Difference between revisions of "2010 AIME II Problems/Problem 15"
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<math>\frac {AN}{AM} = \frac {\frac {AB}{2}}{\frac {AC}{2}} = \frac {15}{13}</math>. Also, <math>P</math> is the center of spiral similarity of segments <math>MD</math> and <math>EN</math>, so <math>\triangle PMD \sim \triangle PNE</math>. Therefore, <math>\frac {PN}{PM} = \frac {NE}{MD}</math>, which can easily be computed by the angle bisector theorem to be <math>\frac {145}{117}</math>. It follows that <math>\frac {BQ}{CQ} = \frac {15}{13} \cdot \frac {145}{117} = \frac {725}{507}</math>, giving us an answer of <math>725 - 507 = \boxed{218}</math>. | <math>\frac {AN}{AM} = \frac {\frac {AB}{2}}{\frac {AC}{2}} = \frac {15}{13}</math>. Also, <math>P</math> is the center of spiral similarity of segments <math>MD</math> and <math>EN</math>, so <math>\triangle PMD \sim \triangle PNE</math>. Therefore, <math>\frac {PN}{PM} = \frac {NE}{MD}</math>, which can easily be computed by the angle bisector theorem to be <math>\frac {145}{117}</math>. It follows that <math>\frac {BQ}{CQ} = \frac {15}{13} \cdot \frac {145}{117} = \frac {725}{507}</math>, giving us an answer of <math>725 - 507 = \boxed{218}</math>. | ||
− | '''Note:''' Spiral similarities may sound complex, but they're really not. The fact that <math>\triangle PMD \sim \triangle | + | '''Note:''' Spiral similarities may sound complex, but they're really not. The fact that <math>\triangle PMD \sim \triangle PNE</math> is really just a result of simple angle chasing. |
Source: [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1831745#p1831745] by Zhero | Source: [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1831745#p1831745] by Zhero |
Revision as of 19:08, 14 March 2011
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