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Fix a number $k$ of colors. For a two-point set whose points are distance $d$ apart, take the vertices of a regular simplex with $k+1$ vertices and side length $d$. The pigeonhole principle gives two vertices of one color, and they form the required congruent copy. Thus every two-point set, equivalently every line segment, is a Euclidean Ramsey set.

For an equilateral triangle of side length $d$, take a regular simplex with $2k+1$ vertices and side length $d$. The pigeonhole principle gives three vertices of one color, and every three vertices of a regular simplex form an equilateral triangle. Hence every equilateral triangle is Euclidean Ramsey.

To prove the product theorem for Euclidean Ramsey sets, let $S_X$ be a finite Ramsey witness for $X$ under $k$ colors. There are at most $k^{|S_X|}$ possible color patterns on $S_X$. Choose a finite Ramsey witness $S_Y$ for $Y$ under that many colors. Given a $k$-coloring of $S_X\times S_Y$, color each $y\in S_Y$ by the complete patternThere is a copy $Y^{\prime }\cong Y$ on which this pattern is constant. The common pattern on $S_X$ contains a monochromatic copy $X^{\prime }\cong X$. Every point of $X^{\prime }\times Y^{\prime }$ then has the same original color, and the orthogonal product is congruent to $X\times Y$.

A rectangle is the Cartesian product of two line segments, so it is Euclidean Ramsey. Three suitable vertices of a rectangle form a right triangle; any subset of a monochromatic set is monochromatic. Thus every right triangle is Euclidean Ramsey.

It remains to show that the collinear set $\{0,1,2\}$ behaves differently. In every ${\mathbb{R}}^m$, use the finite coloringA congruent copy has the form $a-v,a,a+v$ with $\Vert v\Vert =1$ for the Euclidean norm. The parallelogram law givesPut $n_{+}=\lfloor 2\Vert a+v{\Vert }^2\rfloor$, $n_{-}=\lfloor 2\Vert a-v{\Vert }^2\rfloor$, and $n_0=\lfloor 2\Vert a{\Vert }^2\rfloor$. The errors introduced by the three floor functions show thatIf all three points had one color, the integer in the middle would be divisible by ten, which is impossible. This proves the three-term unit arithmetic progression is not Euclidean Ramsey assertion.

Let $X=\{v_1,\dots ,v_m\}$ be the vertices of a regular polygon, indexed cyclically so that $v_i\mapsto v_{i+1}$ is an isometry. We first prove the product lemma needed for the cyclic-symmetry argument.

Suppose $A\subseteq X$ has the property that, for every number of colors, a finite witness forces a copy of $X$ on which the vertices corresponding to $A$ are monochromatic. We claim that for every $N,k$ there is a finite witness forcing an $A$-invariant copy of $X^N$. Proceed by mathematical induction on $N$. The case $N=1$ is the assumption. For the step, choose a finite set $S$ that forces an $A$-invariant $X^{N-1}$ for a coloring with $k^{|X|-|A|+1}$ colors, and choose $T$ that forces an $A$-monochromatic $X$ for a coloring with $k^{|S|}$ colors.

Given a $k$-coloring $c$ of $S\times T$, color $t\in T$ by its complete vector $(c(s,t){)}_{s\in S}$. In the resulting copy of $X$ in $T$, that vector is the same for every member of $A$. Each $s\in S$ can therefore be colored by the vector consisting of $c(s,a)$ for one $a\in A$ and the colors $c(s,x)$ for $x\in X\setminus A$. Applying the choice of $S$ gives an $A$-invariant copy of $X^{N-1}$, and its product with the chosen copy of $X$ in $T$ is the desired $A$-invariant copy of $X^N$. This proves the lemma rather than assuming it.

We now prove by induction on $r$ that a finite witness forces a copy of $X$ whose first $r$ vertices have one color. The case $r=1$ is trivial. For $r=2$, use that a line segment is a Euclidean Ramsey set; each occurrence of a segment congruent to $v_1{v}_2$ can be extended to a copy of the regular polygon, and only finitely many extensions are needed for a finite witness.

Assume the assertion for $r$, and put $A=\{v_1,\dots ,v_r\}$. The product lemma lets us work inside an $A$-invariantly colored copy of $X^N$, with $N$ as large as required. For an increasing $(m-1)$-element subset $H=\{a_1<\cdots <a_{m-1}\}$ of $[N]$ and $1\le i\le r$, define a word $w_i(H)$ by putting $v_{i+j}$ in coordinate $a_j$, with subscripts read modulo $m$, and putting $v_1$ in every other coordinate. Color $H$ by the $r$-tupleBy the Finite Ramsey theorem, for sufficiently large $N$ there are coordinates $b_1<\cdots <b_m$ on which all $(m-1)$-subsets have the same tuple color.

For $1\le i\le r+1$, let $z_i$ be the word whose entries in coordinates $b_1,\dots ,b_m$ arecyclically, and whose other entries are $v_1$. For $i\le r$, the word $z_i$ differs from $w_i(\{b_2,\dots ,b_m\})$ only in coordinate $b_1$, where the two entries $v_i,v_1$ lie in $A$. Similarly $z_{i+1}$ differs from $w_i(\{b_1,\dots ,b_{m-1}\})$ only in coordinate $b_m$, again by two members of $A$. The $A$-invariance and the homogeneous choice of the $b_j$ therefore giveHence $z_1,\dots ,z_{r+1}$ have one color.

The cyclic words $z_1,\dots ,z_m$ form an isometric copy of $\sqrt{m}X$: in each of the $m$ varying coordinates the cyclic shift is an isometry, so every squared distance is multiplied by $m$. Rescaling the finite witness by $1/\sqrt{m}$ gives a copy of $X$. The induction reaches $r=m$, proving that every regular polygon is a Euclidean Ramsey set.

For three consecutive vertices $p_0,p_1,p_2$ of a regular $M$-gon scaled to have side length one,Since the polygon is Euclidean Ramsey, choosing $M$ large enough proves that $\{0,1,2\}$ is an approximately Euclidean Ramsey set.

In fact every finite $X\subseteq {\mathbb{R}}^d$ is approximately Ramsey. First perturb each coordinate of each point onto a sufficiently fine lattice $\delta \mathbb{Z}$, changing all pairwise distances by less than $\epsilon /2$. For each coordinate, map the finitely many required integers to consecutive vertices of a very large regular polygon, scaled so that one angular step has arc length $\delta$. If the step angle is $\theta$, the chord replacing a difference of $q$ lattice steps has lengthas $\theta \to 0$, uniformly over the finitely many differences involved. Taking the orthogonal Cartesian product of $d$ such polygons therefore embeds the perturbed points with a further distance error below $\epsilon /2$. Each polygon is Euclidean Ramsey, and their product is Euclidean Ramsey by the product theorem for Euclidean Ramsey sets. A monochromatic copy of that product contains the corresponding approximate copy of $X$, completing the proof.

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.