Let $ABC$ be an equilateral triangle and $\mathcal{E}$ the set of all points contained in the three segments $AB$, $BC$, and $CA$ (including $A$, $B$, and $C$). Determine whether, for every partition of $\mathcal{E}$ into two disjoint subsets, at least one of the two subsets contains the vertices of a right-angled triangle.
Problem
Source: IMO 1983, Day 2, Problem 4
Tags: geometry, Ramsey Theory, combinatorics, Extremal combinatorics, right triangle, IMO, IMO 1983
grobber
12.11.2005 14:15
The answer is positive; there always a class of the partition will contain the three vertices of a right-angled triangle. First notice that if there are at least two points of the same color on a side, then all the points in $\mathcal{E}$ which project orthogonally onto points of that color on the respective side must have the opposite color if we are to have no monochromatic right-angled triangles. Now assume we can find a side on which there is at most one point bearing one of the colors (blue, say; we take the two colors to be red and blue). Then, by the observation above, it's obvious that the other two sides contain three blue vertices of a right-angled triangle, and we're done (all the points of those two sides are blue, except maybe for the endpoints which they have in common with the other side and the point which projects orthogonally onto the unique blue point on the third side). We can now assume all three sides contain at least two points of each color. Take $U\in BC$ s.t. $BC=3BU$. $U$ projects orthogonally onto $V\in CA$, and $V$ projects orthogonally onto $T\in AB$. We can easily see that $T$ projects orthogonally onto $U$. Again, by the observation in the previous paragraph, the points $U,V,T$ must have different colors if we are to have no monochromatic right-angled triangle, and this is impossible (we only have two colors).
jasperE3
30.03.2022 18:34
Does this work? Let $X\cup Y=\mathcal E$ and $X\cap Y=\emptyset$. Note that one of these subsets contains the neighborhood of either $A$, $B$, or $C$. WLOG $X$ contains the neighborhood of $A$. Let $D,E$ be points in $X$ such that $ADE$ is equilateral. Then, if $F$ is the midpoint of $AE$, $\angle AFD$ is right.
