You start with an infinite Go board. On every point of the board you place one colored stone. There are $n>1$ different colors. Find all natural numbers $n$ that no matter how the stones are colored, three stones of the same color form the vertices of a right-angled triangle. The catheti (legs) of the right triangle must be on the lines of the board.

Any ideas how to solve this kind of problem and to which area of mathematics this question belongs?

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C. Podolski
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  • Any column of the Go board contains an infinite number of stones. Since $n$ is finite, there must be one color of stone such that there are at least three of them in a given column. Three collinear stones don't form a right triangle. – John Douma Aug 29 '18 at 21:45
  • @C.Podolski You might want to make it clear you only need the stones on the verticies of a right triangle to be the same color. I was thinking you wanted the edges filled in too. – Joe Aug 29 '18 at 21:54
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    Do you mean that no matter how the stones are placed, and given a color, we can always find at least one right triangle? – John Douma Aug 29 '18 at 22:08
  • @JohnDouma: No, we could have only a finite number of stones of one color and no right triangle formed. We could also have an infinite set of stones of one color all in the same row or all on a given diagonal. Some other color will have a triangle then. – Ross Millikan Aug 29 '18 at 22:16
  • @RossMillikan I understand what you are saying but that is not how I interpret the statement as it is written. I am sure you are right. – John Douma Aug 29 '18 at 22:22
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    The correct field for this is combinatorics. Within combinatorics, the study of structure that arises in very large (or infinite sets) is called Ramsey theory. When we are specifically talking about infinite sets, the term infinitary combinatorics is also appropriate. I would refer to this problem as Ramsey Theory alone, as it does not share the same character of most of infinitary combinatorics, despite being technically a combinatorial theorem about infinite sets. Infinitary combinatorics is primarily about the combinatorial structure of ordinals. – Stella Biderman Aug 29 '18 at 22:25
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    "Ramsey Theory," "Ramsey's Theorem" (which inspired the field), and "Infinitary combinatorics" all have good Wikipedia articles if you wish to learn more about the field. – Stella Biderman Aug 29 '18 at 22:29
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    Note that the original question did not specify isosceles right triangles The edit has invalidated several good answers. It should have been asked as a separate question. It is unfortunate that you assigned the bounty to the edited version. – Ross Millikan Sep 01 '18 at 15:30
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    I completely agree with Ross Millikan, I performed a rollback and removed the bounty. Chamaleon questions should be avoided, especially if they are bounty questions. Remarkably, you already asked the *isosceles* variant [here](https://math.stackexchange.com/questions/2901855/isosceles-right-angled-triangles-defined-on-an-infinite-go-board-by-same-colored), and you should have been honest about the problem being part of [a contest](https://www.mathe-wettbewerbe.de/bwm/aufgaben/aufgaben-2018/aufgaben-18-2.pdf). – Jack D'Aurizio Sep 02 '18 at 14:15

3 Answers3


Any finite number $n$ will produce a right triangle, in fact an infinite number of them. The secret is that we can strike out lots of rows or columns and still have an infinite board. Given an $n$, pick a row of the board. There is at least one color that has an infinite number of stones in the row, call it red. Now delete all the columns that do not have a red stone in the row we are considering. We still have an infinite board, but the row under consideration has only red stones. If there is a red stone anywhere else on the board it will make an infinite number of red right triangles. Strike out the row under consideration and we have the same problem with $n-1$ colors. We can keep eliminating colors one by one until we get to just one.

