The above applet graphs two segments with a common vertex, and then shows the locus of points such that the segments subtend congruent angles ζ (defined by the blue segments) and ψ (defined by the green segments). It turns out that, in general, the curve is shaped like a Greek letter φ. Strangely, on the red curve between the blue and green points the property changes: the segments now subtend supplementary angles. However, considered as a curve in the projective plane, the properties do not change.

The parametric equations are given, and it is relatively simple (but messy) to find an x-y equation for the curve. In general, it is a cubic.

Henri Picciotto conjectures that the curve also gives the solution to Hero's problem on a circular island. Imagine you are on a circular island centered at the shared vertex. You are sitting with a bucket at, say, the blue vertex and your house, on fire, is at the green vertex. (If the two points are outside of the circle, you may assume that the circle represents a lake; the question remains the same.) You want to run to the shore of the island (to get water) and then to your house as fast as possible. Henri conjectures that you should run to one of the points of intersection of the shorecircle with the φ-curve. True?

It appears to be true so long as the line segment joining the blue and green points does not intersect the circle. In fact, for some choices of (x0,y0), (x1,y1), (x2,y2), and r (radius of circle) there are 4 points of intersection of the φ-curve with the circle, and they do seem to coincide with local maxima and minima of the summed distance function. I offer \$20 for the first proof or counterexample. (I don't know if it is true or not.)

By the way, parametric equations for the red curve are:

• x(t)=(x0 (x1-x2)^2+(x2 (-y0+y1)+x1 (y0-y2)) (y1-y2)-8 t^3 (2 x0-x1-x2) (x2 (y0-y1)+x0 (y1-y2)+x1 (-y0+y2))+4 t^2 (2 x1^2 x2+2 x0^2 (x1+x2)-x0 (x1^2+6 x1 x2+x2^2-2 (y1-y2)^2)+x2 (y0-y1) (2 y0+y1-3 y2)+x1 (2 x2^2+(y0-y2) (2 y0-3 y1+y2)))+2 t (x1^2 (y0-y2)+3 x1 x2 (y1-y2)+x0 (-x2 (2 y0+y1-3 y2)+x1 (2 y0-3 y1+y2))-(y0-y1) (x2^2+2 (y0-y2) (-y1+y2))))/(x1^2-2 x1 x2+x2^2+y1^2-2 y1 y2+y2^2+8 t (-x2 y0-x0 y1+x2 y1+x1 (y0-y2)+x0 y2)+4 t^2 (4 x0^2+x1^2+2 x1 x2+x2^2-4 x0 (x1+x2)+4 y0^2-4 y0 y1+y1^2-4 y0 y2+2 y1 y2+y2^2))
• y(t)=(-x1 x2 y1+x2^2 y1+y0 y1^2+x0 (x1-x2) (y1-y2)+x1^2 y2-x1 x2 y2-2 y0 y1 y2+y0 y2^2+8 t^3 (2 y0-y1-y2) (x2 (-y0+y1)+x1 (y0-y2)+x0 (-y1+y2))-2 t (2 x0^2 (x1-x2)+2 x1^2 x2+x2 (y0-y1) (y1+3 y2)+x0 (-2 x1^2+2 x2^2+(y1-y2) (2 y0+y1+y2))+x1 (-2 x2^2-(y0-y2) (3 y1+y2)))+4 t^2 (x2^2 (2 y0-y1)+2 x0^2 y1+2 y0^2 y1-y0 y1^2+x1^2 (2 y0-y2)+2 x0^2 y2+2 y0^2 y2-6 y0 y1 y2+2 y1^2 y2-y0 y2^2+2 y1 y2^2-x0 x2 (y1+3 y2)-x1 (x0 (3 y1+y2)+x2 (4 y0-3 (y1+y2)))))/(x1^2-2 x1 x2+x2^2+y1^2-2 y1 y2+y2^2+8 t (-x2 y0-x0 y1+x2 y1+x1 (y0-y2)+x0 y2)+4 t^2 (4 x0^2+x1^2+2 x1 x2+x2^2-4 x0 (x1+x2)+4 y0^2-4 y0 y1+y1^2-4 y0 y2+2 y1 y2+y2^2))
and non-parametric equations are:
• ((x1-x0)^2+(y1-y0)^2-(x-x1)^2-(y-y1)^2-(x-x0)^2-(y-y0)^2)^2 ((x-x2)^2+(y-y2)^2)-((x2-x0)^2+(y2-y0)^2-(x-x2)^2-(y-y2)^2-(x-x0)^2-(y-y0)^2)^2 ((x-x1)^2+(y-y1)^2)=0.
• -2 x0 y x^2+2 x0 y^2 y2-2 x0 y0 x^2+2 x0 y0 x x2+2 x0 y0 y^2-2 x0 y0 y y2+x^2 y x2-x y^2 y2+2 y0 x^3-2 x2 x x1 y-2 y0 x^2 x1+x1 y^3+2 y y0 y2 x1-2 x0 x y2 x1-2 x2 x1 x0 y0+2 x2 x1 x0 y+2 x2 y0 x x1+2 x0^2 x y+x0^2 y2 x1-x0^2 x1 y-x0^2 y2 x-2 y^2 y0 x1+2 x0 x^2 y2-y0^2 y2 x1+y0^2 x1 y+y0^2 y2 x-2 x0 y^3+x2 y0^2 y-2 y0 x^2 x2+2 y0 x y^2+2 x0 y0 x x1-2 x0 y0 y y1-y1 x^3-y2 x^3+x2 y^3+y2 x^2 x1-2 y0^2 y x-x2 x0^2 y-2 x2 y0 y^2+y1 x2 x0^2-y1 x2 y0^2+2 y1 x0 x^2-y1 x0^2 x+y1 y0^2 x+2 x0 y^2 y1+x^2 y x1-x y^2 y1+2 y1 x2 y y0-2 y1 x0 x x2+2 y1 x y y2-2 y1 y2 y0 x+2 y1 x0 y2 y0-2 y1 x0 y y2-x2 y^2 y1+y1 x^2 x2-x1 y^2 y2=0.
The first is obtained by using the law of cosines to find the angles ζ and ψ in terms of (x,y) and the (xi,yi), then setting ζ=ψ. The second is derived from the first by removing a linear factor.