2020 CIME I Problems/Problem 9

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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 $15+8=\boxed{023}$.

Video Solution

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

See also

2020 CIME I (ProblemsAnswer KeyResources)
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

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