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

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   <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>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 tangent to the spheres at the green points. Therefore, the blue dashed line segments are the radii of the spheres.</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>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> From here, we can conclude all of the following:
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   <li>We can conclude all of the following:
 
       <ul style="list-style-type:square; margin-left: 1.5em;">
 
       <ul style="list-style-type:square; margin-left: 1.5em;">
 
         <li>The four black dashed line segments all lie in plane <math>\mathcal{R}.</math></li><p>
 
         <li>The four black dashed line segments all lie in plane <math>\mathcal{R}.</math></li><p>
 
         <li>The four green solid line segments all lie in plane <math>\mathcal{P}.</math></li><p>
 
         <li>The four green solid line segments all lie in plane <math>\mathcal{P}.</math></li><p>
         <li>The red point (the foot of the perpendicular from the smallest sphere's center to line <math>\ell</math>), as well as the other points on line <math>\ell,</math> all lie in planes <math>\mathcal{P,Q},</math> and <math>\mathcal{R}.</math></li><p>
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         <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> the three planes are concurrent to line <math>\ell.</math></li><p>
 +
        <li>The red point is the foot of the perpendicular from the smallest sphere's center to line <math>\ell.</math></li><p>
 
       </ul>
 
       </ul>
 
</li><p>
 
</li><p>
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==Solution 1==
 
==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 <math>M</math>, which is 36 away from the midpoint of the 72 side <math>A</math>, and connect these two midpoints.
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The centers of the three spheres form a <math>49</math>-<math>49</math>-<math>72</math> 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 <math>72</math> side of this triangle. Take its midpoint <math>M</math>, which is <math>36</math> away from the midpoint <math>A</math> of the <math>72</math> side, and connect these two midpoints.
  
Now consider the point at which the plane is tangent to the small sphere, and connect <math>M</math> with the small sphere's tangent point <math>B</math>. Extend <math>MB</math> through B until it hits the ray from <math>A</math> through the center of the small sphere (convince yourself that these two intersect). Call this intersection <math>D</math>, the center of the small sphere <math>C</math>, we want to find <math>BD</math>.
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Now consider the point at which the plane is tangent to the small sphere, and connect <math>M</math> with the small sphere's tangent point <math>B</math>. Extend <math>\overline{MB}</math> through <math>B</math> until it hits the ray from <math>A</math> through the center of the small sphere (convince yourself that these two intersect). Call this intersection <math>D</math>, the center of the small sphere <math>C</math>, we want to find <math>BD</math>.
  
By Pythagorus AC= <math>\sqrt{49^2-36^2}=\sqrt{1105}</math>, and we know <math>MB=36,BC=13</math>. We know that <math>MB,BC</math> must be parallel, using ratios we realize that <math>CD=\frac{13}{23}\sqrt{1105}</math>. Apply Pythagorean theorem on triangle BCD; <math>BD=\frac{312}{23}</math>, so 312 + 23 = <math>\boxed{335}</math>
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By Pythagorus, <math>AC=\sqrt{49^2-36^2}=\sqrt{1105}</math>, and we know that <math>MB=36</math> and <math>BC=13</math>. We know that <math>\overline{MB}</math> and <math>\overline{BC}</math> must be parallel, using ratios we realize that <math>CD=\frac{13}{23}\sqrt{1105}</math>. Apply the Pythagorean theorem to <math>\triangle BCD</math>, <math>BD=\frac{312}{23}</math>, so <math>312 + 23 = \boxed{335}</math>.
  
 
-Ross Gao
 
-Ross Gao
  
==Solution 2 (Coord Bash)==
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==Solution 2 (Coordinates 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 <math>\ell</math> because of the symmetry we created.  
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Let's try to see some symmetry. We can use an <math>xyz</math>-plane to plot where the circles are. The two large spheres are externally tangent, so we'll make them at <math>(0,-36,0)</math> and <math>(0,36,0)</math>. The center of the little sphere would be <math>(x,0,-23)</math> since we don't know how much the little sphere will be "pushed" down. We use the 3D distance formula to find that <math>x=-24</math> (since <math>x=24</math> wouldn't make sense). Now, we draw a line through the little sphere and the origin. It also intersects <math>\ell</math> because of the symmetry we created.  
  
<math>\ell</math> 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 = <math>\dfrac{36}{23}</math>. Therefore, they intersect at <math>\left(-\dfrac{864}{23},0,-36\right)</math>. Since the little circle's x coordinate is -24 and the intersection point's x coordinate is <math>\dfrac{864}{23}</math>, we get <math>\dfrac{864}{23}</math> - 24 = <math>\dfrac{312}{23}</math>. Therefore, our answer to this problem is 312 + 23 =  <math>\boxed{335}</math>.  
+
<math>\ell</math> lies on the plane too, so these two lines must intersect. The point at where it intersects is <math>(-24a,0,23a)</math>. We can use the distance formula again to find that <math>a=\dfrac{36}{23}</math>. Therefore, they intersect at <math>\left(-\dfrac{864}{23},0,-36\right)</math>. Since the little circle's <math>x</math>-coordinate is <math>-24</math> and the intersection point's <math>x</math>-coordinate is <math>\dfrac{864}{23}</math>, we get <math>\dfrac{864}{23} - 24 = \dfrac{312}{23}</math>. Therefore, our answer to this problem is <math>312 + 23 =  \boxed{335}</math>.  
  
