We have $101$ tenorist, every two cooperated in exactly one concert, but there is no concert in which all...
$begingroup$
We have $101$ tenorist, every two cooperated in exactly one concert, but there is no concert in which all participated.
Prove that someone participated in at least $11$ concerts.
This is an old problem from Moscow math Olympiad. I did try to solve it but somehow I can't. Any thoughts?
I did try with double counting. Say we have $T_1,T_2,...,T_{101}$ tenorists and $A_1,A_2,...A_n$ concert. So every pair ${T_i,T_j}$ is ''connected'' to exactly one concert. So we have:
$$ sum {deg(A_i)choose 2} = sum deg({T_i,T_j}) = {101choose 2};;;;;(1)$$
We have to prove that the degree of some $T_j$ is at least $11$. Suppose there is no such $j$, then for every $T_j$ the degree is at most $10$ and we have:
$$ sum deg(A_i) = sum deg(T_i) leq 1010 ;;;;;(2)$$
I don't know what to do now.
combinatorics contest-math
asked Oct 8 '17 at 12:04
greedoidgreedoid
38.7k114797
$endgroup$
add a comment |
$begingroup$
We have $101$ tenorist, every two cooperated in exactly one concert, but there is no concert in which all participated.
Prove that someone participated in at least $11$ concerts.
This is an old problem from Moscow math Olympiad. I did try to solve it but somehow I can't. Any thoughts?
I did try with double counting. Say we have $T_1,T_2,...,T_{101}$ tenorists and $A_1,A_2,...A_n$ concert. So every pair ${T_i,T_j}$ is ''connected'' to exactly one concert. So we have:
$$ sum {deg(A_i)choose 2} = sum deg({T_i,T_j}) = {101choose 2};;;;;(1)$$
We have to prove that the degree of some $T_j$ is at least $11$. Suppose there is no such $j$, then for every $T_j$ the degree is at most $10$ and we have:
$$ sum deg(A_i) = sum deg(T_i) leq 1010 ;;;;;(2)$$
I don't know what to do now.
combinatorics contest-math
asked Oct 8 '17 at 12:04
greedoidgreedoid
38.7k114797
$endgroup$
$begingroup$
Is the assumption that there has been no attack involving more than 2 of these 101 terrorists?
$endgroup$
– Bram28
Oct 8 '17 at 12:11
$begingroup$
The model I have in my mind is the following. If there where $p^2$ terrorists, $p$ a prime, then it is possible that they all participated in exactly $p$ attacks. This is because the modulo $p$ plane $Bbb{F}_ptimesBbb{F}_p$ can be partitioned into $p$ parallel lines of respective slopes $infty,0,1,2,ldots,p-1$. Each terrorist is a point on that plane. Each attack was done by one of those lines. Two points determine a unique line, so every pair of terrorists participated in a single attack. But, how to prove that this is "optimal", and how to lift the assumption that $p=10$ is also ok.
$endgroup$
– Jyrki Lahtonen
Oct 8 '17 at 13:18
$begingroup$
To get an idea of what goes on I would first try to prove that with 5 terrorists somebody took part in at least three attacks. Then hope for a light-bulb experience :-/
$endgroup$
– Jyrki Lahtonen
Oct 8 '17 at 13:20
add a comment |
$begingroup$
We have $101$ tenorist, every two cooperated in exactly one concert, but there is no concert in which all participated.
Prove that someone participated in at least $11$ concerts.
This is an old problem from Moscow math Olympiad. I did try to solve it but somehow I can't. Any thoughts?
