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BdMO Online Forum • View topic - BDMO national higher secondery -prb 07

BDMO national higher secondery -prb 07

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BDMO national higher secondery -prb 07

Post Number:#1  Unread postby Md Ashraful Kader » Mon Feb 08, 2016 6:05 pm

In triangle ABC, the points A',B',C' are on sides BC,AC,AB respectively.Also,AA',BB',CC' intersect at the point O (they are concurrent at O).Also, (AO/OA'+BO/OB'+CO/OC')=92.Find the value of (AO/OA' . BO/OB' .CO/OC').
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Re: BDMO national higher secondery -prb 07

Post Number:#2  Unread postby asif e elahi » Mon Feb 15, 2016 8:48 am

Assume triangles $COB,COA,AOB$ have area $p,q,r$ respectively. Then express every ratio in terms of $p,q,r$.
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Re: BDMO national higher secondery -prb 07

Post Number:#3  Unread postby SANZEED » Mon Feb 15, 2016 12:04 pm

You can solve this using Mass point geometry as well. Ideas about mass point geometry required for this problem can be found here:

https://en.wikipedia.org/wiki/Mass_point_geometry
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$
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