From: Steve Gray Subject: Generalizing the three equilateral triangles theorem Date: Sun, 21 Jan 2001 07:11:17 GMT Newsgroups: sci.math On Mon, 15 Jan 2001 18:23:17 +0100, "Panh" wrote: >> "On a given circle (O,R), let A1A2, A3A4, A5A6 be three arcs such >> that: arc(A1A2) = arc(A3A4) = arc(A5A6) = pi/3. If A,B,C are the >> midpoints of the non-equal chords A2A3, A4A5, A6A1 then the triangle >> ABC is equilateral." >www.cut-the-knot.com > GRAY: This theorem is a special case of something much more general, described below. To see that it's related, consider the above theorem stated in terms of three equilateral triangles having a common vertex. The generalizations are: first, the three triangles do not have to be equal in size. Second, they do not have to have a common vertex, but can each have a vertex at a vertex of another equilateral triangle. Third, the triangles do not have to be equilateral, but all must be similar. The final triangle will be similar to all the others. I have a simple purely synthetic proof of this. The actual theorem statement follows. I developed the generalization in connection with some e-mails with Will Self. Definition - Given a polygon P with vertices Vi, to SPAWN P with respect to a particular vertex Vi means to create a duplicate polygon Pi initially congruent to and coincident with P, then rotate Pi around vertex Vi by some angle Gi, then scale Pi with scale factor Fi. (The original P remains unchanged, and obviously Pi will be similar to P and have the same "handedness".) Theorem: Given a triangle (A,B,C), spawn it around A by an arbitrary angle and scale, giving new triangle (A,B1,C1). Spawn (A,B,C) again by a different arbitrary angle and scale around B, giving new triangle (A2,B,C2). Spawn (A,B,C) again by a different arbitrary angle and scale around C, giving (A3,B3,C). Draw segments (C1,A2), (C2,A3), and (B3,A1). Then the triangle formed by the midpoints of these segments is similar to ABC. -- Respect reality. It's all you've got.