2013 AMC 12B Problems/Problem 19
- The following problem is from both the 2013 AMC 12B #19 and 2013 AMC 10B #23, so both problems redirect to this page.
In triangle , , , and . Distinct points , , and lie on segments , , and , respectively, such that , , and . The length of segment can be written as , where and are relatively prime positive integers. What is ?
Since , quadrilateral is cyclic. It follows that . In addition, triangles and are similar. It follows that . By Ptolemy's Theorem, we have . Cancelling , we obtain , so our answer is .
Using the similar triangles in triangle gives and . Quadrilateral is cyclic, implying that = 180°. Therefore, , and triangles and are similar. Solving the resulting proportion gives . Therefore, and our answer is .
If we draw a diagram as given, but then add point on such that in order to use the Pythagorean theorem, we end up with similar triangles and . Thus, and , where is the length of . Using the Pythagorean theorem, we now get and can be found out noting that is just through base times height (since , similar triangles gives ), and that is just . From there, Now, , and squaring and adding both sides and subtracting a 169 from both sides gives , so . Thus, the answer is .