Let $X,Y$ be distinct points in a plane with the same color, say red. Consider a circle $\omega$ with center $X$ that pass through $Y$. Note that any point $N$ on this circle that doesn't lie on $XY$ must be either blue or green. Otherwise, $X,Y,N$ will form a red isoscele. Now, let $ABCDE$ be a regular pentagon such that all of its vertices lie on $\omega$ and doesn't lie on $XY$. By Pigeonhole's Principle, there must be $3$ points from $A,B,C,D,E$ with the same color. Of course, two out of that three points are adjacent because there are only $5$ points. Without loss of generality, let $A,B$ be color in blue. We divide the position of the other blue point into three cases. $C$ is blue. Then $ABC$ will form a blue isoscele$: 108^{\circ}-36^{\circ}-36^{\circ}$. $D$ is blue. Then $ABD$ will form a blue isoscele$:36^{\circ}-72^{\circ}-72^{\circ}$. $E$ is blue. Then $ABE$ will form a blue isoscele$:108^{\circ}-36^{\circ}-36^{\circ}$. Hence, there must be an isoscele whose vertices have the same color. EDIT: 400th post!