SELECTION EXAMINATION

a. There are defined totally seven equilateral triangles from the given points. Since $\Delta E=EZ=Z\Delta =\frac{\alpha }{2}$ the seven equilateral triangles are Figure 7 Figure 8 $AB\Gamma$, $A\Delta Z$, $B\Delta E$, $E\Gamma Z$, $\Delta EZ$, $BHE$ (symmetric of $\Delta BE$ with respect to the line $B\Gamma$), $\Delta H\Gamma$ (it has $\Gamma H=\Gamma \Delta =\frac{\alpha \sqrt{3}}{2}$ = altitude of equilateral triangle, $\Gamma \hat{\Delta }H=60^{\circ }$).

b. Let $B$ and $E$ are colored red. If the points $\Delta$ or $H$ are also red, then we have the wanted triangle. Suppose that $\Delta$ and $H$ are colored blue. If the point $\Gamma$ is blue then we are done. Let the point $\Gamma$ is colored red. If the point $Z$ is colored red, then the triangle $E\Gamma Z$ has its vertices red. Let point $Z$ is colored blue. In that case for any coloring of $A$ one of the triangles $AB\Gamma$, $A\Delta Z$ will have its vertices of the same color. Figure 9

c. In that case we have not the same conclusion. In figure 9 we give a coloring with all equilateral triangles having their vertices with different colors. The points $B$, $\Gamma$, $Z$ have been colored red and the remaining points blue.