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

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==Solution==
 
==Solution==
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We can't have a solution without a problem.
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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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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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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>, and so we're done.
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-Ross Gao
  
 
==See also==
 
==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 10:07, 23 March 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 $m + n$.

Solution

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; $BD=\frac{312}{23}$, and so we're done.

-Ross Gao

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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