A nice problem

For students of class 11-12 (age 16+)
User avatar
Phlembac Adib Hasan
Posts: 1016
Joined: Tue Nov 22, 2011 7:49 pm
Location: 127.0.0.1
Contact:

A nice problem

Unread post by Phlembac Adib Hasan » Tue Jan 03, 2012 10:01 am

(From Geometry Revisited) $ABC$ is a triangle and let $G$ is its centroid.A line through $G$ meets $AB$ and $AC$ at point $P$ and $Q$, respectively.Prove that (with respect to sign) \[ \frac {BP} {PA} + \frac { CQ } {QA} =1 \]
Welcome to BdMO Online Forum. Check out Forum Guides & Rules

User avatar
Nadim Ul Abrar
Posts: 244
Joined: Sat May 07, 2011 12:36 pm
Location: B.A.R.D , kotbari , Comilla

Re: A nice problem

Unread post by Nadim Ul Abrar » Tue Jan 03, 2012 4:26 pm

Note that $PA=BP+2PE \Rightarrow \frac{BP}{PA}+\frac{2PE}{PA}=1 $

Now via menelaus we have $\frac{QA.CG.PE}{CQ.EG.PA}=1 \Rightarrow \frac{CQ}{QA}=\frac{2PE}{AP}$.
$\frac{1}{0}$

User avatar
Phlembac Adib Hasan
Posts: 1016
Joined: Tue Nov 22, 2011 7:49 pm
Location: 127.0.0.1
Contact:

Re: A nice problem

Unread post by Phlembac Adib Hasan » Tue Jan 03, 2012 6:39 pm

My proof was also the same. This problem taught me Manelaus. :D
Welcome to BdMO Online Forum. Check out Forum Guides & Rules

Post Reply