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

Revision as of 05:47, 3 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 $m + n$ .

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 $72.$ Note that the four black dashed line segments all lie in plane $\mathcal{R}.$

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. Note that all green solid line segments lie in plane $\mathcal{P}.$

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 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.

Diagram in progress.

As shown above, 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 $T_3A.$

Diagram in progress.

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

~MRENTHUSIASM

@@ Line 7: / Line 7: @@
 <u><b>Remarks</b></u>
 <ol style="margin-left: 1.5em;">
-   <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> Note that all black dashed line segments lie in plane <math>\mathcal{R}.</math></li><p>
+   <li>Let <math>\mathcal{R}</math> 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 <math>49,49,</math> and <math>72.</math> Note that the four black dashed line segments all lie in plane <math>\mathcal{R}.</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. Note that all green solid line segments lie in plane <math>\mathcal{P}.</math></li><p>
    <li>By symmetry, 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> So, the red point (the foot of the perpendicular from the smallest sphere's center to line <math>\ell</math>) lies in all three planes.</li><p>

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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