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

Revision as of 10:14, 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 $. Using$ \triangle{BON}~\triangle{OPN}$, we get {NP=\frac{2}{5}r^2-\frac{5}{2}$ (Error compiling LaTeX. Unknown error_msg). Therefore, $BP=\frac{5}{2}+\frac{11}{10}=18/5$ . $18+5=\boxed{023}$ .

@@ 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>.
+==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==
 {{AIME box|year=2015|n=II|num-b=10|num-a=12}}
 {{MAA Notice}}

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