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In triangle $PQR,$ $M$ is the midpoint of $\overline{PQ}.$  Let $X$ be the point on $\overline{QR}$ such that $\overline{PX}$ bisects $\angle QPR,$ and let the perpendicular bisector of $\overline{PQ}$ intersect $\overline{PX}$ at $Y.$  If $PQ = 36,$ $PR = 22,$ and $QR = 8,$ then find the area of triangle $PQR.$

 Dec 18, 2023
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Since PX bisects ∠QPR, angles QPX and RPX are congruent. Applying the sine law to triangles QPX and RPX, we find

 

[\frac{5}{\sin\angle QPX}=\frac{4}{\sin\angle RPX}=k]so

 

[\sin\angle QPX=\frac{5}{4k},\qquad \sin\angle RPX=\frac{4}{5k}.]Then

 

[\cos\angle QPX = \sqrt{1-\sin^2\angle QPX} = \sqrt{1-\frac{25}{16k^2}} = \frac{\sqrt{16k^2-25}}{4k}]and

 

[\cos\angle RPX = \sqrt{1-\sin^2\angle RPX} = \sqrt{1-\frac{16}{25k^2}} = \frac{\sqrt{25k^2-16}}{5k}.]

 

Since ∠QRP and ∠QXP, ∠RXP are all

 

supplementary to ∠RPX, they all have the same cosine. Therefore,

 

[\cos\angle QXP = \cos\angle RXP = \cos\angle RPX = \frac{\sqrt{25k^2-16}}{5k}]and

 

[\cos\angle QRP = \cos(\angle QXP+\angle RXP) = \cos\angle QXP\cos\angle RXP-\sin\angle QXP\sin\angle RXP.]

 

Substituting, we get

 

\begin{align*} \cos\angle QRP &= \frac{\sqrt{25k^2-16}}{5k} \cdot \frac{\sqrt{25k^2-16}}{5k} - \frac{5}{4k}

 

cdot \frac{\sqrt{16k^2-25}}{5k} \ &= \frac{3k^2}{25k^2} - \frac{5}{20k^2} \

 

&= \frac{2k^2-5}{20k^2} = \frac{k^2-5}{10k^2}. \end{align*}

 

By the Law of Cosines on triangle PQR,

 

[QR^2 = PQ^2 + PR^2 - 2\cdot PQ\cdot PR\cdot \cos\angle QRP.]Substituting, we get

 

[25 = 9 + 16 - 2\cdot3\cdot4\cdot \frac{k^2-5}{10k^2}]so

 

[10k^2 = 25 \cdot 10 + 5 = 255,]which means k2=10255​. Hence,

 

[\cos\angle QRP = \frac{k^2-5}{10k^2} = \frac{250/10}{255/10} = \frac{50}{51}]and

 

[\sin\angle QRP = \sqrt{1-\cos^2\angle QRP} = \sqrt{1-\frac{2500}{2601}} = \frac{\sqrt{601}}{51}.]

 

Finally, the area of triangle PQR is

 

[\frac{1}{2}PQ\cdot PR \cdot \sin\angle QRP =

 

\frac{1}{2}\cdot 3\cdot 4\cdot \frac{\sqrt{601}}{51} =

 

\boxed{\frac{6\sqrt{601}}{51}}.]

 Dec 18, 2023

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