## BdMO National 2013: Secondary, Higher Secondary 3

BdMO
Posts: 134
Joined: Tue Jan 18, 2011 1:31 pm

### BdMO National 2013: Secondary, Higher Secondary 3

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
Posts: 13
Joined: Sun Mar 17, 2013 3:36 pm

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

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

*Mahi*
Posts: 1175
Joined: Wed Dec 29, 2010 12:46 pm
Location: 23.786228,90.354974
Contact:

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

@Siam:
Posting your approach would be nice.

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

Siam
Posts: 13
Joined: Sun Mar 17, 2013 3:36 pm

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

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\$

samiul_samin
Posts: 1007
Joined: Sat Dec 09, 2017 1:32 pm