Difference between revisions of "2021 AIME II Problems/Problem 10"

Revision as of 19:16, 2 April 2021

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 $m + n$ .

Reference Diagram

File:2021 AIME II Problem 10 Diagram.png

~MRENTHUSIASM

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

@@ Line 1: / Line 1: @@
 ==Problem==
 Two spheres with radii <math>36</math> and one sphere with radius <math>13</math> are each externally tangent to the other two spheres and to two different planes <math>\mathcal{P}</math> and <math>\mathcal{Q}</math>. The intersection of planes <math>\mathcal{P}</math> and <math>\mathcal{Q}</math> is the line <math>\ell</math>. The distance from line <math>\ell</math> to the point where the sphere with radius <math>13</math> is tangent to plane <math>\mathcal{P}</math> is <math>\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.
+==Reference Diagram==
+[[File:2021 AIME II Problem 10 Diagram.png|center]]
+~MRENTHUSIASM
 ==Solution 1==

2021 AIME II (Problems • Answer Key • Resources)
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