BdMO National Secondary: Problem Collection(2016)

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
User avatar
Tasnood
Posts: 73
Joined: Tue Jan 06, 2015 1:46 pm

Re: BdMO National Secondary: Problem Collection(2016)

Unread post by Tasnood » Mon Feb 05, 2018 10:39 pm

5(b)
We can set the Earthlings in $n!$ ways if they are in straight line. But considering rotation, there exists a group of $n$ arrangements, that are actually same if we rotate them.
If $n=3$, rearrangements are: $ABC,BCA,CAB,ACB,CBA,BAC$, where the first three are same and so are the last three.
So, actual arrangements=$2$=$\frac{3!}{3}$. We can say:$\frac{n!}{n}$ [Can check for $n=4$]

The rest is same. $m$ Martians can be placed in $n$ places in $n \choose m$ ways. And, the Martians can be arranged in $m!$ ways.

So, the result=$\frac{n!}{n}.m!.{n \choose m}$=$\frac{n.{(n-1)!}^2}{(n-m)!}$

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

Re: BdMO National Secondary: Problem Collection(2016)

Unread post by samiul_samin » Tue Feb 13, 2018 8:37 am

Where is the problem 7?

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

Re: BdMO National Secondary: Problem Collection(2016)

Unread post by samiul_samin » Thu Feb 15, 2018 9:39 am

Here you can get the full solution of No,7
Zawadx wrote:
Fri Aug 05, 2016 10:15 am
This is actually a well-known problem. There's even a Numberphile video on it:
Problem video: https://www.youtube.com/watch?v=ZWib5olGbQ0
Mathematical explanation: https://www.youtube.com/watch?v=XIOoCKO-ybQ

The BdMO question also outlines all the steps, so you could try to just solve by doing it like an exercise ;/

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

Re: BdMO National Secondary: Problem Collection(2016)

Unread post by samiul_samin » Sun Mar 04, 2018 8:25 am

Problem 3.

prottoy das
Posts: 17
Joined: Thu Feb 01, 2018 11:28 am
Location: Sylhet

Re: BdMO National Secondary: Problem Collection(2016)

Unread post by prottoy das » Mon Feb 18, 2019 1:35 pm

Let $CH\perp AB$. $BI$ & $DO$ meet at $E$. Now $BIOD$ cyclic so, $\angle IBD=\angle EOI$. Again, as $\angle ABC$ is bisected by $BI$ so $\angle IBD= \angle IBH$. So,$\angle IBH=\angle EBH=\angle EOH$, so the quadrangle $EDBH$ is cyclic. So $\angle BEO=\angle BHO=90^\circ$

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

Re: BdMO National Secondary: Problem Collection(2016)

Unread post by samiul_samin » Mon Feb 18, 2019 2:45 pm

samiul_samin wrote:
Sun Mar 04, 2018 8:25 am
Problem 3.
This link is corrupted.Here you can get the solution of problem no 3.

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

Re: BdMO National Secondary: Problem Collection(2016)

Unread post by samiul_samin » Mon Feb 18, 2019 2:49 pm

prottoy das wrote:
Mon Feb 18, 2019 1:35 pm
Let $CH\perp AB$. $BI$ & $DO$ meet at $E$. Now $BIOD$ cyclic so, $\angle IBD=\angle EOI$. Again, as $\angle ABC$ is bisected by $BI$ so $\angle IBD= \angle IBH$. So,$\angle IBH=\angle EBH=\angle EOH$, so the quadrangle $EDBH$ is cyclic. So $\angle BEO=\angle BHO=90^\circ$
Problem no $6(b)$.
Mension the problem number to easily understand the solution.

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

Re: BdMO National Secondary: Problem Collection(2016)

Unread post by samiul_samin » Mon Feb 18, 2019 5:14 pm

Problem 1(a)
Hard solution :o :o :o
$(n-1)×n×(n+1)=(n^2-n)(n+1)$[According to fermat's little theorem given expression is divided by $2$ ]

$(n-1)×n×(n+1)=(n^2-n)(n+1)=n^3-n$[According to fermat's little theorem given expression is divided by $3$ ]

So,$(n-1)×n×(n+1)$ is divided by $2×3$ or $6$. Proved.

Post Reply