Putnam 2025, Problem B-1

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 passing through \( A, B \) and \( C \) is also this color. Prove that all points of the plane are the same color.

Solution:
**First solution.** By way of contradiction, suppose that there exist a red point \( P_1 \) and a green point \( P_2 \). Let \( P_3, P_4 \) be the two points that form equilateral triangles with \( P_1 \) and \( P_2 \). These cannot be both red (as \( P_2 \) is the center of the circle through \( P_1, P_3, P_4 \)) or both green (as \( P_1 \) is the center of the circle through \( P_2, P_3, P_4 \)). Without loss of generality, suppose that \( P_3 \) is red and \( P_4 \) is green. Now define points \( P_5, \ldots, P_8 \) as follows.
- Let \( P_5 \) be the reflection of \( P_2 \) across \( P_1 P_4 \). This point cannot be green because \( P_1 \) is the center of the circle through \( P_2, P_4, P_5 \), so it must be red.
- Let \( P_6 \) be the reflection of \( P_1 \) across \( P_4 P_5 \). Let \( P_7 \) be the reflection of \( P_1 \) across \( P_2 P_4 \). Neither of these points can be red because \( P_4 \) is the center of the circle through \( P_1, P_5, P_6, P_7 \), so they must be green.
- Let \( P_8 \) be the reflection of \( P_1 \) across \( P_2 P_3 \). This point cannot be red because \( P_2 \) is the center of the circle through \( P_1, P_3, P_8 \), so it must be green.

Now \( P_1 \) is red while \( P_6, P_7, P_8 \) are green, a contradiction.

**Second solution.** We prove more generally that for fixed positive integers \( k \) and \( \ell \), if the points of the plane are colored in \( k \) different colors subject to the condition that if \( \ell \) distinct points on a circle are the same color, then so is the center, then again all points must be the same color.

Suppose by way of contradiction that there are points of two distinct colors, say \( P_1 \) (colored red) and \( P_2 \) (colored green). Each circle centered at \( P_i \) then contains at most \( (k - 1)\ell \) points that are not the color of \( P_1 \) (namely, at most \( \ell \) point in each of the other \( k - 1 \) colors). Now draw \( m \) distinct circles around each \( P_i \), each of radius greater than the distance \( P_1 P_2 \); these circles meet in \( 2m^2 \) distinct points. Of these points, at least \( 2m^2 - m(k - 1)\ell \) are red and at least \( 2m^2 - m(k - 1)\ell \) are green, but for \( m \) sufficiently large this accounts for more than \( 2m^2 \) points, a contradiction.