Icosahedron of circumradius=1, inradius=0.79465447, midradius=0.85065080: (1) Viewed face-on, cross section is a regular hexagon with circumradius=sqrt((5+sqrt(5))*2/15)=0.982246946 and inradius=sqrt((5+sqrt(5))/10)=0.850651. (2) Viewed vertex-on, cross section is a regular 10-gon with circumradius=2/sqrt(5)=0.89442719 and inradius=cos(18 degrees)*circumradius i.e. inradius=sqrt((5+sqrt(5)/10)=0.850651. (3) Viewed edge-on, cross section is a hexagon (which is non-regular, with sym group of order 4 not 12) with 4 "far" vertices at radius=1, forming a "golden rectangle" inscribable in the icosahedron, plus 2 "close" vertices at radius=sqrt((5+sqrt(5))/10)=0.850651. These facts hopefully can be seen from https://en.wikipedia.org/wiki/Regular_icosahedron#Orthogonal_projections and doing some trig. (Assuming I have not made an error) these facts fail to show that an icosahedron can be Ruperted (i.e. passed thru a hole drilled in a copy) since none of these 3 orthogonal projections fits strictly inside any other. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)