# Problem Of The Week #440 October 26th, 2020

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#### anemone

##### MHB POTW Director
Staff member
Here is this week's POTW:

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In a convex quadrilateral $PQRS$, $PQ=RS$, $(\sqrt{3}+1)QR=SP$ and $\angle RSP-\angle SPQ=30^{\circ}$. Prove that $\angle PQR-\angle QRS=90^{\circ}$.

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Remember to read the POTW submission guidelines to find out how to submit your answers!

#### anemone

##### MHB POTW Director
Staff member
Hello all!

I am going to give the members another week's time to take a stab at last week's POTW. I am looking forward with hope to receive an answer soon.

#### anemone

##### MHB POTW Director
Staff member
No one answered last two week's POTW. However, for those who are interested, you can check the suggested solution by other as follows:
\begin{tikzpicture}

\coordinate[label=above: P] (P) at (2,3);
\coordinate[label=above:Q] (Q) at (4,6);
\coordinate[label=right:R] (R) at (12, 0);
\coordinate[label=below: S] (S) at (7.333333,0);
\draw (P) -- (S)-- (R)-- (Q)--(P);
\draw (P) -- (R);
\draw (Q) -- (S);

\end{tikzpicture}

Let $[\text{figure}]$ denote the area of figure. We have

$[PQRS]=[PQR]+[RSP]=[QRS]+[SPQ]$

Let $PQ=p,\,QR=q,\,RS=r,\,SP=s$. The above relations reduce to

$pq\sin\angle PQR+rs\sin \angle RSP=qr\sin\angle QRS+sp\sin \angle SPQ$

Using $p=r$ and $(\sqrt{3}+1)q=s$ and dividing by $pq$, we get

$\sin \angle PQR+(\sqrt{3}+1)\sin \angle RSP=\sin \angle QRS+(\sqrt{3}+1)\sin \angle SPQ$

Therefore, $\sin \angle PQR-\sin \angle QRS=(\sqrt{3}+1)(\sin \angle SPQ-\sin \angle RSP)$

This can be written in the form

$2\sin \dfrac{\angle PQR+\angle QRS}{2}\cos \dfrac{\angle PQR+\angle QRS}{2}=(\sqrt{3}+1)2\sin \dfrac{\angle SPQ-\angle RSP}{2}\cos \dfrac{\angle SPQ+\angle RSP}{2}$

Using the relations

$\cos \dfrac{\angle PQR+\angle QRS}{2}=-\cos \dfrac{\angle SPQ+\angle RSP}{2}$

and

$\cos \dfrac{\angle SPQ-\angle RSP}{2}=-\sin 15^{\circ}=- \dfrac{\sqrt{3}-1}{2\sqrt{2}}$

we obtain
$\sin \dfrac{\angle PQR-\angle QRS}{2}=(\sqrt{3}-1)\left(-\dfrac{\sqrt{3}-1}{2\sqrt{2}}\right)=\dfrac{1}{\sqrt{2}}$

This shows that

$\dfrac{\angle PQR-\angle QRS}{2}=\dfrac{\pi}{4}$ or $\dfrac{3\pi}{4}$

Using the convexity of $PQRS$, we can rule out the latter alternative. We obtain

$\angle PQR-\angle QRS=\dfrac{\pi}{2}$

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