Search found 4 matches
- Mon Apr 05, 2021 12:16 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Junior 2020 P2
- Replies: 1
- Views: 2467
Re: BdMO National Junior 2020 P2
$\triangle AGE$ is similar to $\triangle GBC$. Since, $AE = \frac{1}{2} AD = \frac{1}{2} BC$, $AG = \frac{1}{2}BG $ or $AG = AB$. Again, $\triangle FHD$ is similar to $\triangle HBC$ and $FD = \frac{1}{4}AD $. So we have $DH = \frac{1}{3}CD$ Now, $BG = AB+AG = 2AB$ whereas $CH = CD+DH= \frac{4}{3}CD...
- Mon Apr 05, 2021 11:51 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Junior 2020 P7
- Replies: 2
- Views: 2543
Re: BdMO National Junior 2020 P7
Let $n$ be any three digit number formed by $PQR$. The lowest possible value for $STU$ is $100$ whereas the highest possible value is $999$. Thus thee lowest possible sum is $(n+100)$ whereas the highest possible sum is $(n + 999)$. Range of the sum is $$999 - 100 = 899$$. Since $$\frac{899}{37} =24...
- Sun Apr 04, 2021 11:05 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Junior 2020 P9
- Replies: 3
- Views: 3223
Re: BdMO National Junior 2020 P9
For $k$ being the number chosen for elimination, we can denote the number of balls in the piles initially as $1,2,3,...,k,(k+1),...,(k+m)$. Therefore, after eliminating $k$ coins by the given rule we are left with $1,2,...,(k-1),0,1,2,...,n$ balls altogether. Observation: In the ultimate step we are...
- Sun Apr 04, 2021 11:36 am
- Forum: Secondary Level
- Topic: problem from selection olympiad
- Replies: 1
- Views: 5579
Re: problem from selection olympiad
Let us assume you were asleep for $t$ hours. The first candle loses $\frac{1}{9}$ part and the second candle loses $\frac{1}{6}$ part in 1 hour. So if you are asleep for $t$ hours, $1^{st}$ candle would have $$1-\frac{t}{9}$$ part remaining whereas the $2^{nd}$ candle will have $$1-\frac{t}{6}$$ par...