Difference between revisions of "2015 AIME II Problems/Problem 11"
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==Solution 2== | ==Solution 2== | ||
Notice that <math>\angle{CBO}=90-A</math>, so <math>\angle{BQO}=A</math>. From this we get that <math>\triangle{BPQ}\sim \triangle{BCA}</math>. So <math>\dfrac{BP}{BC}=\dfrac{BQ}{BA}</math>, plugging in the given values we get <math>\dfrac{BP}{4}=\dfrac{4.5}{5}</math>, so <math>BP=\dfrac{18}{5}</math>, and <math>m+n=\boxed{023}</math>. | Notice that <math>\angle{CBO}=90-A</math>, so <math>\angle{BQO}=A</math>. From this we get that <math>\triangle{BPQ}\sim \triangle{BCA}</math>. So <math>\dfrac{BP}{BC}=\dfrac{BQ}{BA}</math>, plugging in the given values we get <math>\dfrac{BP}{4}=\dfrac{4.5}{5}</math>, so <math>BP=\dfrac{18}{5}</math>, and <math>m+n=\boxed{023}</math>. | ||
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==See also== | ==See also== | ||
{{AIME box|year=2015|n=II|num-b=10|num-a=12}} | {{AIME box|year=2015|n=II|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:00, 28 March 2015
Contents
Problem
The circumcircle of acute has center
. The line passing through point
perpendicular to
intersects lines
and
and
and
, respectively. Also
,
,
, and
, where
and
are relatively prime positive integers. Find
.
Solution
Call the and
foot of the altitudes from
to
and
, respectively. Let
and let
. Notice that
because both are right triangles, and
. Then,
. However, since
is the circumcenter of triangle
,
is a perpendicular bisector by the definition of a circumcenter. Hence,
. We can use the Pythagorean theorem to find
, so we have
Likewise,
because both are right triangles, and
. Hence, since
as well, we have that
. It follows that
. We add this to
to get
, so
. Our answer is
.
Solution 2
Notice that , so
. From this we get that
. So
, plugging in the given values we get
, so
, and
.
See also
2015 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.