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

(Solution 3)
(Solution 3)
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==Solution 3==
 
==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}</math>, we get <math>r=3</math>. Now let's find <math>NP</math>. Using <math>\triangle{BON}~\triangle{OPN}</math>, <math>{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>.
+
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}</math>, we get <math>r=3</math>. Now let's find <math>NP</math>. After some calculations with <math>\triangle{BON}</math> ~ <math>\triangle{OPN}</math>, <math>{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 10:15, 29 August 2017

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 $OB = r$ and let $OQ = k$. Notice that $\triangle{OMB} \sim \triangle{OQB}$ because both are right triangles, and $\angle{OBQ} \cong \angle{OBM}$. Then, $MB = r\left(\frac{r}{4.5}\right) = \frac{r^2}{4.5}$. 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

Using the diagram given, $r=BO$. Since $O$ bisects any chord, $BM=MC=2$ 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 $\triangle{BON}$ ~ $\triangle{OPN}$, ${NP=\frac{2}{5}r^2-\frac{5}{2}}$. Therefore, $BP=\frac{5}{2}+\frac{11}{10}=18/5$. $18+5=\boxed{023}$.

See also

2015 AIME II (ProblemsAnswer KeyResources)
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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