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

Revision as of 16:37, 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

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 $\boxed{D}$ .

~Nivek

2018 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 ~Nivek
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+==See Also==
+{{AMC10 box|year=2018|ab=A|num-b=14|num-a=16}}
+{{MAA Notice}}

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

Revision as of 16:37, 8 February 2018

Solution

See Also