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 $m+n$.


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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 $m+n = \boxed{023}$.


Video Solution


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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