Difference between revisions of "2015 AIME II Problems/Problem 11"
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==Problem== | ==Problem== | ||
− | 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> | + | 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> at <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== | ==Diagram== |
Latest revision as of 20:59, 26 November 2023
Contents
[hide]Problem
The circumcircle of acute has center
. The line passing through point
perpendicular to
intersects lines
and
at
and
, respectively. Also
,
,
, and
, where
and
are relatively prime positive integers. Find
.
Diagram
Solution
Solution 1
Call and
the feet of the altitudes from
to
and
, respectively. Let
. Notice that
because both are right triangles, and
. By
,
. 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 (fastest)
Minor arc so
. Since
is isosceles (
and
are radii),
.
, so
. From this we get that
. So
, plugging in the given values we get
, so
, and
.
Solution 3
Let . Drawing perpendiculars,
and
. From there,
Thus,
Using
, we get
. Now let's find
. After some calculations with
~
,
. Therefore,
.
Solution 4
Let . Extend
to touch the circumcircle at a point
. Then, note that
. But since
is a diameter,
, implying
. It follows that
is a cyclic quadrilateral.
Let . By Power of a Point,
The answer is
.
Solution 5
Denote the circumradius of to be
, the circumcircle of
to be
, and the shortest distance from
to circle
to be
.
Using Power of a Point on relative to circle
, we get that
. Using Pythagorean Theorem on triangle
to get
. Subtracting the first equation from the second, we get that
and therefore
. Now, set
. Using law of cosines on
to find
in terms of
and plugging that into the extended law of sines, we get
. Squaring both sides and cross multiplying, we get
. Now, we get
using quadratic formula. If you drew a decent diagram,
is acute and therefore
(You can also try plugging in both in the end and seeing which gives a rational solution). Note that
Using the cosine addition formula and then plugging in what we know about
, we get that
. Now, the hard part is to find what
is. We therefore want
. For the numerator, by inspection
will not work for integers
and
. The other case is if there is
. By inspection,
works. Therefore, plugging all this in yields the answer,
. Solution by hyxue
Solution 6
Reflect
,
across
to points
and
, respectively with
on the circle and
collinear. Now,
by parallel lines. From here,
as
collinear. From here,
is cyclic, and by power of a point we obtain
.
~awang11's sol
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.