Hans Walser, [20130411a]
Motivation: M. N. D., N.
Golden Angle
First
we inscribe a pentagon into a square (Fig. 1a) and insert the blue angle .
Then
we draw a diagonal in the Golden Rectangle ([Walser 2001, p. 34], [Walser 2013,
p. 53]) and mark the blue angle (Fig. 1b).
Fig. 1: Two angles
The
conjecture is .
For the pentagon in the square we use the notation of Figure 2. The origin of the coordinate system is the centre of the Pentagon, which is not the centre of the square.
Fig. 2: Coordinate system
We
have ,
, and
.
Therefore:
For the Golden Section we use the usual notation:
(1)
It is easy to prove:
(2)
We can
reduce every square of to a linear
expression in
.
Using ([Walser 2001, p. 54], [Walser 2013, p. 77])
(3)
we get:
(4)
Figure 3 depicts the situation in the Golden Rectangle.
Fig. 3: The Golden Rectangle
We get:
(5)
Because of
(6)
and (2) and (5) we have:
(7)
We compare (4) and (7):
(8)
This is equivalent to:
(9)
On the right hand side we have, using (2):
This
is equal to the left hand side of (9). Hence .
Numeric
result:
References
[Walser 2001] Walser, Hans: The Golden Section. Translated by Peter Hilton and Jean Pedersen. The Mathematical Association of America 2001. ISBN 0-88385-534-8
[Walser 2013] Walser, Hans: Der Goldene Schnitt. 6., bearbeitete und erweiterte Auflage. Edition am Gutenbergplatz, Leipzig 2013. ISBN 978-3-937219-85-1