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Fig. 5.9 The “golden triangle”
Let T be a “golden” isosceles triangle with vertex angle equal to one-tenth of the
full circle angle, that is
5.
The internal triangle of Fig. 5.9 is isosceles too and similar to T because it has
the same angles as T :
α = π /
5 (equal angles at the basis). In the golden
triangle T , if 1 is the length of the basis and
π /
5, 2
π /
5, and 2
π /
φ
is the length of the oblique sides, then
it is easy to find the following proportion:
φ
:1
=
1:
( φ
1
) .
Therefore
φ
is solution of the equation:
2
φ
= φ +
1
.
5
This equation has only one positive solution given by
φ =(
1
+
) /
2, that is, the
golden ratio, approximately equal to 1
.
618. Moreover, from the equation above, it
follows that:
1
φ
therefore by applying iteratively its second member to replace
φ =
1
+
φ
,wegeta continu-
ous fraction approximating to
φ
:
1
φ =
1
+
1
1
+
1
1
+
1
1 +
1
+ ...
It can be showed, by induction, that:
n
n
)= φ
ϕ
F
(
n
φ ϕ
 
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