In modern geometry, given an equilateral triangle, one can't construct a square with the same area with the use of Hilbert tools. Why is this? The claim seems untrue to me, so there must be something wrong with my understanding.

First, given an equilateral triangle of side length $s$, the area of the triangle is $\frac{s^2\sqrt{3}}{4}$, so it seems to construct a square would require that $\frac{s\sqrt[4]{3}}{2}$ be a constructible number. Obviously $2$ is constructible, and so is $s$ since it is a given side. Isn't $\sqrt[4]{3}$ also constructible, since we have $$ \mathbb{Q}\subseteq\mathbb{Q}(\sqrt{3})\subseteq\mathbb{Q}(\sqrt[4]{3}) $$ so $\deg(\mathbb{Q}(\sqrt[4]{3})/\mathbb{Q})=4=2^2$? What am I missing?

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    $\begingroup$ Where did you encounter this claim? $\endgroup$ May 3 '11 at 9:49
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    $\begingroup$ I found it in that Hartshorne Modern Geometry book that was recommended in one of the questions on this site. It's question 28.11, which says "Show that it is not possible with Hilbert's tools to construct a square with area equal to a given equilateral triangle. (Of course, it is possible with ruler and compass by II.14).)" I thought Hilbert's tools were just axioms that played the role of ruler and compass, which is why I'm confused. $\endgroup$
    – Dom
    May 3 '11 at 10:21
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    $\begingroup$ Hilbert's tools don't use Euclid's parallel postulate, which you seem to assume with your statements on triangles. See Section 10 (Hilbert Planes) of Hartshorne's Geometry: Euclid and Beyond for more details. $\endgroup$ May 3 '11 at 11:22

I assume Hartshorne is talking about a Euclidean triangle, so assuming the formula for the area holds, you then want to show that $\sqrt[4]{3}$ is Hilbert constructible.

However, I think the issue here is that the field of Hilbert constructible numbers, $\Omega$, is a proper subset of $K$, the field of constructible numbers, as the square root operation is limited to $a\mapsto\sqrt{1+a^2}$, not $a\mapsto\sqrt{a}$.

Of course $\sqrt[4]{3}\in K$, since it is obtainable from rationals with the $\sqrt{}$ operation. However, take a look at Exercise 28.9 before this, which states a number $\alpha$ is constructible with Hilbert's tools if and only if $\alpha$ is constructible and totally real. Already, we know $\sqrt[4]{3}$ is constructible, but it is not totally real. We see this since the minimal polynomial of $\sqrt[4]{3}$ is $X^4-3$ which has roots $\pm\sqrt[4]{3}$ and $\pm i\sqrt[4]{3}$, so not all the conjugate elements of $\sqrt[4]{3}$ are real. Hence $\sqrt[4]{3}$ is not totally real, and thus not constructible by Hilbert's tools, although it is constructible with ruler and compass.


Are you sure you can't? I seem to be able to do it with a compass and straightedge.

1) Start with an equilateral triangle with points ABC.

2) Bisect the side AB, and call the midpoint D.

3) Points ADC now form three corners of a rectangle with width $1/2$ the base of the triangle and height the altitude of ABC. That is, the area of the rectangle is the same as the area of ABC.

4) Draw a square with the same area as this rectangle.


Apparently, knowing about the areas in this construction requires use of the parallel postulate, which is absent in the context of this question, which is presented here, as Doug pointed out in a comment on the question.

  • $\begingroup$ You can think of this as dissecting the equilateral triangle into two "30-60-90" right triangles, and then rejoining those two triangles at their hypotenuses. $\endgroup$ May 3 '11 at 10:12
  • $\begingroup$ Perhaps you should expand on "4) Draw a square with the same area as this rectangle", as it seems more complicated than 1-3. $\endgroup$
    – Henry
    May 3 '11 at 11:16
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    $\begingroup$ It is more complicated, but that's where the fun would be! It sounds like I misunderstood the question anyway and my answer is wrong. I didn't understand that "Hilbert's Tools" correspond to tools in a system with fewer axioms than Euclidean geometry; I quickly skimmed something and thought it was a more luxurious tool kit in usual Euclidean geometry. My knowledge is too limited to say why this construction doesn't work there, but I'm guessing I used a fact about areas of triangles (or other shapes) which required the parallel postulate (which is absent in the context of this question). $\endgroup$
    – matt
    May 3 '11 at 11:33

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