Difference between revisions of "2020 CIME I Problems/Problem 9"

Latest revision as of 10:18, 3 September 2021

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

@@ Line 42: / Line 42: @@
 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 \sim \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>.
+==Solution 2 (Law of Sines)==
+We look for the ratio <math>\frac{BP}{CP}</math> so thus we use the Law of Sines since it involves ratios.
+By the Law of Sines used on <math>\triangle APB</math> and <math>\triangle DPC</math>,
+<cmath>\frac{6}{\sin \angle APB} = \frac{BP}{\sin A}</cmath><cmath>\frac{2}{\sin \angle DPC} = \frac{CP}{\sin A}</cmath>Since <math>\angle APC = \angle APB</math> implies <math>\sin \angle APB = \sin \angle DPC</math>, this implies
+<cmath>\frac{BP}{CP} = 3 \cdot \frac{\sin A}{\sin D}</cmath>Now we just need to find <math>\frac{\sin A}{\sin D}</math> or its reciprocal to get the answer.
+We use Law of Sines again on <math>\triangle ABD</math> and <math>\triangle ACD</math> as follows:
+<cmath>\frac{\sin A}{5} = \frac{\sin \angle ABD}{AD}</cmath><cmath>\frac{8}{\sin D} = \frac{AD}{\sin \angle ACD}</cmath>Hence <math>\frac{\sin A}{\sin D} = \frac{8}{5}</math>.
+Thus <math>\frac{BP}{CP} = 3 \cdot \frac{5}{8} = \boxed{\frac{15}{8}}</math>.
+The answer is <math>m+n = \boxed{023}</math>.
+~FIREDRAGONMATH16
+==Video Solution==
+https://www.youtube.com/watch?v=atUCE3oSieg&lc=UgwRISSUhBk6GBF9g294AaABAg
+==See also==
 {{CIME box|year=2020|n=I|num-b=8|num-a=10}}
 [[Category:Intermediate Geometry Problems]]
 {{MAC Notice}}

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