We proceed to apply these theorems to sev eral examples.
Consider first the problem of an angle. The given angle' 0 and the required 0 angle may be determined by their cosines. 2 0 Let a = cos 0 and x = cos From trigonomer 2 try cos 0 = 2 cos ' 1, that is, 2x' 1 a.
2 Hence + a Therefore, by Theorem I, 2 the problem is elementary. The formula also indicates a definite 'method of construction.
In the trisection of a given angle we require 0 0 the formula cos ()= 4 cos' 3 cos . Her 3 cus 0 --a is known, and cos is required.
3 The equation of the problem is 3.r a =-- O. When solved by Cardan's formula this leads to cube roots. But before Theorem II can he applied it must he shown that no expression involving only square roots can satisfy the equation. This is true in the present! ease by the following general theorem taken from the theory of equations: Theorem An irreducible equation whose degree is not a power of two cannot have a root expressible by radicals of the second degree. (The term irreducible equation is here employed to describe an equation f (x)= 9 with rational coefficients whose left member cannot be factored rationally).
In general the algebraic questions which arise in this connection require for their com plete discussion the powerful Galois Theory of Equations. See EQUATIONS, GALOIS' THEORY OF.
A second of the so-called famous problems of elementary geometry is the Delian problem, of the duplication of cube. Given a cube with side a, to construct a cube with side x having double the volume. The equation of The problem is Theorem III and then TheoreM II apply. The corresponding problem concerning the square, leading to the equation is easily solved: the side of the required square is simply the diagonal of the given square.
Regular The construction of a regular polygon of n sides is equivalent to the division of a given circumference into n equal arcs. The only cases treated by Greek geom eters and the ordinary text-books are, for prime numbers, n=3 and n = 5; from these constructions of the regular triangle and ,pentagon, combined with the construction for bisecting an angle, the constructions for the cases 2K. where IC is any in
teger, are easily found.
No advance was made, that is, no new con structible polygons were discovered, until Gauss, about a century ago, applied the algehra'c method. The equation of the problem may be pia into the form ? xtt -1- . .. -1- x -1- 1= 0, which is then termed the cyclotomic equation. When n is a prime number the equation is irreducible. Hence by Theorem III the con struction is possible only when n-1 is a power of 2; That is, n must be of the form -I- 1. Prime numbers of this type are necessarily of the form 2'' 1, and are known as Fermat primes. The values v=0 and v a=1 give the familiar cases n=3 and n=5; the first new case, arising from v=2, is n=17. The construction for the regular polygon of 17 sides is complicated, but the steps are indi cated definitely by the algebraic solution of the cyclotomic equation, which is in fact solva ble by square roots.
The general result on regular polygons is as follows: The regular polygon of n sides can he constructed with ruler and compass if, and only if, the prime factors of n are 2 re peated any number of times and distinct Fermat primes.
The first impossible cases are n=7 and Quadrature of the Circle. This most famous problem of geometry requires the con struction of a square having the same area as a given circle. That this is impossible (that is, that the construction cannot be effected with the ruler and compass) was not definitely shown until 1882, although the' failure of innumerable attempts had led many to suspect the true re sult. The rectification of the circle, that is the construction of a straight line having the same length as a given circumference, is an equiva lent problem, and hence also impossible. This is so on account of the theorem that the area of a circle equals one-half of the prod uct of the radius into the circumference.