2021 AIME II Problems/Problem 10

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 $m + n$ .

Diagram

File:2021 AIME II Problem 10 Diagram.png

Remarks

Let $\mathcal{R}$ be the plane that is determined by the centers of the spheres, as shown in the black points. Clearly, the side-lengths of the black dashed triangle are $49,49,$ and $72.$

Plane $\mathcal{P}$ is tangent to the spheres at the green points. Therefore, the blue dashed line segments are the radii of the spheres.

By symmetry, since planes $\mathcal{P}$ and $\mathcal{Q}$ are reflections of each other about plane $\mathcal{R},$ it follows that the three planes are concurrent to line $\ell.$ So, the four black dashed line segments all lie in plane $\mathcal{R};$ the four green solid line segments all lie in plane $\mathcal{P};$ the red point (the foot of the perpendicular from the smallest sphere's center to line $\ell$ ) lies in all three planes.

~MRENTHUSIASM (by Geometry Expressions)

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 (Coord Bash)

Let's try to see some symmetry. We can use a coordinate plane to plot where the circles are. The 2 large spheres are externally tangent, so we'll make them at 0, -36, 0 and 0, 36, 0. The center of the little sphere would be x, 0, and -23 since we don't know how much the little sphere will be "pushed" down. We use the 3D distance formula to find that x is -24 (since 24 wouldn't make sense). Now, we draw a line through the little sphere and the origin. It also intersects $\ell$ because of the symmetry we created.

$\ell$ lies on the plane too, so these 2 lines must intersect. The point at where it intersects is -24a, 0, and 23a. We can use the distance formula again to find that a = $\dfrac{36}{23}$ . Therefore, they intersect at $\left(-\dfrac{864}{23},0,-36\right)$ . Since the little circle's x coordinate is -24 and the intersection point's x coordinate is $\dfrac{864}{23}$ , we get $\dfrac{864}{23}$ - 24 = $\dfrac{312}{23}$ . Therefore, our answer to this problem is 312 + 23 = $\boxed{335}$ .

~Arcticturn

Solution 3 (Illustration of Solution 1)

This solution refers to the Diagram section.

As shown below, let $O_1,O_2,O_3$ be the centers of the spheres (where sphere $O_3$ is the smallest) and $T_1,T_2,T_3$ be their respective points of tangency to plane $\mathcal{P}.$ Suppose $A$ is the foot of the perpendicular from $O_3$ to line $\ell.$ We wish to find $T_3A.$

File:2021 AIME II Problem 10 Solution 1.png

Diagram in progress.

Solution in progress. A million thanks for not editing it. I will finish within today.

~MRENTHUSIASM

2021 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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