Pythagoras and the Pythagoreans
24
α
A
B
C
D
E
P
Q
Pentagon
α
β
180 − β + 2α = 180
β = 72
A
P
Q
C
Divide a line AC into the ‘section’ with respect to both endpoints.
So PC:AC=AP:PC; also AQ:AC=QC:AQ. Draw an arc with center
A
and radius
AQ
. Also, draw an arc with center
C
with radius
P C
.
Define
B
to be the intersection of these arcs. This makes the triangles
AQB
and
CBP
congruent. The triangles
BP Q
and
AQB
are similar,
and therefore
P Q
:
QB
=
QP
:
AB
. Thus the angle
6
P BQ
=
6
QABAB
=
AQ
.
Define
α
:=
6
PAB and
β
:=
6
QPB. Then
180
o
−
β
−
2
α
= 180
o
.
This implies
α
=
1
2
β
, and hence
(2 +
1
2
)
β
= 180
. Solving for
β
we,
get
β
= 72
o
. Since
4
PBQ is isoceles, the angle
6
QBP
= 32
o
. Now
complete the line BE=AC and the line BD=AC and connect edges AE,
ED and DC. Apply similarity of triangles to show that all edges have
the same length. This completes the proof.
6.3
Regular Polygons
The only regular polygons known to the Greeks were the equilaterial
triangle and the pentagon. It was not until about 1800 that C. F. Guass
added to the list of constructable regular polyons by showing that there
are three more, of 17, 257, and 65,537 sides respectively. Precisely, he
showed that the constructable regular polygons must have
2
m
p
1
p
2
. . . p
r