                                 Answers
     (Only look at these after you have tried the activities for yourself

1. Changing the radii makes no difference to the view of a Platonic solid
since the view is scaled to the size of a side.

2. The radii have to be close enough for the polyhedra they make to
intersect one another.  The ex-radius must be bigger than the radius listed
under the size. Look at the truncated tetrahedron, it uses 4 triangles and 4
hexagons, to make regular faces the 1st component has in-radius 2.04 the 2nd
1.22. examples of the kinds of polyhedra you can make are in Draw file
Answer2.

3. See Draw file Answer3. The cube and the tetrahedron have no stellations.

4. See Draw file Tutorials.Pictures.Page6 and the last stellation is the
Great Stellation dodecahedron (all 3 are on the cover of your software
wallet)

5. There are reputed to be 59 stellations of the icosahedron. The exact number
depends on whether you count left and right handed forms as different or not.
There are many more stellations because there are many more sides. 
A new example is in Draw file Answer5 

6. These are shown in Draw files Answer6a,b,c with the original Archimedean
solid, all the stellation diagrams and their selections and the polyhedron.

7. These are stored in Answer7 which is an alternate StellaList for !Stellate
let us know if you find more.

8. Yes, the Cuboctahedron. Others can be made by varying the radii in the
Parameters window of the Own Solid sequence.

9. Our result is stored in NewLine9, and in the Draw file Answer9

10. You need two stellation diagrams to make one Archimedean solid, so to 
make a skeletal one you would need 4 diagrams and only 3 are allowed in 
!Stellate.

13. The compound of two tetrahedra is the stellated octahedron or
Stella Octangula. Compounds of 3 and 4 tetrahedra would have 12 and 16
faces but there are no equilateral triangles in the stellation diagrams of
the dodecahedron or the rhombic dodecahedron so no compound of 3 tetrahedra
is possible with this program. There are no components with 16 sides so a
compound of 4 tetrahedra is also not possible. The compound of 5 tetrahedra
is a stellation of the icosahedron with 20 faces.