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Rangpur Secondary 2011/5
Posted: Wed Feb 02, 2011 6:49 pm
by Moon
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 $OQ$.
Re: Rangpur Secondary 2011/5
Posted: Wed Feb 02, 2011 7:45 pm
by Tahmid Hasan
i applied stewart's thm and came up with $OP.OR=108$
so$OS.OQ=108$
but $OS=6$
so $OQ=18$
Re: Rangpur Secondary 2011/5
Posted: Thu Feb 03, 2011 9:01 pm
by *Mahi*
Stewert's theorem on triangle $ABC$-
$PR(OS^2+OR\times OP)=PS^2\times OR+SR^2\times OP$
$\Rightarrow PR(6^2+OR\times OP)=12^2\times OR+12^2OP$
$\Rightarrow PR(6^2+OR\times OP)=12^2(OR+OP)=12^2\times PR$
$\Rightarrow 6^2+OR\times OP=12^2$
$\Rightarrow OR\times OP=144-36=108$
$\Rightarrow OS\times OQ=108$
$\Rightarrow 6\times OQ=108$
$\Rightarrow OQ=18$
Re: Rangpur Secondary 2011/5
Posted: Thu Feb 03, 2011 9:29 pm
by Tahmid Hasan
did the same mistake again
