2017 AMC 12A Problems/Problem 24
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 ?
~raxu, put in by fuzimiao2013
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
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
Denote to be the intersection between line and circle . Note that , making . Thus, is a cyclic quadrilateral. Using Power of a Point on gives .
Since and , . Using Power of a Point on again, . Plugging in gives: By Law of Cosines, we can find , as in Solution 1. Now, and , making . This gives us as a result.
-Solution by sml1809
You could have also got the relation as follows: From the similarities, . PoP on gives . Plugging in and gives implying that .
By Power of a Point,
By multiplying the equations we get
Let , , , , , and
By Ptolemy's theorem,
Video Solution by Richard Rusczyk
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