Each point of a plane is painted in one of three colors. Show that there exists a triangle such that:
$ (i)$ all three vertices of the triangle are of the same color;
$ (ii)$ the radius of the circumcircle of the triangle is $ 2008$;
$ (iii)$ one angle of the triangle is either two or three times greater than one of the other two angles.
(i) and (ii) are simple.
(i) If there existed no triangle with three vertices of same colour then the all the points coloured with a colour say $ c_1$ lie of a straight line. We may at max have three distinct lines for each of three colours that do not fill a plane.
(ii) We can get a monochromatic triangle with circumradius equal to any positive real number. Pick a arbitrary point in space and construct a circle of radius $ r$, a positive real number. Pick a point on the circumference and let it have the colour $ c_1$ wlog.
If we are not able to find another point (or if we find just one more point $ c_1$) on the circumference that is coloured $ c_1$ then every other point on the circumference is coloured either $ c_2$ or $ c_3$. If we do not find a point with $ c_2$ then we are done as rest of the points are $ c_3$. If we find just either 1 or 2 points coloured $ c_2$, then too we have rest of the points coloured $ c_3$.
No no no no... You should prove that there exist a triangle which fullfils all 3 conditions! These are not three diferent problems, these are conditions for triangle