Proving that there is an element common to all $35$ sets given certain set restrictions












5














Consider the $35$ sets $A_1,A_2,dots,A_{35}$ such that $|A_i|=27$ for all $1leq i leq 35$, and every triplet of sets have one exactly one element in common to all three. Prove that there is at least one element common to all $35$ sets.



This was a problem given to me by a friend - he asked me to help him with this but I am unable to figure out the answer. He suggested something about contradiction and the pigeonhole principle, but I'm not sure how to continue. He mentioned it was from some sort of Olympiad, but I dont remember which one (something middle eastern?)



I found this similar question online: https://artofproblemsolving.com/community/q1h1699161p10908761



but this doesn't include any restrictions on cardinality, and is only a case for a pairwise disjoint set.






combinatorics algebra-precalculus contest-math







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  • 1




    "...such that $| A_i|=27$ for all $1leq ileq 27$..." Is this a typo? Should it be $1leq ileq 35$ instead?
    – YiFan
    Nov 22 '18 at 23:04










  • @YiFan Yes, fixed, thank you
    – user574848
    Nov 22 '18 at 23:14










  • Consider A$_1$. There are $binom{34}{2}$= 561 triplets of sets that include A$_1$. BY the pigeonhole principle there is at least 1 element of A$_1$ that is shared by 20 of these triples. Maybe you can keep reducing the problem accordingly.
    – Joel Pereira
    Nov 23 '18 at 0:09
















5














Consider the $35$ sets $A_1,A_2,dots,A_{35}$ such that $|A_i|=27$ for all $1leq i leq 35$, and every triplet of sets have one exactly one element in common to all three. Prove that there is at least one element common to all $35$ sets.



This was a problem given to me by a friend - he asked me to help him with this but I am unable to figure out the answer. He suggested something about contradiction and the pigeonhole principle, but I'm not sure how to continue. He mentioned it was from some sort of Olympiad, but I dont remember which one (something middle eastern?)



I found this similar question online: https://artofproblemsolving.com/community/q1h1699161p10908761



but this doesn't include any restrictions on cardinality, and is only a case for a pairwise disjoint set.






combinatorics algebra-precalculus contest-math







share|cite|improve this question




















  • 1




    "...such that $| A_i|=27$ for all $1leq ileq 27$..." Is this a typo? Should it be $1leq ileq 35$ instead?
    – YiFan
    Nov 22 '18 at 23:04










  • @YiFan Yes, fixed, thank you
    – user574848
    Nov 22 '18 at 23:14










  • Consider A$_1$. There are $binom{34}{2}$= 561 triplets of sets that include A$_1$. BY the pigeonhole principle there is at least 1 element of A$_1$ that is shared by 20 of these triples. Maybe you can keep reducing the problem accordingly.
    – Joel Pereira
    Nov 23 '18 at 0:09














5












5








5


4





Consider the $35$ sets $A_1,A_2,dots,A_{35}$ such that $|A_i|=27$ for all $1leq i leq 35$, and every triplet of sets have one exactly one element in common to all three. Prove that there is at least one element common to all $35$ sets.



This was a problem given to me by a friend - he asked me to help him with this but I am unable to figure out the answer. He suggested something about contradiction and the pigeonhole principle, but I'm not sure how to continue. He mentioned it was from some sort of Olympiad, but I dont remember which one (something middle eastern?)



I found this similar question online: https://artofproblemsolving.com/community/q1h1699161p10908761



but this doesn't include any restrictions on cardinality, and is only a case for a pairwise disjoint set.






combinatorics algebra-precalculus contest-math







share|cite|improve this question















Consider the $35$ sets $A_1,A_2,dots,A_{35}$ such that $|A_i|=27$ for all $1leq i leq 35$, and every triplet of sets have one exactly one element in common to all three. Prove that there is at least one element common to all $35$ sets.



This was a problem given to me by a friend - he asked me to help him with this but I am unable to figure out the answer. He suggested something about contradiction and the pigeonhole principle, but I'm not sure how to continue. He mentioned it was from some sort of Olympiad, but I dont remember which one (something middle eastern?)