Ross Millikan
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  • One crucial step in this argument is to decide from the beginning only to look for _axis-aligned_ right triangles; otherwise the striking-out of rows or columns wouldn't be safe. – hmakholm left over Monica Aug 29 '18 at 22:00
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    "no matter how the stones colored, three stones of the same color build a right triangle" Doesn't this mean that any three stones of the same color build a right triangle? – John Douma Aug 29 '18 at 22:06
  • @JohnDouma: Um, no. For example, stones at coordinates $(0,0)$, $(2,0)$ and $(1,2)$ build an acute isosceles triangle, not a right triangles. – hmakholm left over Monica Aug 29 '18 at 22:08
  • @HenningMakholm Yes, I see that. That's my point. The way the problem is stated, it says that three stones of the same color form a right triangle. It should say, for each color there are three stones that form a right triangle. – John Douma Aug 29 '18 at 22:10
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    @JohnDouma: Since it neither says "any three stones" nor "for some three stones", I think it is reasonable to refrain from interpreting the problem in a way that would make it completely trivial, when a different meaningful interpretation is available. – hmakholm left over Monica Aug 29 '18 at 22:12
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    @HenningMakholm I disagree. Mathematics is about making precise statements. I should not have to seek an interpretation that makes a statement seem reasonable. – John Douma Aug 29 '18 at 22:20
  • "Any finite number n will produce a right triangle, in fact an infinite number of them." - uh, what? It sounds like you're saying that there would have to be a same-colored right triangle even with an infinite number of colors, but with an infinite number of colors, we could give each stone its own color. – user2357112 Aug 30 '18 at 00:08
  • @user2357112: No, I said any finite number of colors will produce an infinite number of right triangles. That doesn't mean we are guaranteed any right triangles with an infinite number of colors. You have shown that we are not. There are many things that are true for any finite number, no matter how large, but not true for an infinite number. – Ross Millikan Aug 30 '18 at 00:13
  • While this is a suggestive argument, you haven't shown that your argument always produces a right triangle. For example, suppose that in row 0 all even columns are red and odd columns are blue; while in all other rows it's the opposite (odd columns red, even columns blue). If I happen to pick row 0 in your first step, I'll strike out the columns containing all other red stones, and I won't find a red right triangle. – Michael Seifert Aug 30 '18 at 00:25
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    @MichaelSeifert Note the end of the proof is an appeal to induction on $n$. Yes, for the board you describe, this algorithm will strike out all odd columns, fails to show there is any red right triangle (even though there were some), and then strikes out row zero. This leaves us with an infinite board where all the stones are blue - the base case $n=1$. For this board, it's obvious there are infinitely many blue right triangles with legs parallel to the axes, and those are also blue right triangles on the original board. – aschepler Aug 30 '18 at 02:51
  • @JohnDouma I don't think it's ever valid in English to add "any" or "for all" where it's not stated. "Three people are in the next room" does not mean "Any three people you choose are all in the next room". – aschepler Aug 30 '18 at 03:04
  • @aschepler To what are you referring? – John Douma Aug 30 '18 at 03:36
  • @JohnDouma, didn't the question ask to "find *n*, such that there are so and so right triangles", instead of declaring as fact that there are some? It was the problem statement, not a definition. – ilkkachu Aug 30 '18 at 07:59
  • @ilkkachu I do not see your quote anywhere in the question. I would suggest you address any ambiguity to the poster. – John Douma Aug 30 '18 at 08:38

Consider a finite $m\times m$ board where every intersection either has a stone, or does not. If there are at least $2m$ stones on this board, then I claim there will exist three stones which are arranged in an axis-aligned right triangle.

To prove this, number the stones from $1$ to $2m$. For each $1\le k \le 2m$, let $f(k)$ be the number of rows spanned by the stones numbered $1$ to $k$, plus the number of columns spanned by these stones. Note that $f(1)=2$, $f(2m)\le 2m$, and $f(k)$ is a weakly increasing function. It cannot be strictly increasing; if it were strictly increasing, you would have $f(2m)\ge 2m-1+f(1)=2m+1$, contradicting $f(2m)\le 2m$. This means there must be some number $k$ for which $f(k)=f(k-1)$. But this means that stone number $k$ is the same row as a previous stone, and in the same column as a previous stone, so that these three stones form a right triangle.

In particular, if an $m\times m$ board is filled with stones in $n$ colors, then some color will appear on at least $m^2/n$ of the stones, so if $m^2/n\ge 2m$, there will be a right triangle of stones in that color. Therefore, for all $n$, an infinite colored board will have a monochromatic coloring, and to find one, you only need to search in a $2n\times 2n$ sub-square of stones.

This problem seems to fall under the topic of Ramsey theory.

As a side note, you can also prove that an $m\times m$ board with only $2m-1$ stones contains an axis-aligned right triangle. This is tight, since you can place $2m-2$ stones without forming a triangle.

Mike Earnest
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Call a stone h-unique if no other stone on the same row has the same colour, call it v-unique if no other stone on the same column has the same colour. Clearly, every row contains at most $n-1$ h-unique stones and every column at most $n-1$ v-unique stones. Hence in a rectangle of $n$ rows and $n^2-n+1$ columns, there are at most $n^2-n$ h-unique stones. Hence there is a column of this rectangle that does not have a h-unique stone. One stone in this column is not v-unique. This stone, plus a stone witnessing its non-h-uniqeness, plus a stone witnessing its non-v-uniqueness, form a right triangle of same-colour stones as desired.

Hagen von Eitzen
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