Title: Approximate solutions for trisecting the angle or squaring the circle
Tags: mg.metric-geometry;ca.classical-analysis-and-odes;geometric-constructions
Question: Hello all, it is well-known by transcendence results or Galois theory that famous geometric problems such as trisecting an angle or "squaring the circle" (i.e. given a disk of radius 1 construct a square of equal area) are not solvable by compasses-and-ruler only
constructions.
On the other side, it is equally well-known, by approximation results such as Weierstrass', that given any $\varepsilon &gt;0$
there is a definite construction process that yields an approximate solution which is correct up to $\varepsilon$.
Of course, the obvious solution is to compute the coordinates of the points you need up to
the precision you need, and then place the points.
This solution however relies on some classical function tables (cosine, arc cosine or the decimal expansion of $\sqrt{\pi}$)
and I am wondering if there is a more "purely geometric solution" needing no calculator or tables.
Specifically, for angle trisection, one could ask the following :
Define explicitly a compasses-and-ruler only algorithm with the following properties :
Initial data : a  circle with center O and radius 1 cm, two points I and J on that circle such that IOJ is a straight angle, and a point
M on the arc between I and J. Let us call N the point on that arc such that the angle ION is one third of IOM. The algorithm
must return a point N' which is undistinguishable from N to the naked eye, and must not rely on any calculator or tables.
Either that question is interesting or it isn't. If it isn't, the "shortest number of steps" solution has a large number of steps and is only a complicated reformulation of the compute-coordinates-with-calculator method.

Here is the accepted answer: Well, for trisection it's very simple. You could divide angle into $2^n$ parts, then just take $\lfloor\frac{2^n}{3}\rfloor$ parts. Of course it could be made as close to one third as you want, but might be hard to do.
For circling the square - draw the $2^n$-gon, then a rectangle with sides $a_n \cdot 2^n$ and $R/2$ where $a_n$ - is the side of the $2^n$-gon$, and R - is the radius of inscribed circle. then it's easy to transform rectangle into square.
I think it's not harder then constructing the 65537-gon

Here is another answer: Do you know about the (Archimedean) solution to trisecting the angle if you allow a marked ruler?

Here is another answer: You can get approximate trisections using the geometric series $$\frac{1}{2} - \frac{1}{4} + \frac{1}{8} - \frac{1}{16} + \cdots = \frac{1}{3}$$  Geometrically, take your angle and halve it; take the bottom half and halve it again; take the top half of that and halve it again, etc.
Comment for this answer: A nice feature of this solution is that the angle at any even or odd step is resp. a lower or upper approximation of the limit.

Here is another answer: It is possible to trisect an angle exactly with origami.  Based on the Huzita Axioms, the numbers constructible by Origami form a field are a field strictly larger than the field of compass-and-straightedge constructible numbers.  You can also double the cube with origami.
Comment for this answer: Interesting facts, but I'm having difficulty seeing why they're relevant to the question that was asked
Comment for this answer: When my geometry teacher in high school told me it was impossible to trisect an arbitrary angle I came up with the following:  Draw an arc, cut your angle and arc out, and roll this into a cone.  You can easily break a circle into 3 equal parts with compass and straightedge, so doing this and unrolling your cone gives an exact solution.  Then he told me scissors were not allowed.  Grrr!

Here is another answer: For squaring the circle, it seems easiest to first measure the radius $r$ of the circle, and then trying to construct an arbitrarily good approximation of the length $ r \sqrt(\pi)$ using $r$ as the unit length, which must lie in the infinite quadratic extension of the rational. Then use that as the sidelength of the square. An interesting question would be how efficient one could make this process to be. Circling the square can be done using a similar quadratic approximation of $\sqrt(1/\pi)$.

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