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

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This solution refers to the <b>Diagram</b> section.
 
This solution refers to the <b>Diagram</b> section.
  
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As shown below, let <math>O_1,O_2,O_3</math> be the centers of the spheres (where sphere <math>O_3</math> is the smallest) and <math>T_1,T_2,T_3</math> be their respective points of tangency to plane <math>\mathcal{P}.</math> Suppose <math>A</math> is the foot of the perpendicular from <math>O_3</math> to line <math>\ell.</math> We wish to find <math>T_3A.</math>
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As shown below, let <math>O_1,O_2,O_3</math> be the centers of the spheres (where sphere <math>O_3</math> is the smallest) and <math>T_1,T_2,T_3</math> be their respective points of tangency to plane <math>\mathcal{P}.</math> Suppose <math>A</math> is the foot of the perpendicular from <math>O_3</math> to line <math>\ell,</math> so that <math>\overleftrightarrow{O_3A}</math> is the perpendicular bisector of <math>\overline{O_1O_2}.</math> We wish to find <math>T_3A.</math>
  
 
[[File:2021 AIME II Problem 10 Solution 1.png|center]]
 
[[File:2021 AIME II Problem 10 Solution 1.png|center]]

Revision as of 20:31, 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

  1. 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 $72.$

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

  3. 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 $\ell,$ so that $\overleftrightarrow{O_3A}$ is the perpendicular bisector of $\overline{O_1O_2}.$ We wish to find $T_3A.$

File:2021 AIME II Problem 10 Solution 1.png

As planes $\mathcal{R}$ and $\mathcal{P}$ intersect at line $\ell,$ we know that both $\overrightarrow{O_1O_3}$ and $\overrightarrow{T_1T_3}$ must intersect line $\ell.$ Furthermore, since $\overline{O_1T_1}\perp\mathcal{P}$ and $\overline{O_3T_3}\perp\mathcal{P},$ it follows that $\overline{O_1T_1}\parallel\overline{O_3T_3},$ from which $O_1,O_3,T_1,$ and $T_3$ are coplanar.

We will focus on the cross-sections $O_1O_3T_3T_1$ and $\mathcal{R}:$

  1. In the three-dimensional space, the intersection of a line and a plane must be exactly one of the empty set, a point, or a line.

    Clearly, the cross-section $O_1O_3T_3T_1$ intersects line $\ell$ at one point. Let the intersection of $\overrightarrow{O_1O_3}$ and line $\ell$ be $B,$ which must also be the intersection of $\overrightarrow{T_1T_3}$ and line $\ell.$

  2. In cross-section $\mathcal{R},$ let $C$ be the foot of the perpendicular from $O_1$ to line $\ell,$ and $D$ be the foot of the perpendicular from $O_3$ to $\overline{O_1C}.$

We obtain the following diagram:

File:2021 AIME II Problem 10 Solution 2.png

In cross-section $O_1O_3T_3T_1,$ since $\overline{O_1T_1}\parallel\overline{O_3T_3}$ as discussed, we deduce that $\triangle O_3T_3B\sim\triangle O_1T_1B$ by AA, with the ratio of similitude $\frac{O_3T_3}{O_1T_1}=\frac{13}{36}.$ Therefore, we get $\frac{O_3B}{O_1B}=\frac{O_3B}{49+O_3B}=\frac{13}{36},$ or $O_3B=\frac{637}{23}.$

In cross-section $\mathcal{R},$ note that $O_1O_3=49$ and $DO_3=\frac{O_1O_2}{2}=36.$ Applying the Pythagorean Theorem to right $\triangle O_1DO_3,$ we have $O_1D=\sqrt{1105}.$ Furthermore, since $\overline{O_1C}\perp\ell$ and $\overline{O_3A}\perp\ell,$ we deduce that $\overline{O_1C}\parallel\overline{O_3A}$ and $\triangle O_1DO_3\sim\triangle O_1CB$ by AA, with the ratio of similitude $\frac{O_1O_3}{O_1B}=\frac{49}{49+\frac{637}{23}}.$ Therefore, we get $\frac{O_1D}{O_1C}=\frac{\sqrt{1105}}{\sqrt{1105}+DC}=\frac{49}{49+\frac{637}{23}},$ or $DC=\frac{13\sqrt{1105}}{23}.$

Finally, note that $\overline{O_3T_3}\perp\overline{T_3A}$ and $O_3T_3=13.$ Since $DCAO_3$ is a rectangle, we have $O_3A=DC=\frac{13\sqrt{1105}}{23}.$ Applying the Pythagorean Theorem to right $\triangle O_3T_3A$ gives $T_3A=\frac{312}{23},$ from which the answer is $312+23=\boxed{335}.$

~MRENTHUSIASM

See Also

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
All AIME Problems and Solutions

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