Difference between revisions of "2015 AIME II Problems/Problem 11"
m (→Solution) |
Expilncalc (talk | contribs) (Added solution) |
||
Line 45: | Line 45: | ||
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>. | ||
+ | ==Solution 3== | ||
+ | Using the diagram given, <math>r=BO</math>. Since <math>O</math> bisects any chord, <math>BM=MC=2</math> and <math>BN=NA=2.5</math>. From there, <math>OM=\sqrt{r^2-4}</math>. Thus, <math>OQ=\frac{\sqrt{4r^2+9}}{2}</math>. Using <math>\triangle{BOQ}, we get </math>r=3<math>. Now let's find </math>NP<math>. Using </math>\triangle{BON}~\triangle{OPN}<math>, we get {NP=\frac{2}{5}r^2-\frac{5}{2}</math>. Therefore, <math>BP=\frac{5}{2}+\frac{11}{10}=18/5</math>. <math>18+5=\boxed{023}</math>. | ||
==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 10:14, 29 August 2017
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
.
Solution 3
Using the diagram given, . Since
bisects any chord,
and
. From there,
. Thus,
. Using
r=3
NP
\triangle{BON}~\triangle{OPN}$, we get {NP=\frac{2}{5}r^2-\frac{5}{2}$ (Error compiling LaTeX. Unknown error_msg). Therefore,
.
.
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.