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.

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...

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

ReplyDeleteTake 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.

bleaug