that $MB,BC$ must be parallel, using ratios we realize that $CD=\frac{13}{23}\sqrt{1105}$. Apply Pythagorean theorem on triangle BCD; $BD=\frac{312}{23}$, so 312 + 23 = $\boxed{335}$

-Ross Gao

Solution 2

See also

2021 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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