**Problem 5:**

$PQRS$ is a cyclic quadrilateral, where $PS = SR$. $PR$ and $QS$ intersect each other at point $O$. If $PS = 12$ and $OS = 6$. Find $OQ$.

Please

$PQRS$ is a cyclic quadrilateral, where $PS = SR$. $PR$ and $QS$ intersect each other at point $O$. If $PS = 12$ and $OS = 6$. Find $OQ$.

"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

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- Tahmid Hasan
**Posts:**665**Joined:**Thu Dec 09, 2010 5:34 pm**Location:**Khulna,Bangladesh.

i applied stewart's thm and came up with $OP.OR=108$

so$OS.OQ=108$

but $OS=6$

so $OQ=18$

so$OS.OQ=108$

but $OS=6$

so $OQ=18$

Last edited by Tahmid Hasan on Thu Feb 03, 2011 9:29 pm, edited 1 time in total.

বড় ভালবাসি তোমায়,মা

Stewert's theorem on triangle $ABC$-

$PR(OS^2+OR\times OP)=PS^2\times OR+SR^2\times OP$

$\Rightarrow PR(6^2+OR\times OP)=12^2\times OR+12^2OP$

$\Rightarrow PR(6^2+OR\times OP)=12^2(OR+OP)=12^2\times PR$

$\Rightarrow 6^2+OR\times OP=12^2$

$\Rightarrow OR\times OP=144-36=108$

$\Rightarrow OS\times OQ=108$

$\Rightarrow 6\times OQ=108$

$\Rightarrow OQ=18$

$PR(OS^2+OR\times OP)=PS^2\times OR+SR^2\times OP$

$\Rightarrow PR(6^2+OR\times OP)=12^2\times OR+12^2OP$

$\Rightarrow PR(6^2+OR\times OP)=12^2(OR+OP)=12^2\times PR$

$\Rightarrow 6^2+OR\times OP=12^2$

$\Rightarrow OR\times OP=144-36=108$

$\Rightarrow OS\times OQ=108$

$\Rightarrow 6\times OQ=108$

$\Rightarrow OQ=18$

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Use $L^AT_EX$, It makes our work a lot easier!

Nur Muhammad Shafiullah | Mahi

Use $L^AT_EX$, It makes our work a lot easier!

Nur Muhammad Shafiullah | Mahi

- Tahmid Hasan
**Posts:**665**Joined:**Thu Dec 09, 2010 5:34 pm**Location:**Khulna,Bangladesh.