I did try with double counting. Say we have $T_1,T_2,...,T_{101}$ tenorists and $A_1,A_2,...A_n$ concert. So every pair ${T_i,T_j}$ is ''connected'' to exactly one concert. So we have:
$$ sum {deg(A_i)choose 2} = sum deg({T_i,T_j}) = {101choose 2};;;;;(1)$$
We have to prove that the degree of some $T_j$ is at least $11$. Suppose there is no such $j$, then for every $T_j$ the degree is at most $10$ and we have:
$$ sum deg(A_i) = sum deg(T_i) leq 1010 ;;;;;(2)$$
I don't know what to do now.
combinatorics contest-math
asked Oct 8 '17 at 12:04
greedoidgreedoid
38.7k114797
$endgroup$
We have $101$ tenorist, every two cooperated in exactly one concert, but there is no concert in which all participated.
Prove that someone participated in at least $11$ concerts.
This is an old problem from Moscow math Olympiad. I did try to solve it but somehow I can't. Any thoughts?
I did try with double counting. Say we have $T_1,T_2,...,T_{101}$ tenorists and $A_1,A_2,...A_n$ concert. So every pair ${T_i,T_j}$ is ''connected'' to exactly one concert. So we have:
$$ sum {deg(A_i)choose 2} = sum deg({T_i,T_j}) = {101choose 2};;;;;(1)$$
We have to prove that the degree of some $T_j$ is at least $11$. Suppose there is no such $j$, then for every $T_j$ the degree is at most $10$ and we have:
$$ sum deg(A_i) = sum deg(T_i) leq 1010 ;;;;;(2)$$
I don't know what to do now.
combinatorics contest-math
combinatorics contest-math
asked Oct 8 '17 at 12:04
greedoidgreedoid
38.7k114797
asked Oct 8 '17 at 12:04
greedoidgreedoid
38.7k114797
edited Nov 29 '18 at 14:44
greedoid
asked Oct 8 '17 at 12:04
greedoidgreedoid
38.7k114797
asked Oct 8 '17 at 12:04
greedoidgreedoid
38.7k114797
asked Oct 8 '17 at 12:04
greedoidgreedoid
38.7k114797
38.7k114797
$begingroup$
Is the assumption that there has been no attack involving more than 2 of these 101 terrorists?
$endgroup$
– Bram28
Oct 8 '17 at 12:11
$begingroup$
The model I have in my mind is the following. If there where $p^2$ terrorists, $p$ a prime, then it is possible that they all participated in exactly $p$ attacks. This is because the modulo $p$ plane $Bbb{F}_ptimesBbb{F}_p$ can be partitioned into $p$ parallel lines of respective slopes $infty,0,1,2,ldots,p-1$. Each terrorist is a point on that plane. Each attack was done by one of those lines. Two points determine a unique line, so every pair of terrorists participated in a single attack. But, how to prove that this is "optimal", and how to lift the assumption that $p=10$ is also ok.
$endgroup$
– Jyrki Lahtonen
Oct 8 '17 at 13:18
$begingroup$
To get an idea of what goes on I would first try to prove that with 5 terrorists somebody took part in at least three attacks. Then hope for a light-bulb experience :-/
$endgroup$
– Jyrki Lahtonen
Oct 8 '17 at 13:20
add a comment |
$begingroup$
Is the assumption that there has been no attack involving more than 2 of these 101 terrorists?
$endgroup$
– Bram28
Oct 8 '17 at 12:11
$begingroup$
The model I have in my mind is the following. If there where $p^2$ terrorists, $p$ a prime, then it is possible that they all participated in exactly $p$ attacks. This is because the modulo $p$ plane $Bbb{F}_ptimesBbb{F}_p$ can be partitioned into $p$ parallel lines of respective slopes $infty,0,1,2,ldots,p-1$. Each terrorist is a point on that plane. Each attack was done by one of those lines. Two points determine a unique line, so every pair of terrorists participated in a single attack. But, how to prove that this is "optimal", and how to lift the assumption that $p=10$ is also ok.
$endgroup$
– Jyrki Lahtonen
Oct 8 '17 at 13:18
$begingroup$
To get an idea of what goes on I would first try to prove that with 5 terrorists somebody took part in at least three attacks. Then hope for a light-bulb experience :-/
$endgroup$
– Jyrki Lahtonen
Oct 8 '17 at 13:20
$begingroup$
Is the assumption that there has been no attack involving more than 2 of these 101 terrorists?
