Punch 2000 holes in 2000 polygons with 1000 needles

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You have two identical perfectly square pieces of paper. The area of each paper is 1000 units.



On each paper, draw 1000 convex, non-overlapping polygons with all polygons having the same area (exactly 1 unit). Obviously, the polygons are covering both papers completely and edges of paper also serve as edges of some polygons). Polygons may have different shapes and number of sides and the drawing on the first paper is completely different from the drawing on the second paper.



Now put the first paper on top of the second and align paper edges perfectly. Prove that it is always possible to punch a hole in all 2000 polygons with 1000 needles (each needle goes through both papers).



What have I tried?



This problem came from my son who likes to torture his father with difficult problems brought back from his math school. My first try was to steal his clever analysis book while he was sleeping and find the right page in the answer section. Alas, this problem had no solution, which basically means that it's either too simple (and I'm too stupid) or it's too difficult.



So I decided to read some theory and discovered that I had some pretty huge gaps in my math education. This problem is definitely about functions. You have a set of 1000 polygons on one side and a set of 1000 polygons on the other side. I have to prove that there is a bijective function between these two sets. Needles are just lines connecting the dots. However, all my attempts to construct such function ended miserably. I guess there has to be some clever theorem than can be applied to problems like this one but I would have to read a pretty thick book to find it.



Thanks for the hint.










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  • How do the needles go through the papers, exactly? Does each needle pass through each paper at exactly one point, and, for any given needle, does it have to pass through the same point on each paper?
    – Tanner Swett
    19 hours ago










  • @TannerSwett Yes, you are right. Imagine a needle as a line perpendicular to two squares lying in two parallel planes.
    – Oldboy
    15 hours ago






  • 2




    On puzzling.SE a potential answer could be: You only need one needle that you reuse for 2000 wholes :).
    – Surb
    5 hours ago






  • 1




    @Surb I should have used 10.000 needles instead of 1.000. I would have got more upvotes :)
    – Oldboy
    4 hours ago














up vote
23
down vote

favorite
8












You have two identical perfectly square pieces of paper. The area of each paper is 1000 units.



On each paper, draw 1000 convex, non-overlapping polygons with all polygons having the same area (exactly 1 unit). Obviously, the polygons are covering both papers completely and edges of paper also serve as edges of some polygons). Polygons may have different shapes and number of sides and the drawing on the first paper is completely different from the drawing on the second paper.



Now put the first paper on top of the second and align paper edges perfectly. Prove that it is always possible to punch a hole in all 2000 polygons with 1000 needles (each needle goes through both papers).



What have I tried?



This problem came from my son who likes to torture his father with difficult problems brought back from his math school. My first try was to steal his clever analysis book while he was sleeping and find the right page in the answer section. Alas, this problem had no solution, which basically means that it's either too simple (and I'm too stupid) or it's too difficult.



So I decided to read some theory and discovered that I had some pretty huge gaps in my math education. This problem is definitely about functions. You have a set of 1000 polygons on one side and a set of 1000 polygons on the other side. I have to prove that there is a bijective function between these two sets. Needles are just lines connecting the dots. However, all my attempts to construct such function ended miserably. I guess there has to be some clever theorem than can be applied to problems like this one but I would have to read a pretty thick book to find it.



Thanks for the hint.










share|cite|improve this question























  • How do the needles go through the papers, exactly? Does each needle pass through each paper at exactly one point, and, for any given needle, does it have to pass through the same point on each paper?
    – Tanner Swett
    19 hours ago










  • @TannerSwett Yes, you are right. Imagine a needle as a line perpendicular to two squares lying in two parallel planes.
    – Oldboy
    15 hours ago






  • 2




    On puzzling.SE a potential answer could be: You only need one needle that you reuse for 2000 wholes :).
    – Surb
    5 hours ago






  • 1




    @Surb I should have used 10.000 needles instead of 1.000. I would have got more upvotes :)
    – Oldboy
    4 hours ago












up vote
23
down vote

favorite
8









up vote
23
down vote

favorite
8






8





You have two identical perfectly square pieces of paper. The area of each paper is 1000 units.



On each paper, draw 1000 convex, non-overlapping polygons with all polygons having the same area (exactly 1 unit). Obviously, the polygons are covering both papers completely and edges of paper also serve as edges of some polygons). Polygons may have different shapes and number of sides and the drawing on the first paper is completely different from the drawing on the second paper.



Now put the first paper on top of the second and align paper edges perfectly. Prove that it is always possible to punch a hole in all 2000 polygons with 1000 needles (each needle goes through both papers).



What have I tried?



This problem came from my son who likes to torture his father with difficult problems brought back from his math school. My first try was to steal his clever analysis book while he was sleeping and find the right page in the answer section. Alas, this problem had no solution, which basically means that it's either too simple (and I'm too stupid) or it's too difficult.



