Midradius
The radius π rho
of the midsphere of a polyhedron,
also called the interradius. Let π P
be a point on the original polyhedron and π P^'
the corresponding point π P
on the dual. Then because π P
and π P^'
are inverse points, the radii π r_d=OP^'
, π R=OP
, and π rho=OQ
satisfy
| π r_dR=rho^2. |
(1)
|
The above figure shows a plane section of a midsphere.
Let π r_d
be the inradius
the dual polyhedron, π R
circumradius of the original polyhedron, and π a
the side length of the original polyhedron. For a regular
polyhedron with SchlΓ€fli symbol π {q,p}
, the dual
polyhedron is π {p,q}
.
Then
| π r_d^2 | π = | π [acsc(pi/p)]^2+R^2 |
(2)
|
| π Image | π = | π a^2+rho^2 |
(3)
|
| π rho^2 | π = | π [acot(pi/p)]^2+R^2. |
(4)
|
Furthermore, let π theta
be the angle subtended by the polyhedron
edge of an Archimedean solid. Then
| π r_d | π = | π 1/2acos(1/2theta)cot(1/2theta) |
(5)
|
| π rho | π = | π 1/2acot(1/2theta) |
(6)
|
| π R | π = | π 1/2acsc(1/2theta), |
(7)
|
so
(Cundy and Rollett 1989).
For a Platonic or Archimedean solid, the midradius π rho=rho_d
of the solid and dual can be expressed in terms of the circumradius π R
of the solid and inradius π r_d
of the dual gives
| π rho | π = | π 1/2sqrt(2)sqrt(r_d^2+r_dsqrt(r_d^2+a^2)) |
(9)
|
| π Image | π = | π sqrt(R^2-1/4a^2) |
(10)
|
and these radii obey
| π Rr_d=rho^2. |
(11)
|
See also
Archimedean Dual, Archimedean Solid, Circumradius, Inradius, Midsphere, Platonic SolidExplore with Wolfram|Alpha
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References
Cundy, H. and Rollett, A. Mathematical Models, 3rd ed. Stradbroke, England: Tarquin Pub., pp. 126-127, 1989.Referenced on Wolfram|Alpha
MidradiusCite this as:
Weisstein, Eric W. "Midradius." From MathWorld--A Wolfram Resource. https://mathworld.wolfram.com/Midradius.html
