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In particular, in the Computation section of in the Wikipedia page for geometric median, there is this statement:

...but no such formula is known for the geometric median, and it has been shown that no explicit formula, nor an exact algorithm involving only arithmetic operations and kth roots can exist in general.

At the end of the same paragraph, there is a link to this citation:

Bajaj (1986); Bajaj (1988). Earlier, Cockayne & Melzak (1969) proved that the Steiner point for 5 points in the plane cannot be constructed with ruler and compass

Now, if I recall correctly, all constructible numbers can be represented with a finite number of additions/subtractions, multiplications/divisions, and/or square roots of integers. If this isn't the case, then I think they're at least closely related.

Anyway, the geometric median (Steiner point) for two, three, or four points has a known finite, exact formula or algorithm, but the impossibility of such a formula or algorithm for five points has been proved. This seems to me to be remarkably similar to how finite, exact, and general solutions for polynomials of degree 2, 3, or 4 are known, but for degrees 5 and above, no such formulas exist.

What made me wonder was that the geometric median for five points cannot be constructed, and there is no quintic formula that has a finite number of square roots, and constructibility and a finite number of square roots seem to be closely related. Hence, that's why I'm asking if there is some underlying reason for why both of these are impossible.

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The Cockayne & Melzak paper (published in Mathematics Magazine in 1969) is available from JSTOR, though unfortunately behind a paywall. But what they showed was that constructing the Steiner point by straightedge-and-compass methods requires finding the roots of an eighth-order polynomial with coefficients \begin{align} y^8&:1,\\ y^7&:-60a,\\ y^6&:15(90a^2+22b^2+22),\\ y^5&:-15^2(88a+60a^3+44ab^2),\\ y^4&:15^3(15a^4+22a^2 b^2-9b^4+132a^2+60b^2-9),\\ y^3&:15^4(88a^3+120a^2b-36a),\\ y^2&:15^5(22a^4+60a^2b^2+6b^4+6b^2-54a^2),\\ y^1&:15^6\cdot 12a(3a^2-b^2),\\ 1\;&:-15^7(b^2-3a^2)^2,\\ \end{align} where $a,b$ are appropriate integers. So the situation is in fact even worse than the quintic polynomial, and it comes as no surprise that the roots of this octic polynomial are not constructible.

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