So I decided to read some theory and discovered that I had some pretty huge gaps in my math education. This problem is definitely about functions. You have a set of 1000 polygons on one side and a set of 1000 polygons on the other side. I have to prove that there is a bijective function between these two sets. Needles are just lines connecting the dots. However, all my attempts to construct such function ended miserably. I guess there has to be some clever theorem than can be applied to problems like this one but I would have to read a pretty thick book to find it.



Thanks for the hint.










share|cite|improve this question















You have two identical perfectly square pieces of paper. The area of each paper is 1000 units.



On each paper, draw 1000 convex, non-overlapping polygons with all polygons having the same area (exactly 1 unit). Obviously, the polygons are covering both papers completely and edges of paper also serve as edges of some polygons). Polygons may have different shapes and number of sides and the drawing on the first paper is completely different from the drawing on the second paper.



Now put the first paper on top of the second and align paper edges perfectly. Prove that it is always possible to punch a hole in all 2000 polygons with 1000 needles (each needle goes through both papers).



What have I tried?



This problem came from my son who likes to torture his father with difficult problems brought back from his math school. My first try was to steal his clever analysis book while he was sleeping and find the right page in the answer section. Alas, this problem had no solution, which basically means that it's either too simple (and I'm too stupid) or it's too difficult.



So I decided to read some theory and discovered that I had some pretty huge gaps in my math education. This problem is definitely about functions. You have a set of 1000 polygons on one side and a set of 1000 polygons on the other side. I have to prove that there is a bijective function between these two sets. Needles are just lines connecting the dots. However, all my attempts to construct such function ended miserably. I guess there has to be some clever theorem than can be applied to problems like this one but I would have to read a pretty thick book to find it.



Thanks for the hint.







functions






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edited 23 hours ago









amWhy

191k27223437




191k27223437










asked 23 hours ago









Oldboy

5,3431427




5,3431427











  • How do the needles go through the papers, exactly? Does each needle pass through each paper at exactly one point, and, for any given needle, does it have to pass through the same point on each paper?
    – Tanner Swett
    19 hours ago










  • @TannerSwett Yes, you are right. Imagine a needle as a line perpendicular to two squares lying in two parallel planes.
    – Oldboy
    15 hours ago






  • 2




    On puzzling.SE a potential answer could be: You only need one needle that you reuse for 2000 wholes :).
    – Surb
    5 hours ago






  • 1




    @Surb I should have used 10.000 needles instead of 1.000. I would have got more upvotes :)
    – Oldboy
    4 hours ago
















  • How do the needles go through the papers, exactly? Does each needle pass through each paper at exactly one point, and, for any given needle, does it have to pass through the same point on each paper?
    – Tanner Swett
    19 hours ago










  • @TannerSwett Yes, you are right. Imagine a needle as a line perpendicular to two squares lying in two parallel planes.
    – Oldboy
    15 hours ago






  • 2




    On puzzling.SE a potential answer could be: You only need one needle that you reuse for 2000 wholes :).
    – Surb
    5 hours ago






  • 1




    @Surb I should have used 10.000 needles instead of 1.000. I would have got more upvotes :)
    – Oldboy
    4 hours ago















How do the needles go through the papers, exactly? Does each needle pass through each paper at exactly one point, and, for any given needle, does it have to pass through the same point on each paper?
– Tanner Swett
19 hours ago




How do the needles go through the papers, exactly? Does each needle pass through each paper at exactly one point, and, for any given needle, does it have to pass through the same point on each paper?
– Tanner Swett
19 hours ago












@TannerSwett Yes, you are right. Imagine a needle as a line perpendicular to two squares lying in two parallel planes.
– Oldboy
15 hours ago




@TannerSwett Yes, you are right. Imagine a needle as a line perpendicular to two squares lying in two parallel planes.
– Oldboy
15 hours ago




2




2




On puzzling.SE a potential answer could be: You only need one needle that you reuse for 2000 wholes :).
– Surb
5 hours ago




On puzzling.SE a potential answer could be: You only need one needle that you reuse for 2000 wholes :).
– Surb
5 hours ago




1




1




@Surb I should have used 10.000 needles instead of 1.000. I would have got more upvotes :)
– Oldboy
4 hours ago




@Surb I should have used 10.000 needles instead of 1.000. I would have got more upvotes :)
– Oldboy
4 hours ago










1 Answer
1






active

oldest

votes

















up vote
34
down vote



accepted










The clever theorem that can be applied to questions like this one is Hall's theorem.



Construct a bipartite graph, with the vertices on one side being the polygons on the first sheet of paper, and the vertices on the other side being the polygons on the second sheet of paper. Draw an edge between two polygons whenever they overlap.



A perfect matching in this graph is a one-to-one pairing of the polygons on the two sheets such that any two polygons that are paired overlap somewhere. If we find a perfect matching, we can punch the holes: for every pair of polygons in the perfect matching, poke a needle through some part of their overlapping region.