$endgroup$
– Bram28
Oct 8 '17 at 12:11
$begingroup$
Is the assumption that there has been no attack involving more than 2 of these 101 terrorists?
$endgroup$
– Bram28
Oct 8 '17 at 12:11
$begingroup$
The model I have in my mind is the following. If there where $p^2$ terrorists, $p$ a prime, then it is possible that they all participated in exactly $p$ attacks. This is because the modulo $p$ plane $Bbb{F}_ptimesBbb{F}_p$ can be partitioned into $p$ parallel lines of respective slopes $infty,0,1,2,ldots,p-1$. Each terrorist is a point on that plane. Each attack was done by one of those lines. Two points determine a unique line, so every pair of terrorists participated in a single attack. But, how to prove that this is "optimal", and how to lift the assumption that $p=10$ is also ok.
$endgroup$
– Jyrki Lahtonen
Oct 8 '17 at 13:18
$begingroup$
The model I have in my mind is the following. If there where $p^2$ terrorists, $p$ a prime, then it is possible that they all participated in exactly $p$ attacks. This is because the modulo $p$ plane $Bbb{F}_ptimesBbb{F}_p$ can be partitioned into $p$ parallel lines of respective slopes $infty,0,1,2,ldots,p-1$. Each terrorist is a point on that plane. Each attack was done by one of those lines. Two points determine a unique line, so every pair of terrorists participated in a single attack. But, how to prove that this is "optimal", and how to lift the assumption that $p=10$ is also ok.
$endgroup$
– Jyrki Lahtonen
Oct 8 '17 at 13:18
$begingroup$
To get an idea of what goes on I would first try to prove that with 5 terrorists somebody took part in at least three attacks. Then hope for a light-bulb experience :-/
$endgroup$
– Jyrki Lahtonen
Oct 8 '17 at 13:20
$begingroup$
To get an idea of what goes on I would first try to prove that with 5 terrorists somebody took part in at least three attacks. Then hope for a light-bulb experience :-/
$endgroup$
– Jyrki Lahtonen
Oct 8 '17 at 13:20
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Take an arbitrary tenorist, say $T_1$. He is involved in at most $10$ concerts, which combined pairs him once with each of the remaining $100$ tenorists. Hence, at least one of these concerts, say $A_k$, must involve at least $11$ tenorists.
Given concert $A_k$ involving $mge11$ tenorists, and a tenorist $T_p$ not in $A_k$, for each tenorist $T_i$ in $A_k$ there will be a unique concert which includes $T_i$ and $T_p$. This gives a list of $m$ concert, one for each $T_iin A_k$, and these must all be distinct concert different from $A_k$ since no two members of $A_k$ may be in another concert.
So, this gives $mge11$ concert in which $T_p$ is a member.
Would the result also be true with a smaller number of tenorists? Unless I'm missing something, my proof should also work if there are $92$ tenorists.
edited Oct 9 '17 at 20:37
greedoid
38.7k114797
answered Oct 8 '17 at 13:20
Einar RødlandEinar Rødland
6,4761229
$endgroup$
$begingroup$
I believe it does. It would be interesting to extend to arbitrary number of terrorists, $n$, and what lower bound can be put on the one who participated in the most attacks...
$endgroup$
– Nick Pavlov
Oct 8 '17 at 13:26
1
$begingroup$
We see that the condition not all participated in a single attack is crucial for existence of $T_p$.
$endgroup$
– greedoid
Oct 8 '17 at 13:31
add a comment |
Your Answer
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$begingroup$
Take an arbitrary tenorist, say $T_1$. He is involved in at most $10$ concerts, which combined pairs him once with each of the remaining $100$ tenorists. Hence, at least one of these concerts, say $A_k$, must involve at least $11$ tenorists.
