Difference between revisions of "2015 AIME II Problems/Problem 11"
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The circumcircle of acute <math>\triangle ABC</math> has center <math>O</math>. The line passing through point <math>O</math> perpendicular to <math>\overline{OB}</math> intersects lines <math>AB</math> and <math>BC</math> and <math>P</math> and <math>Q</math>, respectively. Also <math>AB=5</math>, <math>BC=4</math>, <math>BQ=4.5</math>, and <math>BP=\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | The circumcircle of acute <math>\triangle ABC</math> has center <math>O</math>. The line passing through point <math>O</math> perpendicular to <math>\overline{OB}</math> intersects lines <math>AB</math> and <math>BC</math> and <math>P</math> and <math>Q</math>, respectively. Also <math>AB=5</math>, <math>BC=4</math>, <math>BQ=4.5</math>, and <math>BP=\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | ||
− | == | + | ==Diagram== |
<asy> | <asy> | ||
unitsize(30); | unitsize(30); |
Revision as of 15:09, 22 June 2016
Contents
[hide]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
.
Diagram
Solution 1
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,
. Since we know
and
, we have
. Thus,
.
.
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.