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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== Solution. == | == Solution. == | ||
− | Let <math>Y = MN \cap AQ</math>. <math>\frac {BQ}{QC} = \frac {NY}{MY</math> | + | Let <math>Y = MN \cap AQ</math>. <math>\frac {BQ}{QC} = \frac {NY}{MY}</math> since <math>\triangle AMN \sim \triangle ACB</math>. Since quadrilateral <math>AMYN</math> is cyclic, <math>\triangle MYA \sim \triangle PYN</math> and <math>\triangle MYP \sim \triangle AYN</math>, yielding <math>\frac {YM}{YA} = \frac {MP}{AN}</math> and <math>\frac {YA}{YN} = \frac {AM}{PN}</math>. Multiplying these together yields <math>\frac {YN}{YM} = \left(\frac {AN}{AM}\right) \left(\frac {PN}{PM}\right)</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>FN</math>, so <math>\triangle PMD \sim \triangle PNF</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 {13}{15} \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>FN</math>, so <math>\triangle PMD \sim \triangle PNF</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 {13}{15} \cdot \frac {145}{117} = \frac {725}{507}</math>, giving us an answer of <math>725 - 507 = \boxed{218}</math>. |
Revision as of 21:06, 21 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 .
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