Difference between revisions of "2018 AMC 10A Problems/Problem 15"
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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 | + | ==Solution== |
<asy> | <asy> | ||
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</asy> | </asy> | ||
| − | 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> | ||
Therefore, our answer is <math>65+4= \boxed{\textbf{D) } 69}</math>. | Therefore, our answer is <math>65+4= \boxed{\textbf{D) } 69}</math>. | ||
Revision as of 12:03, 23 August 2021
Problem
Two circles of radius 
are externally tangent to each other and are internally tangent to a circle of radius 
at points 
and 
, as shown in the diagram. The distance 
can be written in the form 
, where 
and 
are relatively prime positive integers. What is 
?
![[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]](http://latex.artofproblemsolving.com/c/1/d/c1d7b33c122356459463ccf321031d04f8f53752.png)
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]](http://latex.artofproblemsolving.com/0/f/5/0f5cde9eec57d1be48c64499b92a5ed49bb52b18.png)
Let the center of the surrounding circle be 
. The circle that is tangent at point will have point 
as the center. Similarly, the circle that is tangent at point will have point 
as the center. Connect , 
, 
, and 
. Now observe that 
is similar to 
by SAS. Writing out the ratios, we get
![\[\frac{XY}{XA}=\frac{YZ}{AB} \Rightarrow \frac{13-5}{13}=\frac{5+5}{AB} \Rightarrow \frac{8}{13}=\frac{10}{AB} \Rightarrow AB=\frac{65}{4}.\]](http://latex.artofproblemsolving.com/5/f/e/5fe2512bf29a8a348577f7d3ab0ac8bfd57152c5.png)
Therefore, our answer is 
.
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
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
