Tuesday, June 1, 2010

Fox 291

Not much time to update the blog. But, let's throw this to the public knowledge.
Note: when 3 concentric arcs are given, their center can be identified easily.
(which is a nice exercise and easy by itself).
Let us know if you have a solution.

Aaaah, one more detail. More than one equilaterals should be drawn on 3 concentric circles. Take the figure as it looks and do not go for the "other" equilateral. Let the simplicity ring over the land, in the morning... and during the night...

1 comment:

  1. Let r3 > r2 > r1 > 0 be the radii of the 3 given circles (C3, C2, C1 respectively).

    Take an arbitrary point P on any circle, e.g. the outer circle C3, and draw a circle centered in P such that it intersects C3 in Q and R. The line bisectors of PQ and PR intersect in C which is the common center of the 3 circles.

    Take an arbitrary point O on C2 and build an equilateral triangle OCC'. Draw a circle C1' centered in C' of radius r1. This circle is the image of C1 by a rotation r of angle 60° (or -60°) centered in O, i.e. for any M on C1, there is a N on C1' such that vector(ON) is the rotation of vector(OM) by 60° (or -60°). Hence OMN form an equilateral triangle.

    When r3-r2 <= r1, C1' intersects with C3 in two points B1 and B2. Build A1 and A2 such that r(A1)=B1 and r(A2)=B2. OA1B1 and OA2B2 are equilateral and by construction each vertex lie on a distinct circle.

    When r3-r2 > r1 there is no solution.