Oh, sorry about the lack of explanation. Areals (also known as barycentric coordinates) are a way of assigning a triple of values to any point in the (projective) plane, based on some arbitrary reference triangle. The vertices of the triangle (A, B and C) are assigned the areals (1,0,0), (0,1,0) and (0,0,1). All scalar multiples are considered equivalent, so (x,y,z) = (kx,ky,kz). The centroid of the triangle (G) is (1,1,1). The line at infinity is defined by x + y + z = 0. When the projective plane is extended so that (x,y,z) are complex instead of real, it can be shown that all circles pass through two points on the line at infinity, known as 'circular points'. Indeed, a circle can be defined as any conic passing through those two points; note that the degrees of freedom are reduced from five to three. Sincerely, Adam P. Goucher ----- Original Message ----- From: "Dan Asimov" <dasimov@earthlink.net> To: "math-fun" <math-fun@mailman.xmission.com> Sent: Thursday, April 05, 2012 1:34 AM Subject: Re: [math-fun] Areals of circular points
Adam wrote:
<< The circular points at infinity have the following (unnormalised) areals:
x = a^2 (2a^2 - b^2 - c^2) - (c^2 - b^2)(c^2 - b^2 + delta) y = b^2 (2b^2 - c^2 - a^2) - (a^2 - c^2)(a^2 - c^2 + delta) z = c^2 (2c^2 - a^2 - b^2) - (b^2 - a^2)(b^2 - a^2 + delta)
where delta = ± sqrt(a^4 + b^4 + c^4 - a^2 b^2 - b^2 c^2 - c^2 a^2)
= ± sqrt((-a-b-c)(-a+b+c)(a-b+c)(a+b-c))
= ± 4i [ABC], where [ABC] is the area of the reference triangle.
The sign choice for delta determines which of the two circular points is being specified.
Sorry if I missed something, but what is a circular point at infinity? And what is an areal?
Thanks,
Dan
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