Jesús de la Peña Hernández

6- Ang. CLM = Ang. CHG because of the symmetry with former step.
Likewise, in pentagon CLMGH, Angs. in M and G are congruent, also by symmetry.
7- A pentagon has been produced whose angles in L, C and H are congruent and have a meas-
ure of 108º; the other two angles are congruent to each other. Then, the 5 angles measure
108º: we have come across a convex regular pentagon that will have congruent its five sides.
8- Unfolding 7 we get 8 which shows the resultant pentagon CLMGH (and its symmetric).

AN INTERESTING VERIFICATION

The rhomb DHLG was formed in Fig. 5: it is made up by the two isosceles triangles

with base HG. We can see in it that GD is the pentagon diagonal and GH is one side. Recalling

the x value obtained at step 4, we ́ll have:

B

HD

x

=sen

B

x

HD

sen

=

2

(^2) sen
B
HD
GH

B
B x
GH
2 sen
= 2 sen
Substituting values from steps 1 and 4, we have:
HD = 0,6498393 ; GH = 0,4016228
These values satisfy the expression
GH^2 =()HD−GH HD
This way it is verified that in a convex regular pentagon the diagonal is divided by the side in
mean and extreme ratio.
10.1.2 FROM A DIN A4 RECTANGLE
Once developed the former Point, we can see now the inexactitude of present solution. For
that purpose it ́s enough to look at figs. 1 and 2 of Point 10.1.1.
In case rectangle 1 was a DIN A4, we ́ll have:
1
2
g=Arctg = 54,73561º
and in fig. 2:
Ang. DCB = 2g = 109,47122 ≠ 108º
10.1.3 FROM A PAPER STRIP MADE OUT OF ARGENTIC RECTANGLES
Produce six adjacent argentic rectangles after fig.1. Zigzag fold its diagonals, eliminate the
small sides, mountain fold the diagonals and finally glue the end triangle equivalent to fig. 3 in
Point 10.1.1.
We obtain the “perforated” pentagon of fig.2 whose side is congruent with the diagonal of
the argentic rectangle we began with.
1 2