Difference between revisions of "2018 AMC 10A Problems/Problem 15"
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Call center of the largest circle <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>WY</math>. Now observe that <math>\triangle XYZ</math> is similar to <math>\triangle XAB</math>. Writing out the ratios, we get | Call center of the largest circle <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>WY</math>. Now observe that <math>\triangle XYZ</math> is similar to <math>\triangle XAB</math>. 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=</math><math>69 \boxed{D}</math>. |
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+ | ~Nivek |
Revision as of 14:56, 8 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 ?
Solution
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 .
~Nivek