Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2015 AIME II Problems/Problem 11"

(→‎Solution 3)
m (→‎Solution 1)
Line 40: Line 40:
 
</asy>
 
</asy>
 
===Solution 1===
 
===Solution 1===
−
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{OQB}</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>.
+
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

Problem

The circumcircle of acute $\triangle ABC$ has center $O$. The line passing through point $O$ perpendicular to $\overline{OB}$ intersects lines $AB$ and $BC$ and $P$ and $Q$, respectively. Also $AB=5$, $BC=4$, $BQ=4.5$, and $BP=\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Diagram

[asy] unitsize(30); draw(Circle((0,0),3)); pair A,B,C,O, Q, P, M, N; A=(2.5, -sqrt(11/4)); B=(-2.5, -sqrt(11/4)); C=(-1.96, 2.28); Q=(-1.89, 2.81); P=(1.13, -1.68); O=origin; M=foot(O,C,B); N=foot(O,A,B); draw(A--B--C--cycle); label("$A$",A,SE); label("$B$",B,SW); label("$C$",C,NW); label("$Q$",Q,NW); dot(O); label("$O$",O,NE); label("$M$",M,W); label("$N$",N,S); label("$P$",P,S); draw(B--O); draw(C--Q); draw(Q--O); draw(O--C); draw(O--A); draw(O--P); draw(O--M, dashed); draw(O--N, dashed); draw(rightanglemark((-2.5, -sqrt(11/4)),(0,0),(-1.89, 2.81),5)); draw(rightanglemark(O,N,B,5)); draw(rightanglemark(B,O,P,5)); draw(rightanglemark(O,M,C,5)); [/asy]

Solution 1

Call the $M$ and $N$ foot of the altitudes from $O$ to $BC$ and $AB$, respectively. Let $OB = r$ . Notice that $\triangle{OMB} \sim \triangle{QOB}$ because both are right triangles, and $\angle{OBQ} \cong \angle{OBM}$. By $\frac{MB}{BO}=\frac{BO}{BQ}$, $MB = r\left(\frac{r}{4.5}\right) = \frac{r^2}{4.5}$. However, since $O$ is the circumcenter of triangle $ABC$, $OM$ is a perpendicular bisector by the definition of a circumcenter. Hence, $\frac{r^2}{4.5} = 2 \implies r = 3$. Since we know $BN=\frac{5}{2}$ and $\triangle BOP \sim \triangle NBO$, we have $\frac{BP}{3} = \frac{3}{\frac{5}{2}}$. Thus, $BP = \frac{18}{5}$. $m + n=\boxed{023}$.

Solution 2

Notice that $\angle{CBO}=90-A$, so $\angle{BQO}=A$. From this we get that $\triangle{BPQ}\sim \triangle{BCA}$. So $\dfrac{BP}{BC}=\dfrac{BQ}{BA}$, plugging in the given values we get $\dfrac{BP}{4}=\dfrac{4.5}{5}$, so $BP=\dfrac{18}{5}$, and $m+n=\boxed{023}$.

Solution 3

Let $r=BO$. Drawing perpendiculars, $BM=MC=2$ and $BN=NA=2.5$. From there, $OM=\sqrt{r^2-4}$. Thus, $OQ=\frac{\sqrt{4r^2+9}}{2}$. Using $\triangle{BOQ}$, we get $r=3$. Now let's find $NP$. After some calculations with $\triangle{BON}$ ~ $\triangle{OPN}$, ${NP=11/10}$. Therefore, $BP=\frac{5}{2}+\frac{11}{10}=18/5$. $18+5=\boxed{023}$.

Solution 4

Let $\angle{BQO}=\alpha$. Extend $OB$ to touch the circumcircle at a point $K$. Then, note that $\angle{KAB}=\angle{CBK}=\angle{QBO}=90^\circ-\alpha$. But since $BK$ is a diameter, $\angle{KAB}=90^\circ$, implying $\angle{CAB}=\alpha$. It follows that $APCQ$ is a cyclic quadrilateral.

Let $BP=x$. By Power of a Point, \[5x=4\cdot\frac 9 2\implies x=\frac{18}{5}.\]The answer is $18+5=\boxed{023}$.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_11&oldid=106841"