BdMO National Secondary 2011/8

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Moon
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BdMO National Secondary 2011/8

Unread post by Moon » Fri Feb 11, 2011 1:50 pm

Problem 8:
Bhaskaracharya has set up a strange study group. Any member of that group has exactly one immediate teacher (the teacher who teaches him) except for Bhaskaracharya himself, although teacher of teacher is also respected as a teacher. As the chancellor of the group, Bhaskaracharya is not taught by anybody. No two members of that group can be teachers of each other. The study group operates in a pairs where each pair consists of one member and his immediate teacher. If such a pairing is possible, is it unique? Justify your answer.
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Re: BdMO National Secondary 2011/8

Unread post by Shapnil » Thu Jan 12, 2012 11:29 pm

I'm not sure is this correct.But I'm trying to solve it
suppose nubmers of members are n.as no two members of that group can be teachers of each other
then then all members are student (except vaskarachrya).then he's the only teacher of the group.then we can sayit is a uniqe way

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Re: BdMO National Secondary 2011/8

Unread post by *Mahi* » Fri Jan 13, 2012 12:18 am

Read the question again carefully.
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Re: BdMO National Secondary 2011/8

Unread post by nafistiham » Sat Jan 14, 2012 4:21 pm

it is not unique i think.suppose, B(Bhaskaracharya) has one student, he makes a pair with him and there are $2n$ persons in the group apart from B and his student.for $n=2$ i am showing, think the $4$ persons are $a,b,c,d$
now, if the relation is cyclic, i mean,
$a$ teaches $b$
$b$ teaches $c$
$c$ teaches $d$
$d$ teaches $a$
everyone has only one immediate teachers.but, we can make both these pairings
\[\binom{a}{b}\binom{c}{d}\]
\[\binom{d}{a}\binom{b}{c}\]
the pairings are different. :D
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: BdMO National Secondary 2011/8

Unread post by nafistiham » Mon Jan 23, 2012 3:15 pm

[quote=
$a$ teaches Sb$
$b$ teaches Sc$
$c$ teaches Sd$
$d$ teaches Sa$
[/quote]

sorry, it should have been like this

$a$ teaches $b$
$b$ teaches $c$
$c$ teaches $d$
$d$ teaches $a$
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: BdMO National Secondary 2011/8

Unread post by *Mahi* » Thu Feb 09, 2012 3:51 pm

nafistiham wrote:
it is not unique i think.suppose, B(Bhaskaracharya) has one student, he makes a pair with him and there are $2n$ persons in the group apart from B and his student.for $n=2$ i am showing, think the $4$ persons are $a,b,c,d$
now, if the relation is cyclic, i mean,
$a$ teaches $b$
$b$ teaches $c$
$c$ teaches $d$
$d$ teaches $a$
everyone has only one immediate teachers.but, we can make both these pairings
\[\binom{a}{b}\binom{c}{d}\]
\[\binom{d}{a}\binom{b}{c}\]
the pairings are different. :D
Cyclic groups are not allowed, as teacher of teacher is respected as teacher.
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Re: BdMO National Secondary 2011/8

Unread post by nafistiham » Mon Feb 13, 2012 5:46 pm

*Mahi* wrote:
nafistiham wrote:
it is not unique i think.suppose, B(Bhaskaracharya) has one student, he makes a pair with him and there are $2n$ persons in the group apart from B and his student.for $n=2$ i am showing, think the $4$ persons are $a,b,c,d$
now, if the relation is cyclic, i mean,
$a$ teaches $b$
$b$ teaches $c$
$c$ teaches $d$
$d$ teaches $a$
everyone has only one immediate teachers.but, we can make both these pairings
\[\binom{a}{b}\binom{c}{d}\]
\[\binom{d}{a}\binom{b}{c}\]
the pairings are different. :D
Cyclic groups are not allowed, as teacher of teacher is respected as teacher.
I think it is unique then.because, cyclic is the only solution, which is not allowed.
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: BdMO National Secondary 2011/8

Unread post by *Mahi* » Mon Feb 13, 2012 6:23 pm

There is a really cool solution of this with graph theory. I'll post it when I have some time.
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Re: BdMO National Secondary 2011/8

Unread post by nafistiham » Mon Feb 13, 2012 7:12 pm

graph theory !! i thought it can be done by group theory :lol:
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: BdMO National Secondary 2011/8

Unread post by samiul_samin » Tue Jan 23, 2018 8:32 pm

What is the solution using graph theory?

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