Difference between revisions of "2015 AIME II Problems/Problem 11"
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==Solution 4== | ==Solution 4== | ||
− | Let <math>\angle{BQO}=\alpha</math>. Extend <math>OB</math> to touch the circumcircle at a point <math>K</math>. Then, note that <math>\angle{ | + | Let <math>\angle{BQO}=\alpha</math>. Extend <math>OB</math> to touch the circumcircle at a point <math>K</math>. Then, note that <math>\angle{KAC}=\angle{CBK}=\angle{QBO}=90^\circ-\alpha</math>. But since <math>BK</math> is a diameter, <math>\angle{KAB}=90^\circ</math>, implying <math>\angle{CAB}=\alpha</math>. It follows that <math>APCQ</math> is a cyclic quadrilateral. |
Let <math>BP=x</math>. By Power of a Point, <cmath>5x=4\cdot\frac 9 2\implies x=\frac{18}{5}.</cmath>The answer is <math>18+5=\boxed{023}</math>. | Let <math>BP=x</math>. By Power of a Point, <cmath>5x=4\cdot\frac 9 2\implies x=\frac{18}{5}.</cmath>The answer is <math>18+5=\boxed{023}</math>. |
Revision as of 17:40, 18 August 2019
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
. 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
Notice that , 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
.
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.