The Fibonacci Sequence in Successive Partitions of a Golden Triangle

A Golden Triangle is a triangle with two of its sides in the ratio (J>:1, where is the Fibonacci Ratio, i.e., (J> = ^(1 + /J) ^ 1.618. Let AABC be a triangle whose sides are a,b, and o and let a/b = k > 1. Bicknell and Hoggatt [1] have shown that (1) a triangle with a side equal to b can be removed from AABC to leave a triangle similar to AABC if and only if k = cj), and (2) a triangle similar to AABC can be removed from AABC to leave a triangle such that the ratios of the areas of AABC and the triangle remaining is k if and only if & = <(>. Unlike the Golden Rectangle whose adjacent sides are in the ratio cj>:l (or l:cj)), the Golden Triangle does not have a single shape. The diagonal of a Golden Rectangle divides it into two Golden Triangles whose sides are in the ratio 11cj):/cj) + 1. The most celebrated Golden Triangle, which can be found in the regular pentagon and regular decagon, has angles of 36° , 72° , and 72° and sides in the ratio 1K p : ^ . In general, Bicknell and Hoggatt demonstrated that a Golden Triangle can be constructed with sides in the ratio l:cj):c7, where (J)" < G < cj). Figure 1, adapted from their presentation, shows Golden Triangle CGH. Line GH is constructed to be of length r$ (r > 0) and line CG to be of length rcj). Line CG is twice divided in the Golden Section by points E and D, with CE = DG v and ED = r/§. A Golden Triangle is formed whenever H is a point on the circle whose center is G and whose radius is EG. Line DH produces ADGH ~~ ACGH, and ACDH whose area is 1/cf) times the area of ACGH. In general, ACDH is not similar to ACGH. Nonetheless, ACDH is also a Golden Triangle, as CH/DH = [ 1 ] . The present paper will explore the consequences of successively partitioning Golden Triangles. To begin, let us show that ACDH can be partitioned into two triangles, one similar to itself and the other having an area 1/cf) times its own area. If line DJ is drawn parallel to line GH, one can readily verify that ADHJ is similar to ACDH. (Alternatively, we could have chosen point J so that CH/CJ = cj). Lines DJ and GH would then be parallel, because CH/CJ = CG/CD.) We now need to show that the ratio of the area of ACDH to the area of ACDJ is cj). If we designate the area of ACGH by S, the area of ACDH is S/$ [1]. Since DJ is parallel to GH, ACDJ ~ ACGH. The ratio CG/CD = <|>, hence the area of ACDJ is S/$. Accordingly, the ratio of ACDH to ACDJ is S/<$> divided by S/$, or cj). Since 5/cJ) 5/cj) = S/$ , we find that the area of ADHJ is S/cJ). We can note several other relationships. Two additional Golden Triangles, ACDJ and ADHJ, are produced so that ACGH is partitioned into three mutually exclusive Golden Triangles. Moreover, ACDJ is congruent to ADGH. They are similar, as both are similar to ACGH and both have areas equal to 5/cJ). Moving beyond the Bicknell-Hoggatt demonstration and its immediate implications, we can show how successive partitions of Golden Triangles generate Fibonacci sequences. Let us repeat the above partitioning, subdividing all of the larger triangles produced in the previous partition. The partitions can be carried out in a manner analogous to the way in which ACDH was partitioned.