Difference between revisions of "2010 AIME II Problems/Problem 15"
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− | == | + | == 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>AC</math> with <math>AM=MC</math> and <math>\angle ABD = \angle DBC</math>. Points <math>N</math> and <math>E</math> lie on <math>A</math>B with <math>AN=NB</math> and <math>\angle ACE = \angle ECB</math>. Let <math>P</math> be the point, other than <math>A</math>, of intersection of the circumcircles of <math>\triangle AMN</math> and <math>\triangle ADE</math>. Ray <math>AP</math> meets <math>BC</math> at <math>Q</math>. The ratio <math>\frac{BQ}{CQ}</math> can be written in the form <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m-n</math>. | 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>AC</math> with <math>AM=MC</math> and <math>\angle ABD = \angle DBC</math>. Points <math>N</math> and <math>E</math> lie on <math>A</math>B with <math>AN=NB</math> and <math>\angle ACE = \angle ECB</math>. Let <math>P</math> be the point, other than <math>A</math>, of intersection of the circumcircles of <math>\triangle AMN</math> and <math>\triangle ADE</math>. Ray <math>AP</math> meets <math>BC</math> at <math>Q</math>. The ratio <math>\frac{BQ}{CQ}</math> can be written in the form <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m-n</math>. | ||
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== Solution. == | == Solution. == |
Revision as of 20:20, 7 March 2013
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 .
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