Everyone will understand the solution to this competition question
Monday 29 December 2025 at 23:26
Visible to anyone in the world
Edited by Richard Walker, Tuesday 30 December 2025 at 23:28
This question was B1 in the December 2025 Putnam competition, which is the top competition for math undergrads at US universities. It's nice because no specialised mathematical knowledge whatsoever is needed to grasp the question and the solution; both just involve simple ideas and the solution needs just a short chain of reasoning.
I'll give the question first and then clarify it a bit.
Suppose that each point in the plane is colored either red or green, subject to the following condition: For every three noncollinear points A,B,C of the same color, the center of the circle
I'm going to change the colour scheme to red and blue, which will work better for anyone red-green colourblind. Here are a couple of sketches to help me explain.
The question is asking us to suppose we have found a scheme that assigns to each point in the plane a colour, either red or blue, in a way that meets the conditions illustrated in the sketches above, i.e.
If three red points lie on a circle, the centre of that circle will always be red, we will never find three red points on a circle with a blue centre.
If three blue points lie on a circle, the centre of that circle will always be blue, we will never find three blue points on a circle with a red centre.
You might say don't any three points lie of a circle, but no, they might all lie on the the same straight line, so we want to rule that out.)
We are asked to prove that these conditions cannot be met unless every point in the plane is the same colour as every other point; every point is red or else every point is blue.
So here is my proof. The idea to imagine trying to colour the points of the plane, always keeping to the conditions bulleted above, using both both red ands blue. We start with two points, one red and one blue, and then try to extend the configuration, still keeping to the rules. We shall find that we must inevitably reach a position for which it is impossible to extend the colouring further without breaking the rules. So no such colouring can exist.
Consider the following diagram, where I have supposed we have two points of opposite colours. It make no difference which is which, so I have made A red and B blue.
From there we can infer colours for points C, D, E, F, G, in alphabetical order, by applying our rules, as follows. To avoid a whole sequence of diagrams each little different from the last I have coloured them in advance but you need to to imagine the colours of the points are not known until we determine each.
First observe that C and D must be different colours; for if both were red C, A and D would be three red points on a circle whos centre B is blue. If both were blue C, B and D would be three blue points on a circle whose centre A is red. Which of C and D is red which blue makes no material difference so we can make C red and D blue as shown.
Now the circle through C, B, D whose centre A is red has two blue points B and D on it already and another blue point on that circle would break our rules. This forces E and F to be red as shown.
Next the circle through C, A, D whose centre B is blue has two red points A and C on it already and another red point on that circle would break our rules. This forces G to be blue as shown.
Now consider the point H which lies on the intersection of a small circle whose centre C is red and a larger circle whose centre B is blue. What colour should H be? Well the small circle with a red centre already has two blue points on it, so a third blue point is ruled out. On the other hand the larger circle with a blue centre already has two red points on it, so a third red point is ruled out. So H cannot be blue and it cannot be red either and it is impossible to colour it in a way consistent with our rules.
So we have found that if any two points are different colours it is impossible to extend that colouring to the whole plane and this the only colouring that meets the conditions is if every point is coloured red or every point is coloured blue.
Everyone will understand the solution to this competition question
This question was B1 in the December 2025 Putnam competition, which is the top competition for math undergrads at US universities. It's nice because no specialised mathematical knowledge whatsoever is needed to grasp the question and the solution; both just involve simple ideas and the solution needs just a short chain of reasoning.
I'll give the question first and then clarify it a bit.
I'm going to change the colour scheme to red and blue, which will work better for anyone red-green colourblind. Here are a couple of sketches to help me explain.
The question is asking us to suppose we have found a scheme that assigns to each point in the plane a colour, either red or blue, in a way that meets the conditions illustrated in the sketches above, i.e.
You might say don't any three points lie of a circle, but no, they might all lie on the the same straight line, so we want to rule that out.)
We are asked to prove that these conditions cannot be met unless every point in the plane is the same colour as every other point; every point is red or else every point is blue.
So here is my proof. The idea to imagine trying to colour the points of the plane, always keeping to the conditions bulleted above, using both both red ands blue. We start with two points, one red and one blue, and then try to extend the configuration, still keeping to the rules. We shall find that we must inevitably reach a position for which it is impossible to extend the colouring further without breaking the rules. So no such colouring can exist.
Consider the following diagram, where I have supposed we have two points of opposite colours. It make no difference which is which, so I have made A red and B blue.
From there we can infer colours for points C, D, E, F, G, in alphabetical order, by applying our rules, as follows. To avoid a whole sequence of diagrams each little different from the last I have coloured them in advance but you need to to imagine the colours of the points are not known until we determine each.
Now consider the point H which lies on the intersection of a small circle whose centre C is red and a larger circle whose centre B is blue. What colour should H be? Well the small circle with a red centre already has two blue points on it, so a third blue point is ruled out. On the other hand the larger circle with a blue centre already has two red points on it, so a third red point is ruled out. So H cannot be blue and it cannot be red either and it is impossible to colour it in a way consistent with our rules.
So we have found that if any two points are different colours it is impossible to extend that colouring to the whole plane and this the only colouring that meets the conditions is if every point is coloured red or every point is coloured blue.