(In this answer, I do as the questioner did, calling the smallest containing sphere the circumsphere and its center the circumcenter, even if this sphere does not touch all vertices. Similarly, I call the largest enclosed sphere the insphere and its center the incenter, even if this sphere does not touch all faces.)
Assuming that the circumcenters and incenters of both the original polyhedron and the dual all coincide, these ratios should be the same.
A convex polyhedron which contains the origin in its interior can be described either as the convex hull of a set $\{a_i\}\subseteq {\Bbb R}^3$ of vertices or as the intersection of a set of bounding half-spaces $\{S_j\}$, where each bounding half-space is of the form
$$S_j=\{x\mid x\cdot b_j\le 1\}, \qquad b_j\in {\Bbb R}^3. $$
Letting the circumsphere be centered at the origin, the circumradius $R$ will equal the maximum value of any $|a_i|$. Also, since the distance between the origin and the plane $\{x\mid x\cdot b_j=1\}$ is $|b_j|^{-1}$, if the insphere is centered at the origin, the inradius will equal the minimum value of any $|b_j|^{-1}$.
The polar dual of a given polyhedron can be constructed by interchanging the $a_i$s and $b_j$s. So, if the original polyhedron had circumradius $R$ and inradius $r$, and if the circumcenter and incenter of the polar dual are still at the origin, then the polar dual must have circumradius $1/r$ and inradius $1/R$, and the ratio $R/r$ of the circumradius to the inradius remains the same.
For an example, take a cuboctahedron with vertices
$$
(\pm 1, \pm 1, 0), \ (\pm 1, 0, \pm 1), \ (0, \pm 1, \pm 1).
$$
In this case, the circumsphere and insphere are both centered at the origin. The radius $\sqrt{2}$ circumsphere touches all vertices; the insphere has radius $1$ and touches only the square faces; and the ratio $R/r$ is $\sqrt{2}$.
The dual of this polyhedron is the rhombic dodecahedron, with vertices
$$
(\pm 1,0,0), \ (0,\pm 1,0), \ (0,0,\pm 1),\ (\pm\frac12, \pm\frac12, \pm\frac12).
$$
The circumsphere and the insphere are both still centered at the origin. The circumsphere has radius $1$ and touches only those vertices where four faces meet (the first six above); the insphere has radius $1/\sqrt{2}$ and touches all faces; and the ratio $R/r$ is still $\sqrt{2}$.
It will take a bit of thought to realize the result.
First fact: one uses is the sum of the angles around a vertex. Projected to the circle it's obvious - its $360^\circ$, and one should realize that non-projected has to be less than that. This means that the polygons will have corners of less than $120^\circ$.
Second fact: the angle sum of a polygon is $(N-2)180^\circ$, and on a regular polygon the angles are therefore $(1-2/N)180^\circ$. This makes it impossible to have polygons with more than $5$ corners.
First and second fact puts a limit on how many polygons can meet at a vertex. For pentagons you can only meet $3$ polygons at a vertex since the vertex angle is $108^\circ$ and a fourth would exceed a sum of $360^\circ$. For squares you would have the same even though a fourth square would only result in a sum of $360^\circ$. For triangles on the other hand you can have up to $5$ joining at a vertex, for the same reason as the square you can't have $6$ even though it would only add up to $360^\circ$.
After that it's just a matter of verifying the possible candidates, and they are all possible. You have with triangles: three joining at vertices means tetrahedron, four means octahedron, and five means icosahedron. For squares and pentagons there's only one possibility and that will mean the cube and the dodecahedron.
The same reasoning can be done using spherical projection of course, but then you use the fact that the angle sum at vertices is constant $360^\circ$ while the angle sum of a polygon is larger than $(N-2)180^\circ$.
Best Answer
Here are the 5 Platonic polyhedra together with their truncations. The best polyhedron program that I know is Stella (you can find the site with Google). A trial download is possible. The following images were created by my own program, MathpadDraw which you can download from mathokay.com.