Difference between revisions of "2018 AMC 10A Problems/Problem 15"
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The length of <math>\overline{OC}</math> is equal to <math>\sqrt{39}</math> by Pythagorean Theorem, the length of the hypotenuse is 8, and the other leg is 5. Using similar triangles, <math>OB</math> is 13, 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>. | The length of <math>\overline{OC}</math> is equal to <math>\sqrt{39}</math> by Pythagorean Theorem, the length of the hypotenuse is 8, and the other leg is 5. Using similar triangles, <math>OB</math> is 13, 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>. | ||
<math>QED \blacksquare</math> | <math>QED \blacksquare</math> | ||
| + | |||
| + | ==See Also== | ||
{{AMC10 box|year=2018|ab=A|num-b=14|num-a=16}} | {{AMC10 box|year=2018|ab=A|num-b=14|num-a=16}} | ||
Revision as of 22:55, 12 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 
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 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]](http://latex.artofproblemsolving.com/0/f/5/0f5cde9eec57d1be48c64499b92a5ed49bb52b18.png)
Call center of the largest circle 
. 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 
. 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 
.
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]](http://latex.artofproblemsolving.com/7/6/b/76b66aa6f450559156469e71b74c7f607248f17d.png)
Let the center of the large circle be 
. Let the common tangent of the two smaller circles be 
. Draw the two radii of the large circle, 
and 
and the two radii of the smaller circles to point . Draw ray 
. Draw 
. This sets us up with similar triangles, which we can solve.
The length of 
is equal to 
by Pythagorean Theorem, the length of the hypotenuse is 8, and the other leg is 5. Using similar triangles, 
is 13, and therefore half of is 
. Doubling gives 
which results in 
.
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 | ||
