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Difference between revisions of "2020 CIME I Problems/Problem 9"
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==Solution==
 
==Solution==
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{{solution}}
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 +
Let <math>C'</math> be the reflection of <math>C</math> over line <math>AD</math>. Since <math>\angle APB = \angle CPD = \angle C'PD</math>, <math>B, P, C</math> are collinear. Suppose <math>X</math> and <math>Y</math> are the projections of <math>B</math> and <math>C</math> onto line <math>AD</math>, respectively. We want to find <math>\frac{BP}{CP}</math> which by similar triangles is also equal to <math>\frac{BX}{C'Y}</math> from <math>\triangle BPX ~ \triangle C'PY</math>. Since <math>C'Y=CY</math>, this also equals <math>\frac{BX}{CY}</math>. We know that <math>\triangle ABD</math> and <math>triangle ACD</math> each share the same base, so this can also be interpreted as <math>\frac{[ABD]}{[ACD]}</math>. The sine area formula gives <cmath>\frac{[ABD]}{[ACD]} = \frac{\frac{1}{2} \cdot 6 \cdot 5 \sin ABD}{\frac{1}{2} \cdot 8 \cdot 2 \sin ACD}.</cmath> Quadrilateral <math>ABCD</math> is cyclic, so <math>\angle ABD = \angle ACD</math> because both angles subtend arc <math>\widehat{AD}</math> on the circumcircle of Quadrilateral <math>ABCD</math>. We can then replace every <math>\angle ACD</math> with <math>\angle ABD</math>, but realise that if we do that, the <math>\angle ABD</math>s will cancel out. The requested area ratio is thus <cmath>\frac{\frac{1}{2} \cdot 6 \cdot 5}{\frac{1}{2} \cdot 8 \cdot 2} = \frac{15}{8}</cmath>. The answer is <math>15+8=\boxed{023}</math>.
  
 
{{CIME box|year=2020|n=I|num-b=8|num-a=10}}
 
{{CIME box|year=2020|n=I|num-b=8|num-a=10}}
 +
 +
[[Category:Intermediate Geometry Problems]]
 
{{MAC Notice}}
 
{{MAC Notice}}

Revision as of 17:54, 6 September 2020

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 $\triangle BPX ~ \triangle C'PY$. 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 $15+8=\boxed{023}$.

2020 CIME I (Problems • Answer Key • Resources)
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Problem 8 		Followed by
Problem 10
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