## BdMO National 2013: Secondary, Higher Secondary 3

### 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 $X$.

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

I didnt need the $AB=7$ part. My answer is $0$

Am I missing anything?

Am I missing anything?

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

@Siam:
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

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

Nur Muhammad Shafiullah | Mahi

### 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, $[AFR]+[APB]+[BQC]-[PQMR]=0$

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

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

**Answer**

I used this figure and my answer is $\fbox 0$