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Every finite cyclic transitive point set is a Euclidean Ramsey set. In particular every regular polygon is Euclidean Ramsey.

Every line segment is a Euclidean Ramsey set.

Write $X=\{x_1,\dots ,x_s\}$. Since $X$ is not a spherical point set, there are real coefficients ${\lambda }_i$, not all zero, such thatIndeed, take a minimal nonspherical subset; its points are affinely dependent, and centering the proper spherical subset shows that the corresponding quadratic sum is nonzero. The three relations are invariant under isometries, and rescaling the ${\lambda }_i$ lets us assume $b=1/2$.

Choose $\delta <1/(2s)$ and color every $y$ by the $s$ intervals of length $\delta$ containing the fractional parts of ${\lambda }_i\Vert y{\Vert }^2$. This uses finitely many colors. If $y_1,\dots ,y_s$ were a monochromatic isometric copy of $X$, then eachwould lie within $\delta$ of an integer. Their sum is within $s\delta <1/2$ of an integer, but because ${\sum }_i{\lambda }_i=0$ it equalsa contradiction. Hence $X$ is not a Euclidean Ramsey set.

Solved by gpt-5.6-sol high.

Let $S$ be a finite Ramsey witness for $X$ under $k$ colors. Choose a finite witness $T$ for $Y$ under $k^{|S|}$ colors. Given a coloring of $S\times T$, color each $t\in T$ by the complete vectorThere is a copy $Y^{\prime }\subseteq T$ on which this vector is constant. Thus, for each $s$, the color $c(s,t)$ is independent of $t\in Y^{\prime }$. These values define a $k$-coloring of $S$, which has a monochromatic copy $X^{\prime }$. Then $X^{\prime }\times Y^{\prime }$ is a monochromatic isometric copy of $X\times Y$. This proves the product theorem for Euclidean Ramsey sets.

Every nondegenerate triangle and every line segment is a Euclidean Ramsey set. If $T$ is the given acute triangle and $I$ is a segment of length $t$ in a new orthogonal coordinate, then $T\times I$ is exactly the vertex set of the triangular prism with base $T$ and height $t$. The product theorem therefore makes it Euclidean Ramsey.

Solved by gpt-5.6-sol high.

Put ${\rho }^2=28-28\sqrt{5}/5$ and, for $j\in \mathbb{Z}/30\mathbb{Z}$, defineThe shift $v_j\mapsto v_{j+1}$ is an isometry acting transitively on the finite set $V=\{v_j\}$, so $V$ is a cyclic transitive point set and hence a Euclidean Ramsey set by the Kriz theorem for cyclic transitive point sets.

For every $j,k$,For $k=1,5,11,15,4,14$, these squared distances are respectivelyConsequently $v_0,v_1,v_{26},v_{15}$, in that order, have consecutive side lengths $9,5,9,10$ and equal diagonals $\sqrt{131}$. This is an isometric copy of the required isosceles trapezium. Since every monochromatic copy of $V$ contains this four-point subset, the trapezium is Euclidean Ramsey.

Solved by gpt-5.6-sol high.

Every regular polygon is a Euclidean Ramsey set. A Hales-Jewett theorem combinatorial line in a Cartesian power of its vertex set is a scaled regular polygon; including finitely many reciprocal square-root scalings makes one such line isometric to the original polygon.

A finite point set is spherical when it lies on a sphere. Every Euclidean Ramsey set is spherical. Whether every finite spherical point set is Euclidean Ramsey is open.

Every nondegenerate triangle is a Euclidean Ramsey set.