Here is Alexander's solution.

Let's prove that it's possible for any triangle to construct circles with centres at the vertices so that the circles just touch. Let a, b and c be the sides and r₁, r₂, r₃ the radii of the circles. The only constraints on the radii are: r₁ > 0, r₂ > 0, r₃ > 0.

r1 + r2 = a r2 + r3 = b r3 + r1 = c.

The solution for this system is:

r1 = (a + c - b) 2 r2 = (a + b - c) 2 r3 = (b + c - a) 2

and since the sum of the lengths of any 2 sides in a triangle is bigger than the 3rd side, there are solutions for any positive a, b and c.

Now consider a convex polygon with n sides: a₁, a₂, a₃,...a_n. Suppose circles can be drawn with centres at the vertices of the polygon such that the circles just touch each other. Let r₁, r₂, r₃,... be the radii. Then:

r1 + r2 = a1 r2 + r3 = a2 ... rn + r1 = an.

" Let's try to figure out what r₁ equals, and all the other solutions will of course be symmetrical. "

r1 = an - rn = an - an-1 + rn-1 = an - an-1 + an-2 - rn-2 = ... = an - an-1 + an-2 - ...+(-1)n-1a1 + (-1)nr1.

" If n is even then r₁ = a_n - a_n-1 + a_n-2 - ... - a₁ + r₁ which means that the condition for the existence of solutions is:

an - an-1 + an-2 - ... - a1 = 0.

It also means that if this condition holds there are an infinite number of solutions and if it does not hold there are no solutions.

If n is odd then what we get is

r1 = 1 2 (an - an-1 + an-2 - ... + a1)

and r₁ is positive when (a_n - a_n-1 + a_n-2 - ... + a₁) > 0.

For polygons with an odd number of sides solutions always exist but 'negative' values of r_i occur when, instead of touching externally, one circle surrounds its 'neighbour' which touches it internally so the length of the edge is given by the difference of the radii and not the sum of the radii.