Skip to the Main Content

Note:These pages make extensive use of the latest XHTML and CSS Standards. They ought to look great in any standards-compliant modern browser. Unfortunately, they will probably look horrible in older browsers, like Netscape 4.x and IE 4.x. Moreover, many posts use MathML, which is, currently only supported in Mozilla. My best suggestion (and you will thank me when surfing an ever-increasing number of sites on the web which have been crafted to use the new standards) is to upgrade to the latest version of your browser. If that's not possible, consider moving to the Standards-compliant and open-source Mozilla browser.

July 5, 2005

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<1r\lt 1), endowed with the metric ds 2=dr 2+r 2dθ 2(1r 2) 2 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=Rr=R, for some constant R<1R\lt 1. It has area

(1)
A(R)=2π 0 Rrdr(1r 2) 2=πR 21R 2 A(R) = 2\pi \int_0^R \frac{r dr}{(1-r^2)^2} = \frac{\pi R^2}{1-R^2} and diameter D(R)=2 0 Rdr1r 2=log(1+R1R) D(R) = 2\int_0^R \frac{dr}{1-r^2} = \log\left(\frac{1+R}{1-R}\right) which approach the usual Euclidean expressions for small RR, and which diverge for R1R\to 1.
A square in the Poincaré plane

Now consider the “square” shown at right1. 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(θ)=cosθ+sinθ2cosθsinθ r(\theta) = \cos\theta + \sin\theta-\sqrt{2\cos\theta \sin\theta} Thus the area of the square is A=4 0 π/212r 2(θ)1r 2(θ)dθ=π2 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(1/3)A(1/\sqrt{3}), as a circle centered at the origin, with coordinate radius R=1/3R=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, RR, of a circle about the origin in the Poincaré plane which can be squared as R=mm+2n R = \sqrt{\frac{m}{m+2n}} where mm is a positive integer and n=2 tp 1p m,p i=2 2 s i+1prime 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=1m=n=1. To compare, you need to set his parameter, k=1/2k=1/2 and recognize his “RR” as D/2D/2. Also, it’s helpful to solve for my RR in terms of the diameter of the circle, R=tanh(D/2) R = \tanh(D/2) and write the area of the circle as a function of its diameter A(D)=πsinh 2(D/2) 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 π/2\pi/2, while the side-length diverges.

Posted by distler at July 5, 2005 9:56 AM

TrackBack URL for this Entry:   http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/587

12 Comments & 0 Trackbacks

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.

Posted by: Robert on July 5, 2005 10:15 AM | Permalink | Reply to this

Re: Squaring the Circle

My compass and straight-edge are the conventional ones you can buy in a drafting supply shop. What I’ve changed are my definitions of “geodesic” (now drawn with a compass, except for those which pass through the origin, which are still drawn with a straight-edge) and of “area.”

Posted by: Jacques Distler on July 5, 2005 10:26 AM | Permalink | PGP Sig | Reply to this

Re: Squaring the Circle

What you’ve done is brilliant to be sure, but what I see in that circle isn’t a proper square. Assuming that the solution is within geometry- have you ever looked at the alchemical drawings of the squared circle? You might find it interesting.

Posted by: jeannine on March 5, 2008 4:38 PM | Permalink | Reply to this

Re: Squaring the Circle

“For small circles, that problem reduces to the Euclidean one, which even the ancient Greeks knew was impossible.”

I think (though I’m not at all sure) that Bolyai showed how to do it for an arbitrary circle in the hyperbolic plane. As long as the circles are of finite size, the plane doesn’t “become Euclidian”, so to speak. Hence Bolyai’s result doesn’t contradict what the Greeks “knew” (they hadn’t proved it so far as I know but they turned out to be right).

That’s my understanding of the situation but I certainly wouldn’t bet on it. If I’m wrong, clarification would be appreciated. Thanks.

Posted by: Kevin Donoghue on July 5, 2005 11:52 AM | Permalink | Reply to this

Re: Squaring the Circle

Consider a circle, centered at the origin, with coordinate-radius, RR, rational. It has an area, A(R)A(R), which is a rational multiple of π\pi. If what you say is true, I must be able to construct a square whose area is an arbitrary rational multiple of π\pi.

I managed to construct a square whose area is π/2\pi/2. Perhaps, by subdividing it, I can construct squares of area an arbitrary rational multiple of π\pi.

I can see how to get some rational multiples of π\pi, but I don’t see an algorithm to get an arbitrary rational multiple. But, then, I haven’t thought about it for more than a minute.

