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> and <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> 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>. | 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>. | ||
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+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=llMgyOkjNgU&list=PL-27w0UNlunxDTyowGrnvo_T7z92OCvpv&index=3 - amshah | ||
==See Also== | ==See Also== |
Revision as of 20:40, 31 January 2021
Contents
[hide]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
?
Solution 1
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
. 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
and
. 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
, and the other leg is
. Using similar triangles,
is
, and therefore half of
is
. Doubling gives
, which results in
.
Video Solution
https://www.youtube.com/watch?v=llMgyOkjNgU&list=PL-27w0UNlunxDTyowGrnvo_T7z92OCvpv&index=3 - amshah
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.