2017 AMC 12A Problems/Problem 24
Contents
Problem
Quadrilateral is inscribed in circle and has side lengths , and . Let and be points on such that and . Let be the intersection of line and the line through parallel to . Let be the intersection of line and the line through parallel to . Let be the point on circle other than that lies on line . What is ?
Diagram
~raxu, put in by fuzimiao2013
Solution 1
Using the given ratios, note that
By AA Similarity, with a ratio of and with a ratio of , so .
Now we find the length of . Because the quadrilateral is cyclic, we can simply use the Law of Cosines. By Power of a Point, . Thus
-solution by FRaelya
Solution 2
We shall make use of the pairs of similar triangles present in the problem, Ptolemy's Theorem, and Power of a Point. Let be the intersection of and . First, from being a cyclic quadrilateral, we have that , . Therefore, , , and , so we have , , and . By Ptolemy's Theorem, Thus, . Then, by Power of a Point, . So, . Next, observe that , so . Also, , so . We can compute after noticing that and that . So, . Then, .
Multiplying our equations for and yields that
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=JdERP0d0W64&list=PLyhPcpM8aMvLZmuDnM-0vrFniLpo7Orbp&index=4 - AMBRIGGS
See Also
2017 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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 12 Problems and Solutions |
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