I found this similar question online: https://artofproblemsolving.com/community/q1h1699161p10908761



but this doesn't include any restrictions on cardinality, and is only a case for a pairwise disjoint set.






combinatorics algebra-precalculus contest-math




combinatorics algebra-precalculus contest-math






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 '18 at 23:14

























asked Nov 19 '18 at 5:46









user574848

16015




16015








  • 1




    "...such that $| A_i|=27$ for all $1leq ileq 27$..." Is this a typo? Should it be $1leq ileq 35$ instead?
    – YiFan
    Nov 22 '18 at 23:04










  • @YiFan Yes, fixed, thank you
    – user574848
    Nov 22 '18 at 23:14










  • Consider A$_1$. There are $binom{34}{2}$= 561 triplets of sets that include A$_1$. BY the pigeonhole principle there is at least 1 element of A$_1$ that is shared by 20 of these triples. Maybe you can keep reducing the problem accordingly.
    – Joel Pereira
    Nov 23 '18 at 0:09














  • 1




    "...such that $| A_i|=27$ for all $1leq ileq 27$..." Is this a typo? Should it be $1leq ileq 35$ instead?
    – YiFan
    Nov 22 '18 at 23:04










  • @YiFan Yes, fixed, thank you
    – user574848
    Nov 22 '18 at 23:14










  • Consider A$_1$. There are $binom{34}{2}$= 561 triplets of sets that include A$_1$. BY the pigeonhole principle there is at least 1 element of A$_1$ that is shared by 20 of these triples. Maybe you can keep reducing the problem accordingly.
    – Joel Pereira
    Nov 23 '18 at 0:09








1




1




"...such that $| A_i|=27$ for all $1leq ileq 27$..." Is this a typo? Should it be $1leq ileq 35$ instead?
– YiFan
Nov 22 '18 at 23:04




"...such that $| A_i|=27$ for all $1leq ileq 27$..." Is this a typo? Should it be $1leq ileq 35$ instead?
– YiFan
Nov 22 '18 at 23:04












@YiFan Yes, fixed, thank you
– user574848
Nov 22 '18 at 23:14




@YiFan Yes, fixed, thank you
– user574848
Nov 22 '18 at 23:14












Consider A$_1$. There are $binom{34}{2}$= 561 triplets of sets that include A$_1$. BY the pigeonhole principle there is at least 1 element of A$_1$ that is shared by 20 of these triples. Maybe you can keep reducing the problem accordingly.
– Joel Pereira
Nov 23 '18 at 0:09




Consider A$_1$. There are $binom{34}{2}$= 561 triplets of sets that include A$_1$. BY the pigeonhole principle there is at least 1 element of A$_1$ that is shared by 20 of these triples. Maybe you can keep reducing the problem accordingly.
– Joel Pereira
Nov 23 '18 at 0:09










1 Answer
1






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oldest

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3





+50









Lets assume there is no element common to all 35.



Any set $A_i$ is part of $binom{34}{2} = 561$ triplets. $A_i$ has 27 items. $ 27 cdot 20 = 540 < 561 $, so from the pigeonhole principle some item a in $A_i$ must be in 21 of the triplets. This means that at least 7 other sets must include a, since $binom{7}{2} = 21$. In turn, this means that a is in at least 8 sets (including $A_i$). Among these sets, there are $binom{8}{2} = 28$ pairs.



Lets assume some set $A_j$ does not contain a. $A_j$ must have some item in common with every one of the $28$ pairs.



If an item b is common to $A_j$ and two subsets that contain a, it must not be in any other subset that contains a, since each triplet has only one item in common, so no three sets that include a can share the item b.



This means there must be at least 28 unique elements in $A_j$, but this is impossible since $|A_j| = 27$.



So a must be common to all sets $A_j$, in contradiction to the assumption.






share|cite|improve this answer























  • Nice proof ! Allow me to suggest an edit : since at the beginning of your proof what you're really doing is fixing an $i$ and not reasoning on all $i$ at once, I think it would be slightly more readable if you put : "$A_1$ is part of 561 triplets and has 27 items , etc."
    – Ewan Delanoy
    Nov 23 '18 at 11:52











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









3





+50









Lets assume there is no element common to all 35.



Any set $A_i$ is part of $binom{34}{2} = 561$ triplets. $A_i$ has 27 items. $ 27 cdot 20 = 540 < 561 $, so from the pigeonhole principle some item a in $A_i$ must be in 21 of the triplets. This means that at least 7 other sets must include a, since $binom{7}{2} = 21$. In turn, this means that a is in at least 8 sets (including $A_i$). Among these sets, there are $binom{8}{2} = 28$ pairs.



Lets assume some set $A_j$ does not contain a. $A_j$ must have some item in common with every one of the $28$ pairs.



If an item b is common to $A_j$ and two subsets that contain a, it must not be in any other subset that contains a, since each triplet has only one item in common, so no three sets that include a can share the item b.



This means there must be at least 28 unique elements in $A_j$, but this is impossible since $|A_j| = 27$.



So a must be common to all sets $A_j$, in contradiction to the assumption.






share|cite|improve this answer























  • Nice proof ! Allow me to suggest an edit : since at the beginning of your proof what you're really doing is fixing an $i$ and not reasoning on all $i$ at once, I think it would be slightly more readable if you put : "$A_1$ is part of 561 triplets and has 27 items , etc."
    – Ewan Delanoy
    Nov 23 '18 at 11:52
















3





+50









Lets assume there is no element common to all 35.



