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Let $A,B,C$ be three non-collinear points in a plane with rational coordinates. Let $r$ be the circumradius of the triangle $ABC$. A simple argument (see here) shows that $r$ has degree $2^p$ over ${\mathbb Q}$, for some $p$ between $0$ and $6$.

For each such $p$, is there a nice purely geometrical description of when $r$ has degree exactly $2^p$ ?

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$p$ can be reduced to 4. Use $R = \frac {abc}{\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}$, where $a, b, c$ are side lengths of the triangle, and that should give you the characterization. – Calvin Lin Jan 9 '13 at 3:54
If $A=(0,0),B=(1,2),C=(3,5)$, then $a=\sqrt{13},b=\sqrt{34},c=\sqrt{5}$, so ${\mathbb Q}(a,b,c)$ has degree $8$. – Ewan Delanoy Jan 9 '13 at 4:11
Doesn't that mean that $\mathbb{Q}(a,b,c) = 2^3$? And we're introducing at most 1 more square root in the denominator, so $p\leq 4$? Unless I'm wrong, of course. – Calvin Lin Jan 9 '13 at 4:12
No, it is I who’s wrong (not surprising at this hour). – Ewan Delanoy Jan 9 '13 at 4:14
So I'm not sure about the denominator, if it will introduce a bunch more degrees, but I think that it shouldn't. In the original argument from coordinates, we could have shifted $(A_x, A_y)$ to $(0,0)$, which would likely reduce $p$ to 4. Once again, not completely certain. – Calvin Lin Jan 9 '13 at 4:18
up vote 1 down vote accepted

$$R = \frac {abc}{\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}} = \frac {abc} {\sqrt{2a^2b^2 + 2b^2 c^2 + 2c^2 a^2 - a^4 -b^4 - c^4}} = \frac {abc}{4S},$$

Since $a^2, b^2, c^2 \in \mathbb{Q}$ by the distance formula, it follows that $R^2 \in \mathbb{Q}$. Hence, the degree is either 1 or 2.

By pick's theorem, the area of the triangle is rational (by scaling to a lattice grid). Hence, the degree of the field extension is 2 if and only if $abc$ is not a perfect square.

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