# geometric construction of a given angle

Given any angle how can you say that it is constructable or not?

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How are you given the angle? –  Qiaochu Yuan May 1 '11 at 13:25
You have read this, yes? –  Guess who it is. May 1 '11 at 13:27
J.M.'s link will undoubtedly help you move towards your question. But just a reference if you want a nice soft introduction is Howie's Fields and Galois Theory. I find his treatment of constructibility to be very nice. However I notice that Dinesh added the (relevant) tag "field-extensions". Which means I don't know your background. You should edit your question to include more information about your background so that we can help you more than pointers to resources(or at least make sure those pointers are the right pointers). –  BBischof May 1 '11 at 15:25
Please use meaningful titles. –  quanta May 2 '11 at 22:42

## 1 Answer

According to wiki link given by J.M an angle is constructible if its sine, cosine or tangent is constructible. So given an angle if cosine of the angle satisfies an irreducible polynomial of degree which is NOT a power of 2 then it is NOT constructible. The proof of this criteria is standard (For example you may find it in Hersteins Topics in Algebra, section 5.4)

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Not every irreducible polynomial of degree $4$ has constructible roots, e.g., the roots of $x^4-x-1$ are not constructible. The splitting field has to have degree a power of $2$, which it won't if, as in this case, the Galois group is $S_4$. –  Gerry Myerson May 2 '11 at 4:57
let $a$ be a root of $p(x)=x^4-x-1$ then $[\mathbb{Q(a)}:\mathbb{Q}]$ = deg of minimal polynomial of a = deg$p(x)$ = 4 so $a$ is constructible. Did I go wrong anywhere? Please notify. thanks –  Dinesh May 2 '11 at 14:10
Apologies for previous wrong answer. I just realised my mistake. Given a number, if the irreducible polynomial satisfying this has degree which is not a power of 2 then it is not constructible, as Gerry Myerson pointed out we can still have a irreducible polynomial of degree 4 whose roots are not constructible.So if the degree of the irreducible polynomial is a power of 2 then it need not be constructible. –  Dinesh May 2 '11 at 17:42
Just to complete the answer, for angles of the type $\theta=\frac{2k \pi}{n}$ with gcd$(k,n)=1$, is it true that the angle is constructible if and only if the minimal polynomial of $\cos(\theta)$ has degree a power of two (if and only if $n=2^mp_1...p_k$, where $p_i$ are distinct Fermat primes). –  N. S. May 2 '11 at 18:16
@user9176, yes. –  Gerry Myerson May 2 '11 at 23:58