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, $[AFR]+[APB]+[BQC]-[PQMR]=0$
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Re: BdMO National 2013: Secondary, Higher Secondary 3
Answer
I used this figure and my answer is $\fbox 0$
I used this figure and my answer is $\fbox 0$