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

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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>.
 
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==
+
==Diagram==
 
[[File:2021 AIME II Problem 10 Diagram.png|center]]
 
[[File:2021 AIME II Problem 10 Diagram.png|center]]
  
 
<u><b>Remarks</b></u>
 
<u><b>Remarks</b></u>
 
<ol style="margin-left: 1.5em;">
 
<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></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>Plane <math>\mathcal{P}</math> is tangent to the spheres at the green points. Therefore, the blue dashed line segments are the radii of the spheres.</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>
+
   <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 four black dashed line segments all lie in plane <math>\mathcal{R};</math> the four green solid line segments all lie in plane <math>\mathcal{P};</math> 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>
 
</ol>
 
</ol>
  
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~Arcticturn
 
~Arcticturn
  
==See also==
+
==Solution 3 (Illustration of Solution 1)==
 +
This solution refers to the <b>Diagram</b> section.
 +
 
 +
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>
 +
 
 +
[[File:2021 AIME II Problem 10 Solution 1.png|center]]
 +
 
 +
As planes <math>\mathcal{R}</math> and <math>\mathcal{P}</math> intersect at line <math>\ell,</math> we know that both <math>\overrightarrow{O_1O_3}</math> and <math>\overrightarrow{T_1T_3}</math> must intersect line <math>\ell.</math> Furthermore, since <math>\overline{O_1T_1}\perp\mathcal{P}</math> and <math>\overline{O_3T_3}\perp\mathcal{P},</math> it follows that <math>\overline{O_1T_1}\parallel\overline{O_3T_3},</math> from which <math>O_1,O_3,T_1,</math> and <math>T_3</math> are coplanar.
 +
 
 +
We will focus on the cross-sections <math>O_1O_3T_3T_1</math> and <math>\mathcal{R}:</math>
 +
<ol style="margin-left: 1.5em;">
 +
  <li><i><b>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.</b></i><p>
 +
Clearly, the cross-section <math>O_1O_3T_3T_1</math> intersects line <math>\ell</math> at one point. Let the intersection of <math>\overrightarrow{O_1O_3}</math> and line <math>\ell</math> be <math>B,</math> which must also be the intersection of <math>\overrightarrow{T_1T_3}</math> and line <math>\ell.</math></li>
 +
  <li>In cross-section <math>\mathcal{R},</math> let <math>C</math> be the foot of the perpendicular from <math>O_1</math> to line <math>\ell,</math> and <math>D</math> be the foot of the perpendicular from <math>O_3</math> to <math>\overline{O_1C}.</math></li><p>
 +
</ol>
 +
We obtain the following diagram:
 +
 
 +
[[File:2021 AIME II Problem 10 Solution 2.png|center]]
 +
 
 +
In cross-section <math>O_1O_3T_3T_1,</math> since <math>\overline{O_1T_1}\parallel\overline{O_3T_3}</math> as discussed, we deduce that <math>\triangle O_3T_3B\sim\triangle O_1T_1B</math> by AA, with the ratio of similitude <math>\frac{O_3T_3}{O_1T_1}=\frac{13}{36}.</math> Therefore, we get <math>\frac{O_3B}{O_1B}=\frac{O_3B}{49+O_3B}=\frac{13}{36},</math> or <math>O_3B=\frac{637}{23}.</math>
 +
 
 +
In cross-section <math>\mathcal{R},</math> note that <math>O_1O_3=49</math> and <math>DO_3=\frac{O_1O_2}{2}=36.</math> Applying the Pythagorean Theorem to right <math>\triangle O_1DO_3,</math> we have <math>O_1D=\sqrt{1105}.</math> Furthermore, since <math>\overline{O_1C}\perp\ell</math> and <math>\overline{O_3A}\perp\ell,</math> we deduce that <math>\overline{O_1C}\parallel\overline{O_3A}</math> and <math>\triangle O_1DO_3\sim\triangle O_1CB</math> by AA, with the ratio of similitude <math>\frac{O_1O_3}{O_1B}=\frac{49}{49+\frac{637}{23}}.</math> Therefore, we get <math>\frac{O_1D}{O_1C}=\frac{\sqrt{1105}}{\sqrt{1105}+DC}=\frac{49}{49+\frac{637}{23}},</math> or <math>DC=\frac{13\sqrt{1105}}{23}.</math>
 +
 
 +
Finally, note that <math>\overline{O_3T_3}\perp\overline{T_3A}</math> and <math>O_3T_3=13.</math> Since <math>DCAO_3</math> is a rectangle, we have <math>O_3A=DC=\frac{13\sqrt{1105}}{23}.</math> Applying the Pythagorean Theorem to right <math>\triangle O_3T_3A</math> gives <math>T_3A=\frac{312}{23},</math> from which the answer is <math>312+23=\boxed{335}.</math>
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 +
~MRENTHUSIASM
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 +
==See Also==
 
{{AIME box|year=2021|n=II|num-b=9|num-a=11}}
 
{{AIME box|year=2021|n=II|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:01, 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

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 $\ell.$ We wish to find $T_3A.$

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:

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 (ProblemsAnswer KeyResources)
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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