That’s certainly necessary for your rendition of Bolyai’s result to be true, but even that is far from sufficient.

Posted by: Jacques Distler on July 5, 2005 12:16 PM | Permalink | PGP Sig | Reply to this

Re: Squaring the Circle

Indeed, as discussed above, not even all rational values of R 2R^2 are allowed — let alone a continuum. If I read that 1948 paper correctly, Bolyai’s solution consisted of squaring just a single circle, the solution I wrote down above.

Nestorovich’s contribution was to find the rest of the discrete series of rational values for R 2R^2 for which the circle can be squared. Though he doesn’t give much in the way of details.

Posted by: Jacques Distler on July 5, 2005 11:09 PM | Permalink | PGP Sig | Reply to this

Re: Squaring the Circle

You have forgotten the GIF to fall back to. (I took the URL from the source code; it is definitely not there.)

Posted by: Kristjan Kannike on July 6, 2005 12:51 AM | Permalink | Reply to this

Missing GIF

Naw. It’s there. The URL just got spooged when I revised the post to make the image smaller. It’s fixed now.

Thanks.

Posted by: Jacques Distler on July 6, 2005 1:01 AM | Permalink | PGP Sig | Reply to this

Re: Squaring the Circle

Hmm, given that the non-euclidean spaces were invented on the scent of the fifth axiom, does it mean that there is a relationship between non-go theorems for quadratures and the paralell-throught-a-point axiom? Never heard of such beast.

Posted by: Alejandro Rivero on July 8, 2005 8:54 AM | Permalink | Reply to this

Re: Squaring the Circle

I believe the usual formulation of the “squaring the circle” problem is to construct the sidelength of a square with a ruler and compass from a given length in such a way that the area of the square is equal to the area of the circle whose radius is the given length. This is stronger than simply exhibiting a square whose area is equal to that of a given circle. Even in classical geometry there are circles that are squarable in this weaker sense, for example the circle whose radius is the reciprocal of the square-root of pi. Perhaps one should use another name for the still different and non-trivial problem of finding (in whatever geometry) examples of squares and circles with equal areas and such that both the sidelength and the radius are constructible with ruler and compass from a given unit length.

You state that the “even the ancient Greeks” knew that the circle could not be squared in Euclidean geometry. This assertion is somewhat surprising to me. Can you offer a reference evidencing this knowledge? The impossibility of squaring the circle is usually presented as an application of basic field theory together with the fact that pi does not lie in an iterated quadratic extension of the rational numbers. This latter fact is often deduced from the transcendence of pi (which is massive overkill), but the mere irrationality of pi is not enough. However, I’m not aware that the ancient Greeks even knew that pi was irrational, although they certainly might have suspected as much.

Posted by: Anthony on July 9, 2005 1:51 PM | Permalink | Reply to this

What did they know, and when did they know it?

Not all circles are squareable in hyperbolic geometry. Only a discrete set are. Bolyai managed to square a particular one: the one of diameter D=log(3+131)D= \log\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)

The construction I gave, using the representation of hyperbolic space as the Poincaré plane, has the advantage of mapping the hyperbolic problem into a Euclidean one which can be solved with good-old “Euclidean” straight-edge and compass.

The Euclidean problem is: given a circle of radius 1/31/\sqrt{3}, centered at the origin, construct four circles, each of unit radius, centered at (±1,±1)(\pm 1,\pm 1).

This is stronger than simply exhibiting a square whose area is equal to that of a given circle. Even in classical geometry there are circles that are squarable in this weaker sense, for example the circle whose radius is the reciprocal of the square-root of pi.

That’s not what we did here, though perhaps I phrased things badly …

You state that the “even the ancient Greeks” knew that the circle could not be squared in Euclidean geometry. This assertion is somewhat surprising to me.

I was under the impression that Archimedes knew (or strongly suspected) that π\pi was irrational. However, as you say, that’s far from sufficient. And he couldn’t have known that it was transcendental (or even “not an element of an iterated quadratic extension \mathbb{Q}”).

So I withdraw the comment.

Posted by: Jacques Distler on July 9, 2005 11:40 PM | Permalink | PGP Sig | Reply to this

Re: What did they know, and when did they know it?

It is perhaps worth to mention that the impossibility of the cubature of the pyramid (except, of course, in special cases) was settled in the XXth century, being perhaps the easiest of the problems proposed by Hilbert.

Posted by: Alejandro Rivero on July 11, 2005 3:26 PM | Permalink | Reply to this

Post a New Comment