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Source: Mexican Math Olympiad 2012 - problem 1

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#1 h Dec 1, 2013, 6:06 PM • 3 Y

Y by mathsolver_, Adventure10, Mango247

Let $\mathcal{C}_1$ be a circumference with center $O$ , $P$ a point on it and $\ell$ the line tangent to $\mathcal{C}_1$ at $P$ . Consider a point $Q$ on $\ell$ different from $P$ , and let $\mathcal{C}_2$ be the circumference passing through $O$ , $P$ and $Q$ . Segment $OQ$ cuts $\mathcal{C}_1$ at $S$ and line $PS$ cuts $\mathcal{C}_2$ at a point $R$ diffferent from $P$ . If $r_1$ and $r_2$ are the radii of $\mathcal{C}_1$ and $\mathcal{C}_2$ respectively, Prove
$\frac{PS}{SR} = \frac{r_1}{r_2}.$

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#2 h Dec 1, 2013, 6:57 PM • 1 Y

Y by Adventure10

Since $\triangle OPR$ is a right triangle with hypotense $OQ$ , $r_2=\tfrac 12 OQ$ , so it suffices to show that $\dfrac{PS}{SR}=\dfrac{OP}{\tfrac12OQ}$ . Let $X$ denote the projection of $O$ to the segment $PS$ . Remark that since $OP=OS$ and $\triangle OPS\sim\triangle RQS$ , we have $\angle OSP=\angle RSQ=\angle RQS$ . Furthermore, since $PR$ is tangent to $\mathcal{C}_1$ and since $PQRO$ is cyclic, we have $\angle ROQ=\angle SPR=\tfrac12\angle SOP=\angle SOX.$ Therefore $\triangle SXO\sim\triangle QRO$ and $\dfrac{QR}{QO}=\dfrac{SX}{SO}=\dfrac{\tfrac12 PS}{PO}\implies \dfrac{OP}{\tfrac12OQ}=\dfrac{PS}{QR}=\dfrac{PS}{SR},$ as desired. $\blacksquare$

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#3 h Dec 3, 2013, 1:01 AM • 2 Y

Y by Adventure10, Mango247

Already posted http://www.artofproblemsolving.com/Forum/viewtopic.php?f=125&t=538043

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sayantanchakraborty

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sayantanchakraborty

#4 h Mar 3, 2014, 7:39 PM • 2 Y

Y by Adventure10, Mango247

Let $\angle POQ=2\theta$ .Then $\angle{ORS}=90-2\theta=\angle {OQP}$ and $\angle{OSR}=\theta$ .Thus $PS=2r_1sin\theta$ Also from $\triangle{ROS}=SR=\frac{r_1}{2sin\theta}=2r_2sin\theta$ .(Clearly we get the relation between $r_1$ and $r_2$ from $\triangle{OQP}$ ).Thus $\frac{PS}{SR}=\frac{r_1}{r_2}$ and we are done.

Maths is the doctor of science....

Sayantan....

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codyj

#5 h Aug 8, 2014, 7:02 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

$\theta=\angle POQ$ , $PS=r_1\sqrt{2(1-\cos\theta)}$ , $SR=\frac{r_1\sin\frac\theta2}{\cos\theta}$ , $\frac{PS}{SR}=2\cos\theta$ , $\frac{r_1}{r_2}=\frac{2OP}{OQ}=2\cos\theta$

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