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

Revision as of 22:03, 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

Remarks

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

Plane $\mathcal{P}$ is determined by the points of tangency of the spheres, as shown in the green solid triangle. Therefore, the blue dashed line segments are the radii of the spheres.

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 $\ell.$

To assist understanding, I will finish up the remarks after class. A million thanks for no editing.

~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 9: / Line 9: @@
    <li>Let <math>\mathcal{R}</math> be the plane that is determined by the centers of the spheres, as shown in the black dashed triangle. Clearly, the side-lengths of this triangle are <math>49,49,</math> and <math>72.</math></li><p>
    <li>Plane <math>\mathcal{P}</math> is determined by the points of tangency of the spheres, as shown in the green solid triangle. Therefore, the blue dashed line segments are the radii of the spheres.</li><p>
-   <li></li><p>
+   <li>Since planes <math>\mathcal{P}</math> and <math>\mathcal{Q}</math> are reflections of each other about plane <math>\mathcal{R},</math> it follows that the three planes are concurrent to line <math>\ell.</math></li><p>
 </ol>
 <b>To assist understanding, I will finish up the remarks after class. A million thanks for no editing.</b>

2021 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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