Re: A challenging 3 equations and 3 unknowns

In article <b8cbcb94-b30b-4041-950c-212b14f0b115@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
pnachtwey <pnachtwey@xxxxxxxxx> wrote:

eq1: (x1-x0)^2+(y1-y0)^2-(r0+r1)^2;
eq2: (x2-x0)^2+(y2-y0)^2-(r0+r2)^2;
eq3: (x3-x0)^2+(y3-y0)^2-(r0+r3)^2;
eq4: solve([eq1,eq2,eq3],[x0,y0,r0]);

If, as the original post indicates, the three points are on an
equilateral triangle, that could probably simplify the formulas.

I am not sure how. The equations seem pretty simple as they are.

Oh, it helps a lot because the equations simplify. If you have the
luxury of drawing the coordinate system after the diagram is presented
to you, then you can do it with the origin at point #1, the first
axis passing through point #2 and with a unit of measure equal to
half the distance between points 1 and 2. Then the equations are
simpler because x1=y1=y2=0, x2=2, x3=1, y3^2=3 . In this case a
computer-algebra system works much faster to determine the relationship
between the four radii. Moreover, the symmetry of the diagram
assures us that the polynomial relating r0 to the other three radii
will be symmetric in r1, r2, r3, and hence expressible in terms of

S = r1 + r2 + r3
T = r1*r2 + r2*r3 + r3*r1
U = r1*r2*r3

I get the defining relation to be simply 0 =

(S^4-4*S^2-4*S^2*T+6*U*S+T^2+8*T+16) +(2*S^3-7*S*T-4*S+9*U)*X +(S^2-3*T-3)*X^2

where X = 2*r0 . (Of course the actual presentation of the quadratic
relation for r0 is a matter of taste -- there are other sets of
generators for the symmetric polynomials besides {S,T,U} .)

So I think your problem is very easy to solve now:
1. Choose units so the equilateral triangle has sides of length 2
2. Compute S,T,U as above
3. Solve for r0 = X/2 as above.

As has been noted earlier in this thread, once r0 is known, you can solve
for x0, y0 by solving the two linear equations eq2-eq1 = eq3-eq1 = 0.
In the coordinate system I described above, the solution is

x0 = ( (4+r1^2-r2^2) + (r1-r2) *X )/ 4
y0 = ( (4+r1^2+r2^2-2*r3^2) + (r1+r2-2*r3)*X )/(4*sqrt(3))

but you can also easily solve the linear system in a
previously-defined coordinate system, too.

dave

.

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