Re: geometry question
- From: OwlHoot <ravensdean@xxxxxxxxxxxxxx>
- Date: Sat, 7 Mar 2009 09:09:34 -0800 (PST)
On Mar 7, 1:29 pm, OwlHoot <ravensd...@xxxxxxxxxxxxxx> wrote:
[..]
Considering the various ways (p^2 + 1)(q^2 + 1)(r^2 + 1)
can be composed as a sum of two squares, this has a
general parametric solution (up to a scale factor,
which can be considered absorbed in the k_i) :
k1.(x - a1) = p.q.r - p + q + r
k1.(y - a2) = 1 - q.r + r.p + p.q
k2.(x - b1) = p.q.r + p - q + r
k2.(y - b2) = 1 + q.r - r.p + p.q
k3.(x - c1) = p.q.r + p + q - r
k3.(y - c2) = 1 + q.r + r.p - p.q
Darn - The method needs fixing at this point
If there are three numeric values e_i with:
k1.e1 + k2.e2 + k3.e3 = 0 [1]
e1 + e2 + e3 = 0 [2]
then the above system is consistent if and only if:
f1.p + f2.q + f3.r = f4 [3]
f1.q.r + f2.r.p + f3.p.q = f5 [4]
where the f_i are numeric:
One can always solve [1] and [2] to obtain e1:e2:e3
(As can be seen [1] eliminates x, y from the linear
combinations and [2] eliminates p.q.r and 1.)
However, f4 and f5 need not be zero, which makes
the formulae uglier.
No compellingly simple method to solve [3] and [4]
comes to mind, but I'll think about it.
Cheers
John Ramsden
.
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