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

(Solution 1)
(Solution 1)
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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>
  
==Solution 1==
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==Solution 1 (SOMEONE HELP ME ADD A PICTURE PLEASE)==
  
 
Call center of the largest circle <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>WY</math>. Now observe that <math>\triangle XYZ</math> is similar to <math>\triangle XAB</math>. Writing out the ratios, we get
 
Call center of the largest circle <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>WY</math>. Now observe that <math>\triangle XYZ</math> is similar to <math>\triangle XAB</math>. Writing out the ratios, we get

Revision as of 17:29, 8 February 2018

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 (SOMEONE HELP ME ADD A PICTURE PLEASE)

Call center of the largest circle $X$. 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 $WY$. 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{69}$, which is choice $\boxed{D}$.

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 $\overrightarrow{OC}$. Draw $\overline{AB}$. 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 $65+4=\boxed{69}$, which is choice $\boxed{D}$. $QED \blacksquare$

2018 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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All AMC 10 Problems and Solutions
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