BdMO National 2013: Secondary, Higher Secondary 3

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BdMO
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BdMO National 2013: Secondary, Higher Secondary 3

Unread post by BdMO » Fri Jan 10, 2014 1:35 am

Let $ABCDEF$ be a regular hexagon with $AB=7$. $M$ is the midpoint of $DE$. $AC$ and $BF$ intersect at $P$, $AC$ and $BM$ intersect at $Q$, $AM$ and $BF$ intersect at $R$. Find the value of $[APB]+[BQC]+[ARF]-[PQMR]$. Here $[X]$ denotes the area of polygon $X$.

Siam
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Re: BdMO National 2013: Secondary, Higher Secondary 3

Unread post by Siam » Mon Jan 13, 2014 2:57 pm

I didnt need the $AB=7$ part. My answer is $0$
Am I missing anything?

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*Mahi*
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Re: BdMO National 2013: Secondary, Higher Secondary 3

Unread post by *Mahi* » Tue Jan 14, 2014 2:53 pm

@Siam:
The answer is actually $0$.
Posting your approach would be nice.
Please read Forum Guide and Rules before you post.

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

Nur Muhammad Shafiullah | Mahi

Siam
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Re: BdMO National 2013: Secondary, Higher Secondary 3

Unread post by Siam » Tue Jan 14, 2014 3:45 pm

Since $ABCDEF$ is a regular hexagon, $AB \| FC \| DE$
So, $[AFB]=[ACB]$
Again, height of $[AFB]$ is half of $[AMB]$.
So $[AMB]=2[AFB]$
Or, $[AMB]=[AFB]+[ACB]$
Or, $[AFR]+[ARP]+[APB]+[APB]+[APQ]+[BQC]
=[APB]+[ARP]+[BPQ]+[PQMR]$
Or, $[AFR]+[APB]+[BQC]=[PQMR]$
Or, $[AFR]+[APB]+[BQC]-[PQMR]=0$

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samiul_samin
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Re: BdMO National 2013: Secondary, Higher Secondary 3

Unread post by samiul_samin » Sat Feb 24, 2018 12:06 am

Answer
I used this figure and my answer is $\fbox 0$
Screenshot_2018-02-23-23-38-03-1.png

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