 
~Arcticturn
 
~Arcticturn
  
==Solution 3 (Illustration of Solution 1)==
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==Solution 3 (Similar Triangles and Pythagorean Theorem)==
 
This solution refers to the <b>Diagram</b> section.
 
This solution refers to the <b>Diagram</b> section.
  
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As the intersection of planes <math>\mathcal{R}</math> and <math>\mathcal{P}</math> is 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.
 
As the intersection of planes <math>\mathcal{R}</math> and <math>\mathcal{P}</math> is 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 cross-sections <math>O_1O_3T_3T_1</math> and <math>\mathcal{R}:</math>
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Now, we focus on cross-sections <math>O_1O_3T_3T_1</math> and <math>\mathcal{R}:</math>
 
<ol style="margin-left: 1.5em;">
 
<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>
 
   <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>
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[[File:2021 AIME II Problem 10 Solution 2.png|center]]
 
[[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 obtain <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>O_1O_3T_3T_1,</math> since <math>\overline{O_1T_1}\parallel\overline{O_3T_3}</math> as discussed, we obtain <math>\triangle O_1T_1B\sim\triangle O_3T_3B</math> by AA, with the ratio of similitude <math>\frac{O_1T_1}{O_3T_3}=\frac{36}{13}.</math> Therefore, we get <math>\frac{O_1B}{O_3B}=\frac{49+O_3B}{O_3B}=\frac{36}{13},</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 obtain <math>\overline{O_1C}\parallel\overline{O_3A}</math> so that <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>
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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> Moreover, since <math>\ell\perp\overline{O_1C}</math> and <math>\overline{DO_3}\perp\overline{O_1C},</math> we obtain <math>\ell\parallel\overline{DO_3}</math> so that <math>\triangle O_1CB\sim\triangle O_1DO_3</math> by AA, with the ratio of similitude <math>\frac{O_1B}{O_1O_3}=\frac{49+\frac{637}{23}}{49}.</math> Therefore, we get <math>\frac{O_1C}{O_1D}=\frac{\sqrt{1105}+DC}{\sqrt{1105}}=\frac{49+\frac{637}{23}}{49},</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>
 
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>

Revision as of 10:53, 20 June 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. We can conclude all of the following:
    • The four black dashed line segments all lie in plane $\mathcal{R}.$
    • The four green solid line segments all lie in plane $\mathcal{P}.$
    • By symmetry, since planes $\mathcal{P}$ and $\mathcal{Q}$ are reflections of each other about plane $\mathcal{R},$ the three planes are concurrent to line $\ell.$
    • The red point is the foot of the perpendicular from the smallest sphere's center to line $\ell.$

~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 $A$ of the $72$ side, 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 $\overline{MB}$ 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 that $MB=36$ and $BC=13$. We know that $\overline{MB}$ and $\overline{BC}$ must be parallel, using ratios we realize that $CD=\frac{13}{23}\sqrt{1105}$. Apply the Pythagorean theorem to $\triangle BCD$, $BD=\frac{312}{23}$, so $312 + 23 = \boxed{335}$.

-Ross Gao

Solution 2 (Coordinates Bash)

Let's try to see some symmetry. We can use an $xyz$-plane to plot where the circles are. The two 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 $(x,0,-23)$ since we don't know how much the little sphere will be "pushed" down. We use the 3D distance formula to find that $x=-24$ (since $x=24$ 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 two lines must intersect. The point at where it intersects is $(-24a,0,23a)$. 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 (Similar Triangles and Pythagorean Theorem)

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 $\overleftrightarrow{O_3A}$ is the perpendicular bisector of $\overline{O_1O_2}.$ We wish to find $T_3A.$

As the intersection of planes $\mathcal{R}$ and $\mathcal{P}$ is 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.

Now, we focus on 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, cross-section $O_1O_3T_3T_1$ intersects line $\ell$ at exactly 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 have the following diagram:

In cross-section $O_1O_3T_3T_1,$ since $\overline{O_1T_1}\parallel\overline{O_3T_3}$ as discussed, we obtain $\triangle O_1T_1B\sim\triangle O_3T_3B$ by AA, with the ratio of similitude $\frac{O_1T_1}{O_3T_3}=\frac{36}{13}.$ Therefore, we get $\frac{O_1B}{O_3B}=\frac{49+O_3B}{O_3B}=\frac{36}{13},$ 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 $O_1D=\sqrt{1105}.$ Moreover, since $\ell\perp\overline{O_1C}$ and $\overline{DO_3}\perp\overline{O_1C},$ we obtain $\ell\parallel\overline{DO_3}$ so that $\triangle O_1CB\sim\triangle O_1DO_3$ by AA, with the ratio of similitude $\frac{O_1B}{O_1O_3}=\frac{49+\frac{637}{23}}{49}.$ Therefore, we get $\frac{O_1C}{O_1D}=\frac{\sqrt{1105}+DC}{\sqrt{1105}}=\frac{49+\frac{637}{23}}{49},$ 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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