Difference between revisions of "2018 AMC 10A Problems/Problem 15"
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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>. Draw <math>\overline{AB}</math>. This sets us up with similar triangles, which we can solve. | 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>. Draw <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 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 \ | + | <math>QED \square</math> |
+ | |||
+ | (very minor <math>\latex</math> edit by elements2015) | ||
==See Also== | ==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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:33, 3 March 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
?
Solution 1
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
Therefore, our answer is
.
Solution 2
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
.
(very minor $\latex$ (Error compiling LaTeX. Unknown error_msg) edit by elements2015)
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.