112 
Transactions Texas Academy of Science. 
As a first experiment make it the length of the given sect. Various 
expiriments may now be made. J oining the two unconnected end points 
a right-angled triangle is formed. Perhaps half of this hypothenuse is 
x. But, alas, he observes that x, being half of this hypothenuse, satisfies 
the equation 4x 2 =2(x-f-y) 2 , or x 2 =y(2x+y). Hence this x is not the 
one desired and something else must be tried. 
Several other promising experiments may be tried here without suc- 
cess. Our Greek, then, let us say, after a thorough trial of a perpen- 
dicular at an end point equal to the original sect, either abandons (tem- 
porarily) the perpendicular at the end point or changes its length. Let 
us suppose that he first tries a different length. The new length should 
be either half or double the length of the original sect. Suppose he 
spends a week or so on the double length without success, finally trying 
the half length. Again, would he join the two free points as a further 
step. Again, would he try bisecting the new hypothenuse for x and 
again would it fail to be the x desired, as he could easily prove. Per- 
haps two-thirds of the hypothenuse, h, is x ? He can easily prove that it 
is not. Perhaps one-third of the hypothenuse, h, is y? He can easily 
prove that it is not. Perhaps (h — 1) is y? Perhaps one-third (1+h) is 
x? Perhaps h — 11 is x? A variety of simple hypotheses may be made 
and tested by our old geometer, all involving the sides of the triangle or 
simple fractions thereof with their sums and differences. Each hypoth- 
esis, as soon as made, must be tested. If it fails to stand the test, some 
further one must be made; if it does not fail, the problem is solved. 
How our Greek in a few years could test several hundred such hypoth- 
eses. There are, perhaps, two hundred promising ones in this problem, 
certainly not over a thousand. Granting that only one of these is suc- 
cessful, our Greek ought to get his desired division point in a few years, 
or even in less time, if he goes about it in this systematic way. As a mat- 
ter of fact, x is equal to h — 11 as may be easily proved. For 
4h 2 =5l 2 , x = Jl(|/5 — 1), 
y = l— x=41(3 — |/' 5), 
x 2 = jH(3_ 1 /5)=ly=y(x+y). 
Hence a rather simple hypothesis leads to the correct division point. 
It is hoped that this somewhat lengthy discussion of this one problem, 
which i^ quite a typical one, will point toward several conclusions. 
Geometry is in part, or can be treated as, experimental, and the experi- 
ments are not made altogether at random, and, if wisely made in any 
given case, are not unduly numerous nevertheless, it is hardly fair to 
expect an immature student often to succeed in solving originals of any 
degree of difficulty, especially if asked to pay too explicit attention to the 
logic of his operations. But an immature student can be trailed to try 
