### Squaring the Circle

John Quiggin wants to know how to square the circle on the Poincaré plane. I don’t know how Bolyai did it, but the problem is easy enough.

The Poincaré plane is the unit disk ($r\lt 1$), endowed with the metric $ds^2 = \frac{dr^2 + r^2 d\theta^2}{(1-r^2)^2}$ which has constant negative curvature. Geodesics (“straight lines”) in this metric are circles which intersect the unit circle at right angles (among which, we include straight lines through the origin).

Consider a circle centered at the origin. That is, consider the set of points $r=R$, for some constant $R\lt 1$. It has area

Now consider the “square” shown at right^{1}. It consists of 4 quarter-circles (each of unit radius), which meet the unit circle perpendicularly. We can describe these arcs parametrically. The one in the upper-right quadrant is
$r(\theta) = \cos\theta + \sin\theta-\sqrt{2\cos\theta \sin\theta}$
Thus the area of the square is
$A = 4 \int_0^{\pi/2} \frac{1}{2} \frac{r^2(\theta)}{1-r^2(\theta)} d\theta =\frac{\pi}{2}$
This is the same area, $A(1/\sqrt{3})$, as a circle centered at the origin, with coordinate radius $R=1/\sqrt{3}$. I leave it as an exercise for the reader to construct the latter using compass and straight-edge.

Note that we squared a *particular* circle in the Poincaré plane. We couldn’t possibly have squared an *arbitrary* circle. For small circles, that problem reduces to the Euclidean one, which even the ancient Greeks knew was impossible.

#### Update:

A bit of Googling around yielded this Russian paper which gave the allowed values of the coordinate radius, $R$, of a circle about the origin in the Poincaré plane which can be squared as $R = \sqrt{\frac{m}{m+2n}}$ where $m$ is a positive integer and $n= 2^t p_1\dots p_m,\qquad p_i =2^{2^{s_i}} +1\, \text{prime}$ The solution constructed above corresponds to $m=n=1$. To compare, you need to set his parameter, $k=1/2$ and recognize his “$R$” as $D/2$. Also, it’s helpful to solve for my $R$ in terms of the diameter of the circle, $R = \tanh(D/2)$ and write the area of the circle as a function of its diameter $A(D) = \pi \sinh^2(D/2)$^{1} You should think of it as the limiting member of a family of squares, centered at the origin, of increasing area and increasing side-length. In the limit, the area approaches $\pi/2$, while the side-length diverges.

## Re: Squaring the Circle

What is your definition of compass and straight edge in hyperbolic space? Do you use the usual (euclidean) ones to draw on an Escher drawing or are they some generalizations of:

straight edge: given a line (defined by giving two non-equal points on it) can find the intersection of this with another line.

compass: given points A, B, C, D, E, F finds all points X such that distances AB and CX and DE and FX are the same. Similar for points on a line with some given distance to some other point.