Circle Chord Midpoint Intersection: Length Comparison of RS and PM

[anemone]

Here is this week's POTW:

-----Let $l$ be a circle with center $O$ and let $AB$ be a chord of $l$ whose midpoint $M$ is distinct from $O$.

The ray from $O$ through $M$ meets $l$ again at $R$. Let $P$ be a point on the minor arc $AR$ of $l$, let $PM$ meet $l$ again at $Q$, and let $AB$ meets $QR$ at $S$. Which segment is longer, $RS$ or $PM$?

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

No one answered last week's problem. :(

You can find the proposed solution below:

Spoiler

View attachment 5142

Extend $RO$ to meet the circle $l$ at $T$. Note that $\angle RMS$, $\angle RQT$ are right angles.

Let $\angle MRP=\theta$, $\angle MRS=\alpha$. Then we have $RS=\dfrac{RM}{\cos \alpha}$ and $\angle RPQ=\angle RTQ=90^{\circ}-\alpha$.

Consequently, by the Sine Rule we get:

$\dfrac{PM}{\sin \theta}=\dfrac{RM}{\sin (90^{\circ}-\alpha)}=\dfrac{RM}{\cos \alpha}=RS$.

Since $0<\alpha<90^\circ$, we can conclude by now that $PM<RS$.