Any set $A_i$ is part of $binom{34}{2} = 561$ triplets. $A_i$ has 27 items. $ 27 cdot 20 = 540 < 561 $, so from the pigeonhole principle some item a in $A_i$ must be in 21 of the triplets. This means that at least 7 other sets must include a, since $binom{7}{2} = 21$. In turn, this means that a is in at least 8 sets (including $A_i$). Among these sets, there are $binom{8}{2} = 28$ pairs.



Lets assume some set $A_j$ does not contain a. $A_j$ must have some item in common with every one of the $28$ pairs.



If an item b is common to $A_j$ and two subsets that contain a, it must not be in any other subset that contains a, since each triplet has only one item in common, so no three sets that include a can share the item b.



This means there must be at least 28 unique elements in $A_j$, but this is impossible since $|A_j| = 27$.



So a must be common to all sets $A_j$, in contradiction to the assumption.






share|cite|improve this answer























  • Nice proof ! Allow me to suggest an edit : since at the beginning of your proof what you're really doing is fixing an $i$ and not reasoning on all $i$ at once, I think it would be slightly more readable if you put : "$A_1$ is part of 561 triplets and has 27 items , etc."
    – Ewan Delanoy
    Nov 23 '18 at 11:52














3





+50







3





+50



3




+50




Lets assume there is no element common to all 35.



Any set $A_i$ is part of $binom{34}{2} = 561$ triplets. $A_i$ has 27 items. $ 27 cdot 20 = 540 < 561 $, so from the pigeonhole principle some item a in $A_i$ must be in 21 of the triplets. This means that at least 7 other sets must include a, since $binom{7}{2} = 21$. In turn, this means that a is in at least 8 sets (including $A_i$). Among these sets, there are $binom{8}{2} = 28$ pairs.



Lets assume some set $A_j$ does not contain a. $A_j$ must have some item in common with every one of the $28$ pairs.



If an item b is common to $A_j$ and two subsets that contain a, it must not be in any other subset that contains a, since each triplet has only one item in common, so no three sets that include a can share the item b.



This means there must be at least 28 unique elements in $A_j$, but this is impossible since $|A_j| = 27$.



So a must be common to all sets $A_j$, in contradiction to the assumption.






share|cite|improve this answer














Lets assume there is no element common to all 35.



Any set $A_i$ is part of $binom{34}{2} = 561$ triplets. $A_i$ has 27 items. $ 27 cdot 20 = 540 < 561 $, so from the pigeonhole principle some item a in $A_i$ must be in 21 of the triplets. This means that at least 7 other sets must include a, since $binom{7}{2} = 21$. In turn, this means that a is in at least 8 sets (including $A_i$). Among these sets, there are $binom{8}{2} = 28$ pairs.



Lets assume some set $A_j$ does not contain a. $A_j$ must have some item in common with every one of the $28$ pairs.



If an item b is common to $A_j$ and two subsets that contain a, it must not be in any other subset that contains a, since each triplet has only one item in common, so no three sets that include a can share the item b.



This means there must be at least 28 unique elements in $A_j$, but this is impossible since $|A_j| = 27$.



So a must be common to all sets $A_j$, in contradiction to the assumption.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 23 '18 at 22:43

























answered Nov 23 '18 at 10:20









Daphna Keidar

1996




1996












  • Nice proof ! Allow me to suggest an edit : since at the beginning of your proof what you're really doing is fixing an $i$ and not reasoning on all $i$ at once, I think it would be slightly more readable if you put : "$A_1$ is part of 561 triplets and has 27 items , etc."
    – Ewan Delanoy
    Nov 23 '18 at 11:52


















  • Nice proof ! Allow me to suggest an edit : since at the beginning of your proof what you're really doing is fixing an $i$ and not reasoning on all $i$ at once, I think it would be slightly more readable if you put : "$A_1$ is part of 561 triplets and has 27 items , etc."
    – Ewan Delanoy
    Nov 23 '18 at 11:52
















Nice proof ! Allow me to suggest an edit : since at the beginning of your proof what you're really doing is fixing an $i$ and not reasoning on all $i$ at once, I think it would be slightly more readable if you put : "$A_1$ is part of 561 triplets and has 27 items , etc."
– Ewan Delanoy
Nov 23 '18 at 11:52




Nice proof ! Allow me to suggest an edit : since at the beginning of your proof what you're really doing is fixing an $i$ and not reasoning on all $i$ at once, I think it would be slightly more readable if you put : "$A_1$ is part of 561 triplets and has 27 items , etc."
– Ewan Delanoy
Nov 23 '18 at 11:52


















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