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Given a circle with center $O$. Tangents $PM$ and $PN$ are drawn from an external point $P$ to this circle. $Q\in PM$ and $R\in PN$ are points such that $O$ is the midpoint of $QR$. $X\in PM$ and $Y\in PN$ are points such that $XY$ is tangent to the circle. Given $QR=12$, find the value of $QX\cdot RY$ and show that it doesn't depend on $XY$.

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  • $\begingroup$ What have you done so far? Did you at least draw a picture? $\endgroup$ – Andrei Dec 27 '16 at 19:59
  • $\begingroup$ Yes I have. Can't proceed from there. $\endgroup$ – user402283 Dec 27 '16 at 20:10
  • $\begingroup$ I would add after @Andrei: can you show us a picture ? $\endgroup$ – Jean Marie Dec 27 '16 at 20:16
  • $\begingroup$ Added to the post. $\endgroup$ – user402283 Dec 27 '16 at 20:22
  • $\begingroup$ Do we agree that OP is not in general orthogonal to OQ ? (they are on your figure) $\endgroup$ – Jean Marie Dec 27 '16 at 20:33
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Angles are directed modulo $\pi$ to avoid configuration issues.

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Reflect $Y$ over $OP$. Then $\triangle OYR\cong\triangle OY'Q$.

Since $YX$ and $YR$ are both tangents erected from $Y$ we have $\measuredangle XYO=\measuredangle OYR$ and from the congruence, $\measuredangle OYR=\measuredangle QY'O=\measuredangle XY'O$. So $\measuredangle XYO=\measuredangle XY'O$ implying $OYXY'$ is cyclic.

$YY'\parallel QR$ both being perpendicular to $OP$, and implying $\measuredangle YY'X=\measuredangle OQX$.

Now in $\triangle QXO$ and $\triangle OXY$ we have $\measuredangle QXO=\measuredangle OXY$ since $XQ$ and $XY$ both are tangents erected from $X$, and $\measuredangle YOX=\measuredangle YY'X=\measuredangle OQX$. So $\triangle QXO\sim \triangle OXY$. Similarly $\triangle ROY\sim \triangle OXY$.

Hence $\triangle QXO\sim\triangle ROY$, implying $$\dfrac{QX}{OQ}=\dfrac{RO}{YR}~\implies QX\cdot RY = OQ\cdot OR=OQ^2.~\square$$

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It should be clear that (1) all same color marked angle pairs are equal; (2) $\angle PXY = 2\beta$ and $\angle PYX = 2\alpha$.

Through x, we get $\alpha + \beta = \theta$.

∴ $\triangle ORY \sim \triangle XQO$ because they both are similar to $\triangle XOY$.

∴ $\dfrac {OR}{QX} = \dfrac {RY}{OQ}$

Result follows.

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