# 2020 CIME I Problems/Problem 9

## 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

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