Given concert $A_k$ involving $mge11$ tenorists, and a tenorist $T_p$ not in $A_k$, for each tenorist $T_i$ in $A_k$ there will be a unique concert which includes $T_i$ and $T_p$. This gives a list of $m$ concert, one for each $T_iin A_k$, and these must all be distinct concert different from $A_k$ since no two members of $A_k$ may be in another concert.
So, this gives $mge11$ concert in which $T_p$ is a member.
Would the result also be true with a smaller number of tenorists? Unless I'm missing something, my proof should also work if there are $92$ tenorists.
edited Oct 9 '17 at 20:37
greedoid
38.7k114797
answered Oct 8 '17 at 13:20
Einar RødlandEinar Rødland
6,4761229
$endgroup$
$begingroup$
I believe it does. It would be interesting to extend to arbitrary number of terrorists, $n$, and what lower bound can be put on the one who participated in the most attacks...
$endgroup$
– Nick Pavlov
Oct 8 '17 at 13:26
1
$begingroup$
We see that the condition not all participated in a single attack is crucial for existence of $T_p$.
$endgroup$
– greedoid
Oct 8 '17 at 13:31
add a comment |
$begingroup$
Take an arbitrary tenorist, say $T_1$. He is involved in at most $10$ concerts, which combined pairs him once with each of the remaining $100$ tenorists. Hence, at least one of these concerts, say $A_k$, must involve at least $11$ tenorists.
Given concert $A_k$ involving $mge11$ tenorists, and a tenorist $T_p$ not in $A_k$, for each tenorist $T_i$ in $A_k$ there will be a unique concert which includes $T_i$ and $T_p$. This gives a list of $m$ concert, one for each $T_iin A_k$, and these must all be distinct concert different from $A_k$ since no two members of $A_k$ may be in another concert.
So, this gives $mge11$ concert in which $T_p$ is a member.
Would the result also be true with a smaller number of tenorists? Unless I'm missing something, my proof should also work if there are $92$ tenorists.
edited Oct 9 '17 at 20:37
greedoid
38.7k114797
answered Oct 8 '17 at 13:20
Einar RødlandEinar Rødland
6,4761229
$endgroup$
$begingroup$
I believe it does. It would be interesting to extend to arbitrary number of terrorists, $n$, and what lower bound can be put on the one who participated in the most attacks...
$endgroup$
– Nick Pavlov
Oct 8 '17 at 13:26
1
$begingroup$
We see that the condition not all participated in a single attack is crucial for existence of $T_p$.
$endgroup$
– greedoid
Oct 8 '17 at 13:31
add a comment |
$begingroup$
Take an arbitrary tenorist, say $T_1$. He is involved in at most $10$ concerts, which combined pairs him once with each of the remaining $100$ tenorists. Hence, at least one of these concerts, say $A_k$, must involve at least $11$ tenorists.
Given concert $A_k$ involving $mge11$ tenorists, and a tenorist $T_p$ not in $A_k$, for each tenorist $T_i$ in $A_k$ there will be a unique concert which includes $T_i$ and $T_p$. This gives a list of $m$ concert, one for each $T_iin A_k$, and these must all be distinct concert different from $A_k$ since no two members of $A_k$ may be in another concert.
So, this gives $mge11$ concert in which $T_p$ is a member.
Would the result also be true with a smaller number of tenorists? Unless I'm missing something, my proof should also work if there are $92$ tenorists.
edited Oct 9 '17 at 20:37
greedoid
38.7k114797
answered Oct 8 '17 at 13:20
Einar RødlandEinar Rødland
6,4761229
$endgroup$
Take an arbitrary tenorist, say $T_1$. He is involved in at most $10$ concerts, which combined pairs him once with each of the remaining $100$ tenorists. Hence, at least one of these concerts, say $A_k$, must involve at least $11$ tenorists.
