Golden Section in Two Equilateral Triangles, II
Quang Tuan Bui
November 7, 2011
ABC and AMN are two equilateral triangles where M is midpoint of side BC. Arc 60° centered at B passing through A, C intersects side MN by golden ratio.
Construct two circles: circle A(P) centered at A passing through P and circle C(A) centered at C passing through A. These two circles intersect at two points. Let F denote the one that is in the same side of AC as N.
Suppose D is symmetric to A in M. Two triangles ABP and ACF are congruent because
The rotation R(A, 60°) also maps ΔAMP onto ΔANF; therefore,
|(1)||MP = NF|
and ∠FNA = ∠PMA = ∠MAN = 60°, implying
∠APD = 120° and ∠APF = 60°, so that
|(3)||D, P, F are collinear.|
From (2) and (3) two triangles PMD, PNF are similar. Therefore:
NF/PN = MD/PM
MP/PN = NF/PN by (1)
Now we can calculate:
|x||= MP/PN = MD/PM = AM/PM = NM/PM|
|= (NP + PM)/PM = NP/PM + 1 = PN/MP + 1|
or x = 1/x + 1. From this x = φ, the Golden ratio.
(There is another proof for the same construction.)