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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