Hall's theorem says that a perfect matching is guaranteed to exist if, for every set $S$ of vertices on one side, the set $N(S)$ of vertices adjacent to some vertex in $S$ satisfies $|N(S)| ge |S|$. In other words, if you pick any $k$ polygons on one sheet of paper, there will be at least $k$ polygons on the other sheet of paper adjcent to at least one of the ones you picked.



This follows from looking at the areas. An individual polygon has area $1$. So the $k$ polygons you chose on one sheet have total area $k$. On the other sheet, that same region needs at least $k$ polygons to cover.






share|cite|improve this answer




















  • Do you use the convexity of the polygons somewhere? Or can this assumption be dropped?
    – Janik
    9 hours ago










  • @Janik: Convexity is not required.
    – Henning Makholm
    9 hours ago










  • @HenningMakholm "On each paper, draw 1000 convex, non-overlapping polygons (...)"
    – Janik
    9 hours ago






  • 3




    @Janik: Yes, the polygons in the question happen to be convex, but this is not necessary for the result to hold.
    – Henning Makholm
    9 hours ago











  • @HenningMakholm Ah, I misunderstood what you were saying, sorry. Thanks for your clarification.
    – Janik
    9 hours ago










Your Answer





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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
34
down vote



accepted










The clever theorem that can be applied to questions like this one is Hall's theorem.



Construct a bipartite graph, with the vertices on one side being the polygons on the first sheet of paper, and the vertices on the other side being the polygons on the second sheet of paper. Draw an edge between two polygons whenever they overlap.



A perfect matching in this graph is a one-to-one pairing of the polygons on the two sheets such that any two polygons that are paired overlap somewhere. If we find a perfect matching, we can punch the holes: for every pair of polygons in the perfect matching, poke a needle through some part of their overlapping region.



Hall's theorem says that a perfect matching is guaranteed to exist if, for every set $S$ of vertices on one side, the set $N(S)$ of vertices adjacent to some vertex in $S$ satisfies $|N(S)| ge |S|$. In other words, if you pick any $k$ polygons on one sheet of paper, there will be at least $k$ polygons on the other sheet of paper adjcent to at least one of the ones you picked.



This follows from looking at the areas. An individual polygon has area $1$. So the $k$ polygons you chose on one sheet have total area $k$. On the other sheet, that same region needs at least $k$ polygons to cover.






share|cite|improve this answer




















  • Do you use the convexity of the polygons somewhere? Or can this assumption be dropped?
    – Janik
    9 hours ago










  • @Janik: Convexity is not required.
    – Henning Makholm
    9 hours ago










  • @HenningMakholm "On each paper, draw 1000 convex, non-overlapping polygons (...)"
    – Janik
    9 hours ago






  • 3




    @Janik: Yes, the polygons in the question happen to be convex, but this is not necessary for the result to hold.
    – Henning Makholm
    9 hours ago











  • @HenningMakholm Ah, I misunderstood what you were saying, sorry. Thanks for your clarification.
    – Janik
    9 hours ago














up vote
34
down vote



accepted










The clever theorem that can be applied to questions like this one is Hall's theorem.



Construct a bipartite graph, with the vertices on one side being the polygons on the first sheet of paper, and the vertices on the other side being the polygons on the second sheet of paper. Draw an edge between two polygons whenever they overlap.



A perfect matching in this graph is a one-to-one pairing of the polygons on the two sheets such that any two polygons that are paired overlap somewhere. If we find a perfect matching, we can punch the holes: for every pair of polygons in the perfect matching, poke a needle through some part of their overlapping region.



Hall's theorem says that a perfect matching is guaranteed to exist if, for every set $S$ of vertices on one side, the set $N(S)$ of vertices adjacent to some vertex in $S$ satisfies $|N(S)| ge |S|$. In other words, if you pick any $k$ polygons on one sheet of paper, there will be at least $k$ polygons on the other sheet of paper adjcent to at least one of the ones you picked.



This follows from looking at the areas. An individual polygon has area $1$. So the $k$ polygons you chose on one sheet have total area $k$. On the other sheet, that same region needs at least $k$ polygons to cover.






share|cite|improve this answer




















  • Do you use the convexity of the polygons somewhere? Or can this assumption be dropped?
    – Janik
    9 hours ago










  • @Janik: Convexity is not required.
    – Henning Makholm
    9 hours ago










  • @HenningMakholm "On each paper, draw 1000 convex, non-overlapping polygons (...)"
    – Janik
    9 hours ago






  • 3




    @Janik: Yes, the polygons in the question happen to be convex, but this is not necessary for the result to hold.
    – Henning Makholm
    9 hours ago











  • @HenningMakholm Ah, I misunderstood what you were saying, sorry. Thanks for your clarification.
    – Janik
    9 hours ago












up vote
34
down vote



accepted







up vote
34
down vote



accepted






The clever theorem that can be applied to questions like this one is Hall's theorem.



