assume the contrary -- that there is no monochromatic right triangle. label the equilateral triangle $ ABC$. trisect the sides of the triangle to get 6 points that form a regular hexagon. if any two opposite vertices of this hexagon are colored the same color, then the other four vertices must all be of the opposite color, which give us a right triangle of this color. so opposite vertices must be of opposite colors. it's easy to see that no matter how we color the vertices of the hexagon so that opposite vertices are oppositely colored, two adjacent vertices which lie on a side of the triangle must have opposite colors. call these two points $ P, Q$ and assume they lie on side $ AB$, with $ P$ red and $ Q$ blue. let $ P'\in BC$ be $ P$'s opposite vertex in the hexagon, and $ Q'\in AC$ $ Q$'s opposite. then $ P, Q'$ are both red and form a right triangle with $ A$, while $ Q, P'$ are both blue and form a right triangle with $ A$ as well. thus $ A$ can't be colored red or blue - a contradiction.