Difference between revisions of "2010 AIME II Problems/Problem 15"
James digol (talk | contribs) (Created page with '== '''Problem 15.''' == In triangle <math>ABC</math>, <math>AC = 13</math>, <math>BC = 14</math>, and <math>AB=15</math>. Points <math>M</math> and <math>D</math> lie on <math…') |
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Revision as of 21:42, 14 May 2010
Problem 15.
In triangle , , , and . Points and lie on with and . Points and lie on B with and . Let be the point, other than , of intersection of the circumcircles of and . Ray meets at . The ratio can be written in the form , where and are relatively prime positive integers. Find .
Solution.
Let . $\frac {BQ}{QC} = \frac {NY}{MY$ (Error compiling LaTeX. Unknown error_msg)} since . Since quadrilateral is cyclic, and , yielding and . Multiplying these together yields .
. Also, is the center of spiral similarity of segments and , so . Therefore, , which can easily be computed by the angle bisector theorem to be . It follows that , giving us an answer of .
Note: Spiral similarities may sound complex, but they're really not. The fact that is really just a result of simple angle chasing.
Source: [1] by Zhero