@nm Feel free to use my example in your bug report.  Also note vv's remark on the problem with the And() version.

I can't tell where the problem originates, but here is a stripped down variant that manifests the same issue.

eq := f(x)*A = piecewise(0 < a and b < 1, 10);
solve(eq, A);

@Ronan Here is an example of the use of anonymous functions.  We wish to evaluate expr for a couple of different choices of the function u :

> restart;

> expr := u(x) + u(x-1) + D(u)(x+1);
                     expr := u(x) + u(x - 1) + D(u)(x + 1)

> eval(expr, { u = (x->x^2) });
                             2          2
                            x  + (x - 1)  + 2 x + 2

> eval(expr, { u = (x->sin(x)) });
                       sin(x) + sin(x - 1) + cos(x + 1)

@Kitonum and @vv, you are correct in observing that dividing the data by 10^2 is not justified.  Things work out by a lucky accident in that case, but if the equations are replaced with

eq1 := x^2 + y^2 + z^2 = 81*3^2;
eq2 := 2*x*y + 2*x*z + 2*y*z = 88*3^2;

then upon modifying the worksheet that I posted earlier, we obtain the solution {x = 23, y = 14, z = 2} (among other solutions).  We see that those numbers are not divisible by 3.

@Ronan In your worksheet the value of a is obtained by solving eq2 for a.  The solution turns out to be an integer as desired, although that's not built into the algorithm.  Could that be a lucky accident?

@Kitonum That's a clever solution.  Just for fun, here is a Maple implementation of your work.

Download mw.mw

@Alfred_F Okay, that's a good proof.  In fact, the statement of the problem may be generalized a bit.

Let S be any set of points in R³.  Suppose that for any plane P, the points in the intersection S ∩ P lie on a circle.  Then the points in S lie on a sphere.

The attached worksheet illustrates your proof. 

Download mw.mw

@Earl Let r be the bead's position vector relative to a reference point, and let v be velocity vector.  Then the bead's angular momentum is L = m r x v, where m is the mass and "x" is the cross product.

Hello Earl, I looked through your worksheet but couldn't make heads or tails of it.  It seems to me that you wish to analyze the motion of a bead along a wire track.  That's pretty much the same problem that was discussed in an earlier question. How is this one any different?

You may want to state the problem clearly, in words, without an appeal to a worksheet.

@sand15 Also note that his equations involve both lambda and lambda[1].  Then he sets lambda=1 and lambda[1]=0.1. 

@acer Ah, that explains why implicitplot does not complain in nm's answer.

@nm Needn't we worry about the imaginary 7/5*Pi*I term?

@Earl You have asked an interesting question and I am glad to have been able to provide a solution that you have found satisfactory.  Cheers!

@C_R That's a pretty curve and works nicely in the context of this problem.  And you are correct in pointing out that the the curve's parametrization need not be arclength parametrization.

This is extracted from your worksheet:

Earl, in your worksheet you write:

    Both bobs slide down according to the downward tangential
    force of gravity and the opposing tangential force of friction.
    The latter is the product of the component of the gravitational
    force normal to the arc and the coefficient of friction.

That does not look correct.  The normal component of the reaction force should be the sum of two forces:

1. The component of the gravitational  force normal to the arc, as you have correctly stated;
2. The centripetal force due to the bob's curved path.

To see the significance of the #2 above, imagine turning off the force of gravity and letting the bob move solely due to imparted initial velocity.  According to your formulation, the force of friction will be zero.  But as the bob slides along the curve, the centripetal force will press it against the curve, making it feel the force of friction. That's where #2 comes in.