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 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>. Nice | + | 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>. Nice |
<math>QED \square</math> | <math>QED \square</math> | ||
Revision as of 20:19, 1 January 2020
Contents
[hide]Problem
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
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 . Nice
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 |
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