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Difference between revisions of "2018 AMC 10A Problems/Problem 15"

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== Problem ==
 
== Problem ==
  
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Two circles of radius 5 are externally tangent to each other and are internally tangent to a circle of radius 13 at points  <math>A</math> and <math>B</math>, as shown in the diagram. The distance <math>AB</math> can be written in the form <math>\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?
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Two circles of radius <math>5</math> are externally tangent to each other and are internally tangent to a circle of radius <math>13</math> at points  <math>A</math> and <math>B</math>, as shown in the diagram. The distance <math>AB</math> can be written in the form <math>\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?
  
 
<asy>
 
<asy>
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<math>\textbf{(A) }  21  \qquad    \textbf{(B) }  29  \qquad    \textbf{(C) }  58  \qquad  \textbf{(D) } 69 \qquad  \textbf{(E) }  93 </math>
 
<math>\textbf{(A) }  21  \qquad    \textbf{(B) }  29  \qquad    \textbf{(C) }  58  \qquad  \textbf{(D) } 69 \qquad  \textbf{(E) }  93 </math>
  
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==Solution 1==
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==Solution==
 +
 
 
<asy>
 
<asy>
 
draw(circle((0,0),13));
 
draw(circle((0,0),13));
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label("$Z$", (5,-6.25),NE);
 
label("$Z$", (5,-6.25),NE);
 
</asy>
 
</asy>
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Let the center of the surrounding circle be <math>X</math>. The circle that is tangent at point <math>A</math> will have point <math>Y</math> as the center. Similarly, the circle that is tangent at point <math>B</math> will have point <math>Z</math> as the center. Connect <math>AB</math>, <math>YZ</math>, <math>XA</math>, and <math>XB</math>. Now observe that <math>\triangle XYZ</math> is similar to <math>\triangle XAB</math>. Writing out the ratios, we get
+
 
 +
Let the center of the surrounding circle be <math>X</math>. The circle that is tangent at point <math>A</math> will have point <math>Y</math> as the center. Similarly, the circle that is tangent at point <math>B</math> will have point <math>Z</math> as the center. Connect <math>AB</math>, <math>YZ</math>, <math>XA</math>, and <math>XB</math>. Now observe that <math>\triangle XYZ</math> is similar to <math>\triangle XAB</math> by SAS.
 +
 
 +
Writing out the ratios, we get
 
<cmath>\frac{XY}{XA}=\frac{YZ}{AB} \Rightarrow \frac{13-5}{13}=\frac{5+5}{AB} \Rightarrow \frac{8}{13}=\frac{10}{AB} \Rightarrow AB=\frac{65}{4}.</cmath>
 
<cmath>\frac{XY}{XA}=\frac{YZ}{AB} \Rightarrow \frac{13-5}{13}=\frac{5+5}{AB} \Rightarrow \frac{8}{13}=\frac{10}{AB} \Rightarrow AB=\frac{65}{4}.</cmath>
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Therefore, our answer is <math>65+4= \boxed{\textbf{D) } 69}</math>.
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Therefore, our answer is <math>65+4= \boxed{\textbf{(D) } 69}</math>.
 +
 
 +
==Note==
 +
Let the circle of radius 13 have center <math>O</math>. Let the circle tangent to <math>\odot O</math> at <math>A</math> have center <math>O_1</math> and that tangent at <math>B</math> have center <math>O_2</math>.
 +
 
 +
Here we prove by contradiction that the intersection of <math>\overleftrightarrow{OO_1} </math> and <math>\odot O</math> must be the point of tangency of <math>\odot O</math> and <math>\odot O_1</math>.
 +
 
 +
Assume that <math>A</math> is not the point of tangency of <math>\odot O</math> and <math>\odot O_1</math>. Furthermore, let <math>A</math> be the intersection of <math>\overleftrightarrow{OO_1} </math> and <math>\odot O</math>. Then, let <math>\odot O</math> and <math>\odot O_1</math> be tangent at <math> T </math>.
 +
 
 +
It follows that <math>\angle ATO_1 </math> is a right angle, so if we continue <math>TO_1</math> to hit <math> \odot O </math> at <math>I</math>, we have that <math>\angle ATI</math> must intercept a semicircle. <math>I</math> therefore has to be the intersection of the diameter through <math>A</math> and <math>O</math> and <math>\odot O</math>. We previously assumed that <math>O_1</math> had to be on <math>\overleftrightarrow{OA} </math>, so <math>O_1</math> is point <math>I</math>. Clearly, however, if point <math>O_1</math> is point <math>I</math>, <math>\odot O_1 </math> will intersect <math>\odot O</math> at more than one point, and therefore will not be tangent, leading to a contradiction.
 +
 
 +
Therefore, we have proved that <math>A</math>, the intersection of <math>\overleftrightarrow{OO_1} </math> and <math>\odot O</math>, is the point of tangency between <math>\odot O</math> and <math>\odot O_1</math>, so <math>A</math>, <math>O</math>, and <math>O_1</math> are collinear. Therefore, there exists a line through all three points. Likewise, we can perform the same proof on the circle tangent at B, and our solution above is valid.
 +
 
 +
 
 +
~LeonQS
  
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==Solution 2==
+
==Video Solution (HOW TO THINK CREATIVELY!)==
 +
https://youtu.be/xFnLbr-qt6I
  
−
<asy>
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~Education, the Study of Everything
−
draw(circle((0,0),13));
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draw(circle((5,-6.25),5));
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==Video Solution 1==
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draw(circle((-5,-6.25),5));
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−
label("$A$", (-8.125,-10.15), S);
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https://youtu.be/HJALwsbHZXc
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label("$B$", (8.125,-10.15), S);
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label("$C$", (0,-6.25), NE);
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- Whiz
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draw((0,0)--(-8.125,-10.15));
+
 
