Rangpur Secondary 2011/5

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Moon
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Rangpur Secondary 2011/5

Unread post by Moon » Wed Feb 02, 2011 6:49 pm

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 $OQ$.
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Tahmid Hasan
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Re: Rangpur Secondary 2011/5

Unread post by Tahmid Hasan » Wed Feb 02, 2011 7:45 pm

i applied stewart's thm and came up with $OP.OR=108$
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.
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*Mahi*
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Re: Rangpur Secondary 2011/5

Unread post by *Mahi* » Thu Feb 03, 2011 9:01 pm

Stewert's theorem on triangle $ABC$-
$PR(OS^2+OR\times OP)=PS^2\times OR+SR^2\times OP$
$\Rightarrow PR(6^2+OR\times OP)=12^2\times OR+12^2OP$
$\Rightarrow PR(6^2+OR\times OP)=12^2(OR+OP)=12^2\times PR$
$\Rightarrow 6^2+OR\times OP=12^2$
$\Rightarrow OR\times OP=144-36=108$
$\Rightarrow OS\times OQ=108$
$\Rightarrow 6\times OQ=108$
$\Rightarrow OQ=18$
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Tahmid Hasan
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Re: Rangpur Secondary 2011/5

Unread post by Tahmid Hasan » Thu Feb 03, 2011 9:29 pm

did the same mistake again :P
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