Difference between revisions of "2015 AIME II Problems/Problem 11"
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− | Call the <math>M</math> and <math>N</math> foot of the altitudes from <math>O</math> to <math>BC</math> and <math>AB</math>, respectively. Let <math>OB = r</math> . Notice that <math>\triangle{OMB} \sim \triangle{ | + | Call the <math>M</math> and <math>N</math> foot of the altitudes from <math>O</math> to <math>BC</math> and <math>AB</math>, respectively. Let <math>OB = r</math> . Notice that <math>\triangle{OMB} \sim \triangle{QOB}</math> because both are right triangles, and <math>\angle{OBQ} \cong \angle{OBM}</math>. By <math>\frac{MB}{BO}=\frac{BO}{BQ}</math>, <math>MB = r\left(\frac{r}{4.5}\right) = \frac{r^2}{4.5}</math>. However, since <math>O</math> is the circumcenter of triangle <math>ABC</math>, <math>OM</math> is a perpendicular bisector by the definition of a circumcenter. Hence, <math>\frac{r^2}{4.5} = 2 \implies r = 3</math>. Since we know <math>BN=\frac{5}{2}</math> and <math>\triangle BOP \sim \triangle NBO</math>, we have <math>\frac{BP}{3} = \frac{3}{\frac{5}{2}}</math>. Thus, <math>BP = \frac{18}{5}</math>. <math>m + n=\boxed{023}</math>. |
===Solution 2=== | ===Solution 2=== |
Revision as of 18:21, 22 June 2019
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
. 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 |
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