MYSTERIES OF THE EQUILATERAL TRIANGLE - HIKARI Ltd

40 Mathematical Properties
– Combining by pairs in all possible, three, ways and connecting midpoints
as above, we obtain 9 equilateral triangles.
– Also, the centroids of the three triangles A1B1C1, A2B2C2, A3B3C3
themselves form an equilateral triangle. Moreover, the same is true
if we fix the Ai and successively permute the Bi and Ci, for a total
of 9 ways in which the Ai, Bi, Ci can be arranged so that the
centroids of the three resulting triangles form an equilateral triangle.
Incidentally, the centroid of each of the nine equilateral triangles
thus formed is the same point!
• If the above three positively equilateral triangles be now so placed in the
plane that A1B1C1 forms a positively equilateral triangle, then there exist
fifty-five equilateral triangles associated with this configuration! This we
establish as follows.
– First of all, the centroids of the three triangles B1C1A1, B2C2B1,
B3C3C1 form a positively equilateral triangle. To obtain the 9 equilateral
triangles that can be thus formed, keep the last letters, A1,
B1, C1, fixed and cyclically permute the remaining Bi and Ci. (E.g.,
the centroid of B1C2B1 is the point on the segment B1C2 one third
of the distance B1C2 from B1.)
– Secondly, the midpoints of B3C2, A2C3, A3B2 form an additional
equilateral triangle.
– Finally, the existence of fifty-five equilateral triangles connected
with the figure of three equilateral triangles, so placed in a plane
that one vertex of each also forms an equilateral triangle, is now
apparent. Taking the four given equilateral triangles in (six) pairs,
each pair produces 3 additional equilateral triangles, or 18 in all. If
we take the equilateral triangles in (four) triples, each triple yields
9 additional equilateral triangles, or 36 in all. Finally, there is the
additional equilateral triangle just noted, thus making a total of 55
equilateral triangles which be easily constructed.
• Given two positively equilateral triangles, A1B1C1 and A2B2C2, of any
size or position in the plane, if we construct the positively/negatively
equilateral triangles A1A2A3, B1B2B3 and C1C2C3 then A3B3C3 is itself
a positively equilateral triangle.
Property 19 (Distances from Vertices). The symmetric equation
3(a 4 + b 4 + c 4 + d 4 ) = (a 2 + b 2 + c 2 + d 2 ) 2
relates the side of an equilateral triangle to the distances of a point from its
three corners [129, p. 65]. Any three variables can be taken for the three
distances and solving for the fourth then gives the triangle’s side.