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Problem Of The Week #440 October 26th, 2020

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
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!
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
Hello all!(Smile)

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

MHB POTW Director
Staff member
Feb 14, 2012
3,894
No one answered last two week's POTW.(Sadface) 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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