Let A and B the midpoints of the sides EF and ED of an equilateral triangle
DEF. Extend AB to meet the circumcircle of DEF at C. Show that B divides AC
according to the golden section.
-- George Odom, AMM 90(1983) 482; Solution: 93(1986) 572
cf. : The dramatically simple construction of the extreme and mean ratio
(...), was published, perhaps for the first time, in 1984; see Odom, E.P.
David Fowler: The Mathematics of Plato;s Academy. A New Reconstruction.
Second Edition. Clarendon Press, Oxford, 1999, p. 102.
The appearance of the Golden Ratio (phi) in the equilateral triangle is
somehow unexpected (the sqrt(5) is the "regular" number of r. 5*n-gons,
while the sqrt(3) of the 3*n-gons).
I was wondering if there are known more such phi appearances in the
equilateral triangles.
Possible geometrical objects to search is the regular polyhedra whose the
faces are equilateral triangles.
cf.:
Choose on all edges of a regular octahedron one of the points dividing the
edge in ratio 1/phi in such a way that passing around the four edges starting
from one vertex, the nearer and the farther dividing points are chosen
alternately. Show that:
(a) It is possible to choose the dividing points according to the above
principle.
(b) The 12 dividing points chosen in this way determine a regular icosahedron
(Kozepiskolai Matematikai Lapok, 60(1980) 120)
Note:
This problem was recently solved in sci.math NG by Donald T. Davis:
http://x15.dejanews.com/getdoc.xp?AN=461006891
G. Candella (1502 - 1594) in his added-to-Euclid's-Elements book, Proposition
4, shows how to get the triangle of the icosahedron by dividing in golden
ratio the vertices of a r. tetrahedron.
See: Roger Herz-Fischler: A Mathematical History of the Golden Number.
Dover, 1998, p. 156.
In the plane geometry, the only other one appearance I located is in
the following problem:
Lat ABC be an equilateral triangle and O the midpoint of BC. Draw a line
passing through the O, cutting AB at M, and AC at N, so that:
area(OMB) + area(ONC) = area(ABC)
Supplement of the Bull. of the Greek Math. Society, June 1935, #908
[Answer: M divides AB in golden ratio]
Question: Does anyone know more such "dramatically simple" (to quote David F.)
constructions of the phi in the equilateral triangles?
Antreas