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

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</asy>
 
</asy>
 
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.
 
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.
−
The length of  <math>\overline{OC}</math> is equal to <math>\sqrt{39}</math> by Pythagorean Theorem, the length of the hypotenuse is 8, and the other leg is 5. Using similar triangles, <math>OB</math> is 13, 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>.
+
The length of  <math>\overline{OC}</math> is equal to <math>\sqrt{39}</math> by Pythagorean Theorem, the length of the hypotenuse is 8, and the other leg is 5. Using similar triangles, <math>OB</math> is 13, 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>
 
<math>QED \square</math>
  

Revision as of 20:06, 2 September 2019

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 1

[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 $\triangle XAB$. 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 $65+4= \boxed{\textbf{D) } 69}$.

Solution 2

[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); label("$C$", (0,-6.25), NE); draw((0,0)--(-8.125,-10.15)); draw((0,0)--(8.125,-10.15)); draw((-5,-6.25)--(5,-6.25)); draw((0,0)--(0,-13)); draw((-8.125,-10.15)--(8.125,-10.15)); label("$O$", (0,0), N); [/asy] Let the center of the large circle be $O$. Let the common tangent of the two smaller circles be $C$. Draw the two radii of the large circle, $\overline{OA}$ and $\overline{OB}$ and the two radii of the smaller circles to point $C$. Draw ray $\overrightarrow{OC}$ and $\overline{AB}$. This sets us up with similar triangles, which we can solve. The length of $\overline{OC}$ is equal to $\sqrt{39}$ by Pythagorean Theorem, the length of the hypotenuse is 8, and the other leg is 5. Using similar triangles, $OB$ is 13, and therefore half of $AB$ is $\frac{65}{8}$. Doubling gives $\frac{65}{4}$, which results in $65+4=\boxed{\textbf{D) }69}$. Nice $QED \square$

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

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