# Constructing $\pi^2$ and e with straightedge and compass

Is there a way to construct two curves with lengthratios $\pi^2$, or two areas of ratio $\pi^2$, on a plane surface, with a straightedge and compass?

And is e=2.71... possible?

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The answer is no. I checked whether I could write out a proof for lengths. Can be done, the approach is the natural one. But it is fairly unpleasant to do the details. –  André Nicolas Feb 11 '12 at 19:56

No. These cannot be constructed.

First, $\pi^2$ and $e$ cannot be constructed because they are transcendental numbers. [We say $t$ is a transcendental number if there is no non-zero polynomial $p(x) = a_nx^n+\cdots+a_1x+a_0$ with integer coefficients $a_i$ such that $p(t)=0$.] Any constructable number is algebraic (i.e. not transcendental). [Note: All constructable numbers are algebraic, but not all algebraic numbers are constructable.]

As for a ratio of two lengths, if we construct a circle of radius $r$ and line of length $\ell$ (this can only be done if $r$ and $\ell$ are constructable numbers). Then the length (i.e. circumference) of the circle is $2\pi r$. So the ratio of these lengths is $2 \pi r /\ell$. Next, $2r/\ell$ is constructable (since $r$ and $\ell$ are). Thus we can construct curves whose length's ratios are of the form $c\pi$ or $c/\pi$ where $c$ is a constructable number.

The ratio of the lengths of 2 lines or 2 circles will just be a constructable number.

Since $\pi^2$ is not of the form $c$, $c\pi$, or $c/\pi$ for any constructable number $c$ (since $\pi$ itself is not constructable), we cannot construct curves whose length's ratio is $\pi^2$.

A similar argument will rule out $\pi^2$ appearing as an area (areas of constructable rectangles are constructable numbers and areas of constructable circles are $\pi r^2$ for some constructable number $r$).

Edit: Ok. There are some gaps. I hadn't considered more complicated regions. I'm not sure exactly what one could cook up in terms of areas. But I don't think it'll be anything more complicated than $a+b\pi$ for constructable numbers $a$ and $b$ (which would preclude $\pi^2$).

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What about, say, constructible segments of circles? –  Chris Eagle Feb 11 '12 at 20:01
@ChrisEagle If I remember correctly, one can only construct angles of the form $c\pi$ for some constructable number $c$. So this does not yield anything new. [I do acknowledge there are some gaps in my discussion above.] –  Bill Cook Feb 11 '12 at 20:06
And the area of the intersection of circles? –  LeneS Feb 11 '12 at 20:06
@LeneS good point. –  Bill Cook Feb 11 '12 at 20:07
Any area bounded by circle arcs and straight lines can be expressed as a sum/difference of circle segments and polygons, so I think Bill's answer still holds. –  Lopsy Feb 11 '12 at 20:09