 Solution

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.