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

Revision as of 17:40, 18 August 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 $\angle{KAC}=\angle{CBK}=\angle{QBO}=90^\circ-\alpha$ . 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 $18+5=\boxed{023}$ .

@@ Line 49: / Line 49: @@
 ==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{KAB}=\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>\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>.

2015 AIME II (Problems • Answer Key • Resources)
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