Search found 13 matches
- Tue Jan 14, 2014 5:55 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2013: Secondary 8, Higher Secondary 6
- Replies: 8
- Views: 12033
Re: BdMO National 2013: Secondary 8, Higher Secondary 6
We induct on n. The result is trivial for $n=1,2$. We assume that it is true for $n=m$. Now we prove that it is true for $n=m+1$. Among $m+1$ cities, if there exists a city which is connected with 0 or 1 city, then there are $m+1$ or m roads among the rest m cities.But by our induction hypothesis, ...
- Tue Jan 14, 2014 4:55 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2013: Secondary 7
- Replies: 3
- Views: 4205
Re: BdMO National 2013: Secondary 7
Since $MX \| AP$, $ \triangle MXD \sim \triangle APD$ Similarly, $ \triangle RSQ \sim \triangle AQB$ and$ \triangle NYX \sim \triangle NPB$ Let the perpendicular distance between $AB$ and $CD$ be $h$ and $MN$ to $CD$ be $i$ Now, we have, $[APD]+[PBC]=.5(h \times AP)+(h \times PB)=.5 \times h\times (...
- Tue Jan 14, 2014 3:45 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2013: Secondary, Higher Secondary 3
- Replies: 4
- Views: 6736
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, $[AFR]+[APB]+[BQC]-[PQMR]=0$
- Mon Jan 13, 2014 3:43 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2013: Secondary 7
- Replies: 3
- Views: 4205
Re: BdMO National 2013: Secondary 7
Is the question correct? Will it be $MX+NY=RS$ or $MY+NX=RS$ ?
- Mon Jan 13, 2014 2:57 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2013: Secondary, Higher Secondary 3
- Replies: 4
- Views: 6736
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?
- Mon Jan 13, 2014 2:26 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2013: Secondary 5
- Replies: 4
- Views: 5130
Re: BdMO National 2013: Secondary 5
I also got 5/12
- Wed Dec 11, 2013 1:28 pm
- Forum: News / Announcements
- Topic: Combinatorics Workshop: Day 8 (11.12.13)
- Replies: 8
- Views: 6925
Re: Combinatorics Workshop: Day 8 (11.12.13)
Thats not what I meant. Its corrected now. Sorry
- Wed Dec 11, 2013 11:54 am
- Forum: News / Announcements
- Topic: Combinatorics Workshop: Day 8 (11.12.13)
- Replies: 8
- Views: 6925
Re: Combinatorics Workshop: Day 8 (11.12.13)
Day after tomorrow 2 exams in school :/
- Sun Sep 01, 2013 11:14 pm
- Forum: News / Announcements
- Topic: Online Geometry Camp 2014 Phase 1
- Replies: 19
- Views: 15788
Re: Online Geometry Camp 2014 Phase 1
I tried something like fusing brahmagupta theorem and sine law. But it got me nowhere. @ mursalin
- Sun Sep 01, 2013 10:19 pm
- Forum: News / Announcements
- Topic: Online Geometry Camp 2014 Phase 1
- Replies: 19
- Views: 15788
Re: Online Geometry Camp 2014 Phase 1
Anyone solved problem no. 4 of day 5?