2015 AIME II Problems/Problem 11

Revision as of 21:21, 24 August 2019 by Hyxue (talk | contribs) (Solution 5)

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 $18+5=\boxed{023}$.

Solution 5

Note: This is not a very good solution, but it is relatively natural and requires next to no thinking.

Denote the circumradius of $ABC$ to be $R$, the circumcircle of $ABC$ to be $O$, and the shortest distance from $Q$ to circle $O$ to be $x$.

Using Power of a Point on $Q$ relative to circle $O$, we get that $x(x+2r) = 0.5 \cdot 4.5 = \frac{9}{4}$. Using Pythagorean Theorem on triangle $QOB$ to get $(x + r)^2 + r^2 = \frac{81}{4}$. Subtracting the first equation from the second, we get that $2r^2 = 18$ and therefore $r = 3$. Now, set $\cos{ABC} = y$. Using law of cosines on $ABC$ to find $AC$ in terms of $y$ and plugging that into the extended law of sines, we get $\frac{\sqrt{4^2 + 5^2 - 2 \cdot 4 \cdot 5 x}}{\sqrt{1 - x^2}} = 2R = 6$. Squaring both sides and cross multiplying, we get $36x^2 - 40x + 5 = 0$. Now, we get $x = \frac{10 \pm \sqrt{55}}{18}$ using quadratic formula. If you drew a decent diagram, $B$ is acute and therefore $x = \frac{10 + \sqrt{55}}{18}$(You can also try plugging in both in the end and seeing which gives a rational solution). Note that $BP = 3\frac{1}{\sin{OPB}} = \frac{3}{\cos{\angle ABC - \angle QBO}}.$ Using the cosine addition formula and then plugging in what we know about $QBO$, we get that $BP = \frac{162}{2\cos{B} + \sqrt{5}\sin{B}}$. Now, the hard part is to find what $\sin{B}$ is. We therefore want $\frac{\sqrt{324 - (10 + \sqrt{55})^2}}{18} = \frac{\sqrt{169 - 20\sqrt{55}}}{18}$. For the numerator, by inspection $(a + b\sqrt{55})^2$ will not work for integers $a$ and $b$. The other case is if there is $(a\sqrt{5} + b\sqrt{11})^2$. By inspection, $5\sqrt{5} - 2\sqrt{11}$ works. Therefore, plugging all this in yields the answer, $\frac{18}{5} \rightarrow \boxed{23}$. Solution by hyxue

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS