We will show that the conclusion holds when $r > \frac{\pi}{2}$.

We begin with a circle $C(r)$ with center $C$ and radius $r > \frac{\pi}{2}$. Now, a regular polygon $P$ can be inscribed in the circle to have any odd number of sides $2k + 1$. Because the number of sides is odd, each vertex $A$ is opposite a pair of vertices which determine the opposite side $YZ$, and it is such longest diagonals $AY$ and $AZ$ that we are interested in. The greater $2k + 1$ is taken, the smaller $YZ$ will get; by taking $k$ large enough, we can make $YZ$ so small that $AY$ will be so close to being a diameter (of length $2r$ which is bigger than $\pi$) that the length of $AY$ will also exceed $\pi$. Suppose, then, that $k$ is chosen big enough to make $AY > \pi$.

Clearly all such longest diagonals $AY$ are tangents to a small circle $C(s)$ in the center of $C(r)$.

Figure 1:
Now, it is vital to our solution that the length $AY$ of these tangents be exactly $\pi$ units. Therefore,

let the figure be shrunk toward the center $C$ in the ratio $\pi : AY$; this will carry everything into the interior of the given circle, implying that all the image points will be colored, whereas the boundary of $C(r)$ was not colored to begin with. For simplicity, let us keep the same names for the images under this transformation, bearing in mind that now the length of every diagonal like $AY$ is $\pi$.

Now a tangent to $C(s)$ from a vertex of $P$ meets the circumcircle in one of the two opposite vertices of $P$. Consequently, the sequence of tangents $AY, YA_1, A_1Y_1, Y_1A_2, \dots$ carries one around $P$ from $A$ through the vertices $A, Y, A_1, Y_1, A_2, \dots$ and after $2k$ such steps, we reach the opposite vertex $Z$. Clearly, every adjacent pair of vertices in the above list of vertices are at a distance $\pi$ units and starting with $A$ and ending with $Z$, we traverse through $2k$ tangents. Thus if $A$ is red, then $Y$ is blue, $A_1$ is red, $Y_1$ is blue and so on. This forces $Z$ to be colored red and both ends of the tangent $AZ$ have the same color. We are done. $\square$