−
draw((0,0)--(8.125,-10.15));
+
https://www.youtube.com/watch?v=llMgyOkjNgU&list=PL-27w0UNlunxDTyowGrnvo_T7z92OCvpv&index=3 - amshah
−
draw((-5,-6.25)--(5,-6.25));
+
 
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draw((0,0)--(0,-13));
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== Video Solution 2 by OmegaLearn ==
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draw((-8.125,-10.15)--(8.125,-10.15));
+
https://youtu.be/NsQbhYfGh1Q?t=1328
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label("$O$", (0,0), N);
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−
</asy>
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~ pi_is_3.14
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Let the center of the large circle be <math>O</math>. Let the common tangent of the two smaller circles be <math>C</math>. Draw the two radii of the large circle, <math>\overline{OA}</math> and <math>\overline{OB}</math> and the two radii of the smaller circles to point <math>C</math>. Draw ray <math>\overrightarrow{OC}</math> and <math>\overline{AB}</math>. This sets us up with similar triangles, which we can solve.
 
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The length of  <math>\overline{OC}</math> is equal to <math>\sqrt{39}</math> by Pythagorean Theorem, the length of the hypotenuse is <math>8</math>, and the other leg is <math>5</math>. Using similar triangles, <math>OB</math> is <math>13</math>, and therefore half of <math>AB</math> is <math>\frac{65}{8}</math>. Doubling gives <math>\frac{65}{4}</math>, which results in <math>65+4=\boxed{\textbf{D) }69}</math>. Nice
 
−
<math>QED \square</math>
 
  
 
==See Also==
 
==See Also==

Latest revision as of 12:52, 18 August 2025

Problem

Two circles of radius $5$ are externally tangent to each other and are internally tangent to a circle of radius $13$ at points $A$ and $B$, as shown in the diagram. The distance $AB$ can be written in the form $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?

[asy] draw(circle((0,0),13)); draw(circle((5,-6.2),5)); draw(circle((-5,-6.2),5)); label("$B$", (9.5,-9.5), S); label("$A$", (-9.5,-9.5), S); [/asy]

$\textbf{(A) } 21 \qquad \textbf{(B) } 29 \qquad \textbf{(C) } 58 \qquad \textbf{(D) } 69 \qquad \textbf{(E) } 93$

Solution

[asy] draw(circle((0,0),13)); draw(circle((5,-6.25),5)); draw(circle((-5,-6.25),5)); label("$A$", (-8.125,-10.15), S); label("$B$", (8.125,-10.15), S); draw((0,0)--(-8.125,-10.15)); draw((0,0)--(8.125,-10.15)); draw((-5,-6.25)--(5,-6.25)); draw((-8.125,-10.15)--(8.125,-10.15)); label("$X$", (0,0), N); label("$Y$", (-5,-6.25),NW); label("$Z$", (5,-6.25),NE); [/asy]

Let the center of the surrounding circle be $X$. The circle that is tangent at point $A$ will have point $Y$ as the center. Similarly, the circle that is tangent at point $B$ will have point $Z$ as the center. Connect $AB$, $YZ$, $XA$, and $XB$. Now observe that $\triangle XYZ$ is similar to $\triangle XAB$ by SAS.

Writing out the ratios, we get \[\frac{XY}{XA}=\frac{YZ}{AB} \Rightarrow \frac{13-5}{13}=\frac{5+5}{AB} \Rightarrow \frac{8}{13}=\frac{10}{AB} \Rightarrow AB=\frac{65}{4}.\] Therefore, our answer is $65+4= \boxed{\textbf{(D) } 69}$.

Note

Let the circle of radius 13 have center $O$. Let the circle tangent to $\odot O$ at $A$ have center $O_1$ and that tangent at $B$ have center $O_2$.

Here we prove by contradiction that the intersection of $\overleftrightarrow{OO_1}$ and $\odot O$ must be the point of tangency of $\odot O$ and $\odot O_1$.

Assume that $A$ is not the point of tangency of $\odot O$ and $\odot O_1$. Furthermore, let $A$ be the intersection of $\overleftrightarrow{OO_1}$ and $\odot O$. Then, let $\odot O$ and $\odot O_1$ be tangent at $T$.

It follows that $\angle ATO_1$ is a right angle, so if we continue $TO_1$ to hit $\odot O$ at $I$, we have that $\angle ATI$ must intercept a semicircle. $I$ therefore has to be the intersection of the diameter through $A$ and $O$ and $\odot O$. We previously assumed that $O_1$ had to be on $\overleftrightarrow{OA}$, so $O_1$ is point $I$. Clearly, however, if point $O_1$ is point $I$, $\odot O_1$ will intersect $\odot O$ at more than one point, and therefore will not be tangent, leading to a contradiction.

Therefore, we have proved that $A$, the intersection of $\overleftrightarrow{OO_1}$ and $\odot O$, is the point of tangency between $\odot O$ and $\odot O_1$, so $A$, $O$, and $O_1$ are collinear. Therefore, there exists a line through all three points. Likewise, we can perform the same proof on the circle tangent at B, and our solution above is valid.


~LeonQS

Video Solution (HOW TO THINK CREATIVELY!)

https://youtu.be/xFnLbr-qt6I

~Education, the Study of Everything

Video Solution 1

https://youtu.be/HJALwsbHZXc

- Whiz

https://www.youtube.com/watch?v=llMgyOkjNgU&list=PL-27w0UNlunxDTyowGrnvo_T7z92OCvpv&index=3 - amshah

Video Solution 2 by OmegaLearn

https://youtu.be/NsQbhYfGh1Q?t=1328

~ pi_is_3.14

See Also

2018 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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