Jesús de la Peña Hernández
an irrational number by means of the addition of rational summands which looks rather
contradictory, but in line with our experiment.
To finish, and in accordance with the Euclidean inequality 3 < π < 4, we shall now con-
sider the circumference as the limit of n sides polygons ́ perimeters: first, as the lower limit of a
circumscribed polygon, and second, as the upper limit of an inscribed one when n tends to in-
finity.
In Fig. 2 we find that AB is the side of an inscribed polygon of n sides and CD is that of
a circumscribed one; in both cases the radius of the circumference is r = OA = OB and the cen-
tral angle is 2α, being
2 n
360
α=. Therefore we ́ll have:
AB= 2 rsenα ; CD= 2 rtgα
As 2πr is the length of the circumference, it will be:
n
r rn
n
rn
2
360
2 2 tg
2
360
2 sen < π < ;
n
n
n
n
2
360
tg
2
360
sen <π<
Let ́s see how this inequation looks for different polygons.
n = 10 3,0901699 < π < 3,2491970
n = 50 3,1395260 < π < 3,1457334
n = 100 3,1410759 < π < 3,1426266
n = 1.000 3,1415875 < π < 3,1416030
n = 10.000 3,1415926 < π < 3,1415928
n = 50.000 3,1415927 < π < 3,1415927
(n = 50.000 produces overflow in a calculator screen)
Having all previous considerations in sight, it does not look so discouraging the result
that may be obtained for π by means of a rolled paper tube.
2
A
B
C
D
O