Polyhedron inside Sphere
Date: 5/24/96 at 21:54:49 From: Keays Subject: Polyhedron Inside Sphere Dear Dr. Math, A mate of mine is trying to build a 12-faced shape, each face being a perfect pentagon. (I have forgotten the correct name for it). He wants to know how long each side of the pentagon should be, to fit it in a sphere of diameter 2.9 m. My calculations say about 0.8 m, but I estimate the answer to be closer to 1 m. What is the real answer? Hope this all makes sense to you, Thanks heaps, Roger Keays
Date: 5/28/96 at 19:38:39 From: Doctor Pete Subject: Re: Polyhedron Inside Sphere The polyhedron you are referring to is called the dodecahedron. It has many interesting properties, one of which is the fact that you can inscribe a cube in it. That is, you can pick eight vertices (corners) of the dodecahedron which are also vertices of a cube. Then 12 edges of the cube will touch the 12 faces of the dodecahedron. Since this cube is also inscribed in the same sphere as that of the dodecahedron, it gives an easy way to find the length of the dodecahedron's edge in terms of the radius of the circumscribed sphere. Let R be the radius of the given sphere (R = 1.49m in your case), and let x be the edge length of the inscribed cube. Then x*sqrt(3) = 2*R (use the Pythagorean Theorem twice), so x = 2*R/sqrt(3). But since the edge is coplanar with the face of a pentagon, we have the following picture: A _________________ C /\ _. - / \ _. - / _. - / _. - D / - /- B where BC = x, the edge of the cube, AB = BC, two sides of a pentagonal face, and BAC = 3*pi/5 = 108 degrees. AD is the angle bisector of BAC, so ADC is a right angle. Then sin(DAC) = DC/AC = (x/2)/AC, so AC = x/(2*sin(3*pi/10)) = R/(sin(3*pi/10)*sqrt(3)) = 4*R/(sqrt(15)+sqrt(3)), or R*0.7136. So for your case, it should be about 1.06333 m, which agrees very closely with your result. -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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