Jesús de la Peña Hernández
Let ́s pass to demonstrations. For x =
2(^1) we have:
8
4
x=
8
5
y=
8
3
z=
6
4
a=
6
5
b=
6
3
c=
24
4
d=
24
5
e=
24
3
f =
Those values prove that respective sides keep the ratios 3, 4, 5.
It should be noted that the hypotenuses (the greater sides) are in proportion to 5. On the
other hand it has been shown that for x =
2
(^1) it is a =
3
2
In connection with Point 3.2, it can be checked that for the two values assigned to x, we
have:
3
1
x=
9
5
y=
2
1
a=
4
1
x=
32
17
y=
5
2
a=
As the perimeter of the square amounts to 4, the second part of Point 3.1 will be demon-
strated if, once a b c expressed in function of x, the following hypothesis is verified:
2 = a + b + c , i.e., if:
2 =
()
()^2
2 2
0. 251
1
x x
x x
- −
−
()()
()^2
2 2
2
- 251
- 51 1
x x
x x
- −
- −
- 1 – x
Developing denominators, we have:
- 25 + 0. 25 x^4 + 0. 5 x^2 −x^2 = 0. 25 () 1 +x^4 − 2 x^2 = 0. 5 () 1 −x^2
Therefore:
()
x
x
x
x
x
- −
= 1
1
1
- 51
2
2
0.5 (1 + x) = x + 0.5 (1 +x^2 ) – x 0.5 (1 + x) ,, 0 = 0 which proves the theorem
In a similar way can be proved the second part of Point 3.1.