Construct a bipartite graph, with the vertices on one side being the polygons on the first sheet of paper, and the vertices on the other side being the polygons on the second sheet of paper. Draw an edge between two polygons whenever they overlap.



A perfect matching in this graph is a one-to-one pairing of the polygons on the two sheets such that any two polygons that are paired overlap somewhere. If we find a perfect matching, we can punch the holes: for every pair of polygons in the perfect matching, poke a needle through some part of their overlapping region.



Hall's theorem says that a perfect matching is guaranteed to exist if, for every set $S$ of vertices on one side, the set $N(S)$ of vertices adjacent to some vertex in $S$ satisfies $|N(S)| ge |S|$. In other words, if you pick any $k$ polygons on one sheet of paper, there will be at least $k$ polygons on the other sheet of paper adjcent to at least one of the ones you picked.



This follows from looking at the areas. An individual polygon has area $1$. So the $k$ polygons you chose on one sheet have total area $k$. On the other sheet, that same region needs at least $k$ polygons to cover.






share|cite|improve this answer












The clever theorem that can be applied to questions like this one is Hall's theorem.



Construct a bipartite graph, with the vertices on one side being the polygons on the first sheet of paper, and the vertices on the other side being the polygons on the second sheet of paper. Draw an edge between two polygons whenever they overlap.



A perfect matching in this graph is a one-to-one pairing of the polygons on the two sheets such that any two polygons that are paired overlap somewhere. If we find a perfect matching, we can punch the holes: for every pair of polygons in the perfect matching, poke a needle through some part of their overlapping region.



Hall's theorem says that a perfect matching is guaranteed to exist if, for every set $S$ of vertices on one side, the set $N(S)$ of vertices adjacent to some vertex in $S$ satisfies $|N(S)| ge |S|$. In other words, if you pick any $k$ polygons on one sheet of paper, there will be at least $k$ polygons on the other sheet of paper adjcent to at least one of the ones you picked.



This follows from looking at the areas. An individual polygon has area $1$. So the $k$ polygons you chose on one sheet have total area $k$. On the other sheet, that same region needs at least $k$ polygons to cover.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 22 hours ago









Misha Lavrov

41.3k555101




41.3k555101











  • Do you use the convexity of the polygons somewhere? Or can this assumption be dropped?
    – Janik
    9 hours ago










  • @Janik: Convexity is not required.
    – Henning Makholm
    9 hours ago










  • @HenningMakholm "On each paper, draw 1000 convex, non-overlapping polygons (...)"
    – Janik
    9 hours ago






  • 3




    @Janik: Yes, the polygons in the question happen to be convex, but this is not necessary for the result to hold.
    – Henning Makholm
    9 hours ago











  • @HenningMakholm Ah, I misunderstood what you were saying, sorry. Thanks for your clarification.
    – Janik
    9 hours ago
















  • Do you use the convexity of the polygons somewhere? Or can this assumption be dropped?
    – Janik
    9 hours ago










  • @Janik: Convexity is not required.
    – Henning Makholm
    9 hours ago










  • @HenningMakholm "On each paper, draw 1000 convex, non-overlapping polygons (...)"
    – Janik
    9 hours ago






  • 3




    @Janik: Yes, the polygons in the question happen to be convex, but this is not necessary for the result to hold.
    – Henning Makholm
    9 hours ago











  • @HenningMakholm Ah, I misunderstood what you were saying, sorry. Thanks for your clarification.
    – Janik
    9 hours ago















Do you use the convexity of the polygons somewhere? Or can this assumption be dropped?
– Janik
9 hours ago




Do you use the convexity of the polygons somewhere? Or can this assumption be dropped?
– Janik
9 hours ago












@Janik: Convexity is not required.
– Henning Makholm
9 hours ago




@Janik: Convexity is not required.
– Henning Makholm
9 hours ago












@HenningMakholm "On each paper, draw 1000 convex, non-overlapping polygons (...)"
– Janik
9 hours ago




@HenningMakholm "On each paper, draw 1000 convex, non-overlapping polygons (...)"
– Janik
9 hours ago




3




3




@Janik: Yes, the polygons in the question happen to be convex, but this is not necessary for the result to hold.
– Henning Makholm
9 hours ago





@Janik: Yes, the polygons in the question happen to be convex, but this is not necessary for the result to hold.
– Henning Makholm
9 hours ago













@HenningMakholm Ah, I misunderstood what you were saying, sorry. Thanks for your clarification.
– Janik
9 hours ago




@HenningMakholm Ah, I misunderstood what you were saying, sorry. Thanks for your clarification.
– Janik
9 hours ago

















 

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