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

Revision as of 11:07, 3 August 2021

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

Video Solution 1

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

Video Solution 2

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

~ pi_is_3.14

@@ Line 33: / Line 33: @@
 <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>
 Therefore, our answer is <math>65+4= \boxed{\textbf{D) } 69}</math>.
-==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 <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 <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>.
 ==Video Solution 1==

2018 AMC 10A (Problems • Answer Key • Resources)
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