carlson@lance.tis.llnl.gov (John Carlson) (09/01/89)
I spent last night puzzling over this one, I still don't have an answer. Given: A, B circles in R2 and an two points P & Q, one on each circle (P on A, Q on B). Construct a curve C with endpoints P & Q such that the curve is normal to A & B at P & Q. C only intersects A & B at P & Q. The function describing the curve should be differentiable at all points on the curve except P & Q. C is made up of at most 2 segments (3 parametric quadratics won't do! :-). Remember, C only intersects A & B at P & Q (Otherwise the problem is cake). John Carlson carlson@tis.llnl.gov
hallett@shoreland.uucp (Jeff Hallett x4-6328) (09/01/89)
In article <439@ncis.tis.llnl.gov> carlson@lance.tis.llnl.gov (John Carlson) writes: > >Given: > A, B circles in R2 and an two points P & Q, one on >each circle (P on A, Q on B). > >Construct a curve C with endpoints P & Q such that the curve is >normal to A & B at P & Q. C only intersects A & B at P & Q. The >function describing the curve should be differentiable at all >points on the curve except P & Q. C is made up of at most 2 >segments (3 parametric quadratics won't do! :-). > > >Remember, C only intersects A & B at P & Q (Otherwise the >problem is cake). The problem needs some more definition, I think. Of what form should C be, quadratic, transcendental, etc.? If we don't determine the curve format, there are potentially an infinite number of solutions. Maybe another bound would be to minimize the length of C in R2? Also, I'll make the assumption that the curve should be right or left differentiable at P & Q (even though it isn't fully differentiable, by its termination) Good problem. I like it. f o d d e r -- Jeffrey A. Hallett, PET Software Engineering GE Medical Systems, W641, PO Box 414 Milwaukee, WI 53201 (414) 548-5163 : EMAIL - hallett@positron.gemed.ge.com
greg@herb-ox.berkeley.edu (Greg) (09/03/89)
In article <439@ncis.tis.llnl.gov> carlson@lance.tis.llnl.gov (John Carlson) writes: >Given: A,B circles in R2 and an two points P & Q, one on >each circle (P on A, Q on B). >Construct a curve C with endpoints P & Q such that the curve is >normal to A & B at P & Q. C only intersects A & B at P & Q. The >function describing the curve should be differentiable at all >points on the curve except P & Q. Here is an analytic solution: Interpreting R^2 as the complex plane, there is a fractional linear transformation (a function f:z->(a*z+b)/(c*z+d)) which maps A and B to concentric circles; we may take A to be the unit circle and B to have radius r_B<1. We may rotate the picture so that P has an angle of alpha and Q has an angle of -alpha with the real axis. Now choose a curve in polar coordinates with the desired properties, say theta(t) = alpha*2/pi*arcsin(t), r(t) = (t+1)/2*(1-r_B)+r_B. Set z(t) = r(t)*e^(i*theta(t)); the final answer is f^-1(z(t)). Some notes on fractional linear transformations: These are circle-preserving (if you count lines as circles) transformations of the complex plane. The composition of two flt's is an flt. (Indeed, there is a rule of composition: Write the four coefficients a,b,c, and d as 2 x 2 matrices and multiply the matrices.) We can build the flt we want for this problem in stages: First, translate the circles so that A is centered at the origin. Second, scale (multiply by a complex scalar) so that A has radius 1. Third, apply the map z->1/z to put B inside A. Fourth, rotate so that B's center is on the real axis. B now intersects the real axis in two point b_1 and b_2. Fifth and finally, apply the flt (a*z+b)/(c*z+d) with coefficients satisfying a+b = c+d, a-b = c-d, and (a*b_1+b)/(c*b_1+d) = -(a*b_2+b)/(c*b_2+d), with a,b,c, and d real. This flt will not move A but will shift B so that it too is centered at the origin. This is probably not the easiest solution to carry out, but it carries a moral: You can often simplify a problem about circles using flt's. --- Greg
mitchell@cbmvax.UUCP (Fred Mitchell - QA) (09/03/89)
In article <439@ncis.tis.llnl.gov> carlson@lance.tis.llnl.gov (John Carlson) writes: >I spent last night puzzling over this one, I still don't have an answer. > >Given: > A, B circles in R2 and an two points P & Q, one on >each circle (P on A, Q on B). > >Construct a curve C with endpoints P & Q such that the curve is >normal to A & B at P & Q. C only intersects A & B at P & Q. The >function describing the curve should be differentiable at all >points on the curve except P & Q. C is made up of at most 2 >segments (3 parametric quadratics won't do! :-). You might try a cubic spline. A simpler solution eludes me at the moment. What type of curve, anyway? -- |*******************************************| -Compliments of /// |* All thoughts and comments are soley *| Fred Mitchell \\\/// |* thoses of The Author and has nothing to *| \XX/ |* do with Commodore-Amiga. *| Software QA - Commodore-Amiga |*******************************************|