Difference between revisions of "2010 AIME II Problems/Problem 15"
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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 | ||
+ | ==See Also== | ||
+ | {{AIME box|year=2010|n=II|num-b=14|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:19, 13 March 2015
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
See Also
2010 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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