2018 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 15 Medium

Two circles of radius 5 are externally tangent to each other and are internally tangent to a circle of radius 13 at points A and B, as shown in the diagram. The distance AB can be written in the form \frac{m}{n}, where m and n are relatively prime positive integers. What is m+n? (2018 AMC 10A Problem, Question#15)

  • A.

    21

  • B.

    29

  • C.

    58

  • D.

    69

  • E.

    93

Answer:D

Let the center of the surrounding circle be X. The circle that is tangent at point A will have point Y as the center. Similarly, the circle that is tangent at point B will have point Z as the center. Connect AB, YZ, XA, and XB. Now observe that \triangle XYZ is similar to \triangle XAB. Writing out the ratios, we get \dfrac{XY}{XA}=\dfrac{YZ}{AB}\Rightarrow\dfrac{13-5}{13}=\dfrac{5+5}{AB}\Rightarrow\dfrac{8}{13}=\dfrac{10}{AB}\Rightarrow AB=\dfrac{65}{4}. Therefore, our answer is 65+4=\boxed{\rm D)~69}.

Let the center of the large circle be O. Let the common tangent of the two smaller circles be C. Draw the two radii of the large circle, \overline{OA} and \overline{OB} and the two radii of the smaller circles to point C. Draw ray \overline{OC} and \overline{AB}. This sets us up with similar triangles, which we can solve. The length of \overline{OC} is equal to \sqrt{39} by Pythagorean Theorem, the length of the hypotenuse is 8, and the other leg is 5. Using similar triangles, OB is 13, and therefore half of AB is \frac{65}{8}. Doubling gives \frac{65}{4}, which results in 65+4=\boxed{\rm D)~69}.

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