[math-fun] triangoops, solid angle from three wedges
Nick Baxter points out that in answering
what is the smallest constant c so that the graph of the function f(x) = x^3 - c x contains the vertices of an equilateral triangle?
I somehow edited sin((%pi/9)) into sin(pi)/9. Please imagine c = sqrt(8 - 8 sqrt(3) sin(pi/9)) ~ 1.80578 . Claim: The midpoint of one triangle side is the origin. Otherwise, centering some side and rotating its endpoints back onto the curve will intersect the other two sides with the curve, indicating that c is non-minimal. So let y(x) := x^3 - c x, and the vertices be z := x + i y(x), -z, and z sqrt(-3) = w + i y(w). Eliminate w from the last eqn, leaving a polynomial p(x,c) = 0 biquartic in x and cubic in c. For minimality, dc/dx = 0. This is equivalent to resultant(p,dp/dx,x) = 0, which gives a bicubic in c. Experimentally, c=1 gives no real roots x. c=2 gives two pairs. The critical c should be where these merge into double roots, i.e., when p and dp/dx have a common factor. Which is what the resultant finds. ------- A useful formula I can't find in MathWorld is for the solid angle Omega of a trihedral vertex with wedge (or sphere arc) angles a, b, c: 2 (cos(c) + cos(b) + cos(a) + 1) Omega = acos(-------------------------------------- - 1) (cos(a) + 1) (cos(b) + 1) (cos(c) + 1) With this you can readily solve for the edge or inradius of regular and nearly regular n-sided solids by setting Omega = 4 pi/n. Sucker bet: Which is greater, the edge of a regular icosahedron or its circumradius? Answer: e/r = sqrt(2-2/sqrt(5)) . I had to stare at a picture to believe this. --rwg
At 01:41 PM 12/6/02 -0800, R. William Gosper wrote:
Sucker bet: Which is greater, the edge of a regular icosahedron or its circumradius? Answer: e/r = sqrt(2-2/sqrt(5)) . I had to stare at a picture to believe this. --rwg
The near equality of these two numbers has tripped up many an amateur geometer (and some professionals too). If the two numbers were equal, five non-interpenetrating regular tetrahedra could just be arranged around a common edge. In fact, the five barely fail to go all the way around the edge. So if you try to make an icosahedron with edges equal to its outradius, the edges will be too tight: they have to be a little longer than the outradius in order for everything to meet. Bonnie M. Stewart recounts how he was tricked by the same false identity at the beginning of _Adventures Among the Toroids_. If the identity were true, space could be tiled with tetrahedra. But if you try, you find yourself having to fold the tetrahedra toward each other in 4-space in order to force the last two faces to touch. The result is a regular 4-dimensional polytope. Because the identity is so close to true, very little folding is necessary, so the "curvature" of the resulting polyhedron is gentle, and it takes an astonishing 600 tetrahedra to complete the figure. -A
A long time ago JHC gave (on math-fun) a nice way to see that the central angle of the icosahedron is atan(2) but I can't find or reproduce it :-( Does anyone still have that by any chance? Incidentally, space can (sort of) be tiled with tetrahedra, if they're truncated in a certain strange way -- you truncate in the "normal way" yielding a figure with 4 hexagonal faces and 4 triangular faces, and then place a tetrahedron on each triangular face, whose base matches the triangular face, and whose apex is the center of the regular tetrahedron removed from that vertex of the original tetrahedron... At 04:57 PM 12/7/2002 -0500, you wrote:
At 01:41 PM 12/6/02 -0800, R. William Gosper wrote:
Sucker bet: Which is greater, the edge of a regular icosahedron or its circumradius? Answer: e/r = sqrt(2-2/sqrt(5)) . I had to stare at a picture to believe this. --rwg
The near equality of these two numbers has tripped up many an amateur geometer (and some professionals too). If the two numbers were equal, five non-interpenetrating regular tetrahedra could just be arranged around a common edge. In fact, the five barely fail to go all the way around the edge. So if you try to make an icosahedron with edges equal to its outradius, the edges will be too tight: they have to be a little longer than the outradius in order for everything to meet.
Bonnie M. Stewart recounts how he was tricked by the same false identity at the beginning of _Adventures Among the Toroids_.
If the identity were true, space could be tiled with tetrahedra. But if you try, you find yourself having to fold the tetrahedra toward each other in 4-space in order to force the last two faces to touch. The result is a regular 4-dimensional polytope. Because the identity is so close to true, very little folding is necessary, so the "curvature" of the resulting polyhedron is gentle, and it takes an astonishing 600 tetrahedra to complete the figure.
-A
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participants (3)
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Allan C. Wechsler -
R. William Gosper -
Shel Kaphan