Given concert $A_k$ involving $mge11$ tenorists, and a tenorist $T_p$ not in $A_k$, for each tenorist $T_i$ in $A_k$ there will be a unique concert which includes $T_i$ and $T_p$. This gives a list of $m$ concert, one for each $T_iin A_k$, and these must all be distinct concert different from $A_k$ since no two members of $A_k$ may be in another concert.
So, this gives $mge11$ concert in which $T_p$ is a member.
Would the result also be true with a smaller number of tenorists? Unless I'm missing something, my proof should also work if there are $92$ tenorists.
edited Oct 9 '17 at 20:37
greedoid
38.7k114797
answered Oct 8 '17 at 13:20
Einar RødlandEinar Rødland
6,4761229
edited Oct 9 '17 at 20:37
greedoid
38.7k114797
edited Oct 9 '17 at 20:37
greedoid
38.7k114797
edited Oct 9 '17 at 20:37
greedoid
38.7k114797
38.7k114797
answered Oct 8 '17 at 13:20
Einar RødlandEinar Rødland
6,4761229
answered Oct 8 '17 at 13:20
Einar RødlandEinar Rødland
6,4761229
answered Oct 8 '17 at 13:20
Einar RødlandEinar Rødland
6,4761229
6,4761229
$begingroup$
I believe it does. It would be interesting to extend to arbitrary number of terrorists, $n$, and what lower bound can be put on the one who participated in the most attacks...
$endgroup$
– Nick Pavlov
Oct 8 '17 at 13:26
1
$begingroup$
We see that the condition not all participated in a single attack is crucial for existence of $T_p$.
$endgroup$
– greedoid
Oct 8 '17 at 13:31
add a comment |
$begingroup$
I believe it does. It would be interesting to extend to arbitrary number of terrorists, $n$, and what lower bound can be put on the one who participated in the most attacks...
$endgroup$
– Nick Pavlov
Oct 8 '17 at 13:26
1
$begingroup$
We see that the condition not all participated in a single attack is crucial for existence of $T_p$.
$endgroup$
– greedoid
Oct 8 '17 at 13:31
$begingroup$
I believe it does. It would be interesting to extend to arbitrary number of terrorists, $n$, and what lower bound can be put on the one who participated in the most attacks...
$endgroup$
– Nick Pavlov
Oct 8 '17 at 13:26
$begingroup$
I believe it does. It would be interesting to extend to arbitrary number of terrorists, $n$, and what lower bound can be put on the one who participated in the most attacks...
$endgroup$
– Nick Pavlov
Oct 8 '17 at 13:26
1
1
$begingroup$
We see that the condition not all participated in a single attack is crucial for existence of $T_p$.
$endgroup$
– greedoid
Oct 8 '17 at 13:31
$begingroup$
We see that the condition not all participated in a single attack is crucial for existence of $T_p$.
$endgroup$
– greedoid
Oct 8 '17 at 13:31
add a comment |
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$begingroup$
Is the assumption that there has been no attack involving more than 2 of these 101 terrorists?
$endgroup$
– Bram28
Oct 8 '17 at 12:11
$begingroup$
The model I have in my mind is the following. If there where $p^2$ terrorists, $p$ a prime, then it is possible that they all participated in exactly $p$ attacks. This is because the modulo $p$ plane $Bbb{F}_ptimesBbb{F}_p$ can be partitioned into $p$ parallel lines of respective slopes $infty,0,1,2,ldots,p-1$. Each terrorist is a point on that plane. Each attack was done by one of those lines. Two points determine a unique line, so every pair of terrorists participated in a single attack. But, how to prove that this is "optimal", and how to lift the assumption that $p=10$ is also ok.
$endgroup$
– Jyrki Lahtonen
Oct 8 '17 at 13:18
$begingroup$
To get an idea of what goes on I would first try to prove that with 5 terrorists somebody took part in at least three attacks. Then hope for a light-bulb experience :-/
$endgroup$
– Jyrki Lahtonen
Oct 8 '17 at 13:20