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

2020 CIME I Problems/Problem 9

Revision as of 10:18, 3 September 2021 by Firedragonmath16 (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem 9

Let $ABCD$ be a cyclic quadrilateral with $AB=6, AC=8, BD=5, CD=2$. Let $P$ be the point on $\overline{AD}$ such that $\angle APB = \angle CPD$. Then $\frac{BP}{CP}$ can be expressed in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

[asy] size(4cm); /* draw figures */ draw((0,0)--(1,0)); draw(circle((0.5,-0.12046156424028276), 0.5143063177321622)); draw((0.29472641365670604,0.35110364144073225)--(0,0)); draw((0.29472641365670604,0.35110364144073225)--(0.7999672347109121,0.297307087718076)); draw((0.7999672347109121,0.297307087718076)--(1,0)); draw((0.29472641365670604,0)--(0.29472641365670604,0.35110364144073225)); draw((0.7999672347109121,0.297307087718076)--(0.7999672347109121,-0.297307087718076)); draw((0.7999672347109121,-0.297307087718076)--(0.29472641365670604,0.35110364144073225)); draw(arc((0.5683059275348462,0),0.13393442314476192,127.92567650750303,180)); draw((0.4620043965252797,0.05193209576548634)--(0.4339247468246395,0.06565000785448262)); draw(arc((0.5683059275348462,0),0.13393442314476192,307.925676507503,360)); draw((0.6746074585444126,-0.05193209576548634)--(0.7026871082450528,-0.06565000785448288)); draw((0.7999672347109121,0.297307087718076)--(0.5683059275348462,0)); draw(arc((0.5683059275348462,0),0.13393442314476192,360,412.074323492497)); draw((0.6746074585444126,0.05193209576548634)--(0.702687108245053,0.06565000785448288)); /* dots and labels */ dot((0,0)); label("$A$", (0,0), NW); dot((1,0)); label("$D$", (1,0), NE); dot((0.29472641365670604,0.35110364144073225)); label("$B$", (0.29472641365670604,0.35110364144073225), N); dot((0.7999672347109121,0.297307087718076)); label("$C$", (0.7999672347109121,0.297307087718076), N); dot((0.29472641365670604,0)); label("$X$", (0.29472641365670604,0), SW); dot((0.7999672347109121,0)); label("$Y$", (0.7999672347109121,0), SE); dot((0.7999672347109121,-0.297307087718076)); label("$C'$", (0.7999672347109121,-0.297307087718076), S); dot((0.5683059275348462,0)); label("$P$", (0.5683059275348462,0), SW); [/asy]

Let $C'$ be the reflection of $C$ over line $AD$. Since $\angle APB = \angle CPD = \angle C'PD$, $B, P, C$ are collinear. Suppose $X$ and $Y$ are the projections of $B$ and $C$ onto line $AD$, respectively. We want to find $\frac{BP}{CP}$ which by similar triangles is also equal to $\frac{BX}{C'Y}$ from $\triangle BPX \sim \triangle C'PY$. Since $C'Y=CY$, this also equals $\frac{BX}{CY}$. We know that $\triangle ABD$ and $\triangle ACD$ each share the same base, so this can also be interpreted as $\frac{[ABD]}{[ACD]}$. The sine area formula gives \[\frac{[ABD]}{[ACD]} = \frac{\frac{1}{2} \cdot 6 \cdot 5 \sin ABD}{\frac{1}{2} \cdot 8 \cdot 2 \sin ACD}.\] Quadrilateral $ABCD$ is cyclic, so $\angle ABD = \angle ACD$ because both angles subtend arc $\widehat{AD}$ on the circumcircle of Quadrilateral $ABCD$. We can then replace every $\angle ACD$ with $\angle ABD$, but realise that if we do that, the $\angle ABD$s will cancel out. The requested area ratio is thus \[\frac{\frac{1}{2} \cdot 6 \cdot 5}{\frac{1}{2} \cdot 8 \cdot 2} = \frac{15}{8}\]. The answer is $15+8=\boxed{023}$.

Solution 2 (Law of Sines)

We look for the ratio $\frac{BP}{CP}$ so thus we use the Law of Sines since it involves ratios.

By the Law of Sines used on $\triangle APB$ and $\triangle DPC$,

\[\frac{6}{\sin \angle APB} = \frac{BP}{\sin A}\]\[\frac{2}{\sin \angle DPC} = \frac{CP}{\sin A}\]Since $\angle APC = \angle APB$ implies $\sin \angle APB = \sin \angle DPC$, this implies

\[\frac{BP}{CP} = 3 \cdot \frac{\sin A}{\sin D}\]Now we just need to find $\frac{\sin A}{\sin D}$ or its reciprocal to get the answer.

We use Law of Sines again on $\triangle ABD$ and $\triangle ACD$ as follows:

\[\frac{\sin A}{5} = \frac{\sin \angle ABD}{AD}\]\[\frac{8}{\sin D} = \frac{AD}{\sin \angle ACD}\]Hence $\frac{\sin A}{\sin D} = \frac{8}{5}$.

Thus $\frac{BP}{CP} = 3 \cdot \frac{5}{8} = \boxed{\frac{15}{8}}$.

The answer is $m+n = \boxed{023}$.

~FIREDRAGONMATH16

Video Solution

https://www.youtube.com/watch?v=atUCE3oSieg&lc=UgwRISSUhBk6GBF9g294AaABAg

See also

2020 CIME I (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All CIME Problems and Solutions

The problems on this page are copyrighted by the MAC's Christmas Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_CIME_I_Problems/Problem_9&oldid=161490"