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$ABCD$ is a cyclic quadrilateral with $AB=11$ and $CD= 19$. Points $P$ and $Q$ are on $AB$ and $CD$ respectively such that $AP=6,\, BP=5\, DQ=7$ and $CQ=12,\, PQ=27$. Extend $PQ$ till it meets the circle at point $R$ and point $S$.

Find $RS$. Given Picture!

This is what I have so far:

  • I constructed chords $DS,\, CS,\, AD,\, BC,\, AR,\, RB$ and $BC$
  • Ptolemy in ABCD $AB\times DC+ AD \times BC= AC \times BD$
  • I think Ptolemy needs to be done multiple times in all the cyclic quads i just don't know where to start or how to do that

Thank you so much!

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  • $\begingroup$ Such a nice diagram! I just recently found out about the software called "Geogeba". Is that what you used to make the diagram? $\endgroup$ – quasi Apr 18 '17 at 17:24
  • $\begingroup$ @Olivia Ryan How did you end up with all integers for segments? It would not be by chance ... Maybe remaining sides are also integers? $\endgroup$ – Narasimham Apr 18 '17 at 19:14
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Ptolemy's theorem is not really needed to solve the problem.

Let $PR=x$ and $SQ=y$. We have $7\cdot 12=y(27+x)$ and $5\cdot 6=x(27+y)$.
Can you guess why and what $x+y+27$ is, as a consequence?

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  • $\begingroup$ Nice. I posted an answer (now deleted). I started the same way, but I missed the trick that one only needs to find the value of $x+y$, not the values of $x,y$. $\endgroup$ – quasi Apr 18 '17 at 17:06
  • $\begingroup$ Thank you! but how did you get those equations? what properties did you use? $\endgroup$ – Parley Apr 18 '17 at 17:06
  • $\begingroup$ @Olivia Ryan -- Power of a Point. $\endgroup$ – quasi Apr 18 '17 at 17:07
  • $\begingroup$ @quasi for point S and point R? $\endgroup$ – Parley Apr 18 '17 at 17:08
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    $\begingroup$ @Olivia Ryan -- en.wikipedia.org/wiki/Intersecting_chords_theorem $\endgroup$ – quasi Apr 18 '17 at 17:08
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No need for Ptolemy's Theorem.

Instead, use Power of a Point . . .

Let $x = RP$ and $y = QS$.

Applying Power of a Point to the intersecting chords $AB$, $RS$, we get

$$(x)(y+27) = (5)(6)$$

Applying Power of a Point to the intersecting chords $CD$, $RS$, we get

$$(y)(x + 27) = (7)(12)$$

So now you have two equations in two unknowns. I'll leave the rest as an outline ...

  • Subtract the equations to get a linear equation.$\\[4pt]$
  • Solve for one of the unknowns in terms of the other.$\\[4pt]$
  • Substitute the result into one of the original equations, thus getting a quadratic equation in one unknown.$\\[4pt]$
  • By symmetry, $x,y$ are both roots of the same quadratic.$\\[4pt]$
  • Since you only need the value of $x+y$ (thanks to Jack D'Aurizio for that observation), just use Vieta's formula to get the sum of the roots.$\\[4pt]$
  • Finally, add $27$.
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  • $\begingroup$ is there any way you can explain where to use Ptolemy? $\endgroup$ – Parley Apr 18 '17 at 17:23
  • $\begingroup$ I didn't use Ptolemy's Theorem. Power of a Point is a natural choice for this question. It can be proved using similar triangles -- it's not hard. In a course that discusses cyclic quadrilaterals, I'm surprised it hasn't already been introduced. $\endgroup$ – quasi Apr 18 '17 at 17:26

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