Difference between revisions of "2018 AMC 10A Problems/Problem 15"
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| − | Two circles of radius 5 are externally tangent to each other and are internally tangent to a circle of radius 13 at points <math>A</math> and <math>B</math>, as shown in the diagram. The distance <math>AB</math> can be written in the form <math>\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>? | + | == Problem == |
| + | |||
| + | Two circles of radius <math>5</math> are externally tangent to each other and are internally tangent to a circle of radius <math>13</math> at points <math>A</math> and <math>B</math>, as shown in the diagram. The distance <math>AB</math> can be written in the form <math>\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>? | ||
<asy> | <asy> | ||
| Line 13: | Line 15: | ||
==Solution== | ==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> | ||
| + | |||
| + | 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= | + | Therefore, our answer is <math>65+4= \boxed{\textbf{(D) } 69}</math>. |
| + | |||
| + | ==Video Solution 1== | ||
| + | |||
| + | https://youtu.be/HJALwsbHZXc | ||
| + | |||
| + | - Whiz | ||
| + | |||
| + | https://www.youtube.com/watch?v=llMgyOkjNgU&list=PL-27w0UNlunxDTyowGrnvo_T7z92OCvpv&index=3 - amshah | ||
| + | |||
| + | == Video Solution 2 by OmegaLearn == | ||
| + | https://youtu.be/NsQbhYfGh1Q?t=1328 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC10 box|year=2018|ab=A|num-b=14|num-a=16}} | ||
| + | {{MAA Notice}} | ||
| − | + | [[Category:Introductory Geometry Problems]] | |
Revision as of 03:58, 23 January 2023
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
- Whiz
https://www.youtube.com/watch?v=llMgyOkjNgU&list=PL-27w0UNlunxDTyowGrnvo_T7z92OCvpv&index=3 - amshah
Video Solution 2 by OmegaLearn
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. 
