Gothic Arc

I've been putting a stack of magazines in order when I ran into D. Schattschneider's very positive review of a book by J. L. Heilborn. The review and the book end with a problem of Gothic Arc:

The Gothic Arc OBC is constructed with two circular arcs of radius OB, one centered on O and one on B. The sequence of circles is constructed as shown in the applet below so that each circle is tangent to the semicircle on diameter OB, the arc OBC, and the adjacent circles in the sequence. Show that the radius of each of these tangent circles is a rational fraction of OD. (This is implies that the right triangles with one vertex at the center of a circle from the sequence, another at O, and the base on OB have rational sides. In particular, ODP is a 3-4-5 triangle.)

The book is said to have a complete solution. I do not have the book to check the solution but the configuration is clearly embeddable into a more conventional Pappus' circle chain inscribed into Archimedes' Arbelos. For arbelos, the fact is very well known and is based on the properties of the Pappus' chain and the arbelos as partially demonstrated by Archimedes himself.


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at, download and install Java VM and enjoy the applet.

What if applet does not run?

I shall just show how the "Gothic Arc" may be embedded into a more standard arbelos with a Pappus chain. Arbelos is a shape bounded by three mutually tangent semicircles with a common base. In case the two small semicircles are equal, an Archimedes' formula (with r = 1) implies that the radius of the biggest circle (centered at K in the applet) inscribed into such an arbelos equals 1/3 of the radius of the big semicircle (OB/3, or, R/3, with R = OB.)

There is in fact a general formula, which has been derived elsewhere. If we denote the radius of the circle centered at K as R1, that of the circle with center at P as R2 and so on, then Archimedes' formula admits a generalization:

  Rn = Rr/(r2 + r + n2).

For the case at hand, where r = 1, R1 = R/3 and R2 = R/6. Since R/3 + R/6 = R/2, we can claim that

  R1 + R2 = OD,

which means that KP||AB and D, the center of the semicircle, is the foot of the perpendicular from P to AB. In other words, the circle centered at P is sitting directly on top of the right semicircle. The line PD of their centers is the common axis of symmetry, which implies the possibility of the embedding.

Arbelos being one of geometric wonders, the configuration has many more marvelous properties (see [Bankoff, Bankoff and Trigg]). Triangle ODP is not the only 3-4-5 triangle in this configuration. But to find another, I'll have to introduce more general notations. So let denote the centers of the circles in Pappus' chain as On, their radii as Rn, and the projection of On on AB as Pn, this in agreement with the previous usage of R1 and R2, so that K = O1 and P = O2.

As we know from Pappus' theorem,

  OnPn = 2nRn.

Since in the arc, r = 1, we may simplify the expression for Rn to

  Rn = R/(2 + n2).

Combining the two gives

  OnPn = 2nR/(2 + n2).

We can also show that

  DPn = (n2 - 4)R/[2(2 + n2)].

It then follows by the Pythagorean theorem that

  DOn = (n2 + 4)R/[2(2 + n2)],

Thus all triangles DPnOn are Pythagorean. In particular ΔDP1O1 is 3-4-5. Wait, so far there is no big deal as ΔDP1O1 is none other than ΔDOK. But take n = 6. Then

O6P6= 24·R/76,
DP6= 32·R/76 and
DO6= 40·R/76,

which shows that ΔDP6O6 is also 3-4-5. As a matter of fact, ΔOP3O3 is also 3-4-5.


  1. L. Bankoff, Are the Twin Circles of Archimedes Really Twin?, Mathematics Magazine, Vol. 47, No. 4 (Sept., 1974), 214-218
  2. L. Bankoff, C. W. Trigg, The Ubiquitous 3-4-5 Triangle, Mathematics Magazine, Vol. 47, No. 2 (Mar., 1974), 61-70
  3. J. L. Heilborn, Geometry Civilized. History, Culture, and Technique, Oxford University Press, 2000
  4. D. Schattschneider, Reviwew of Heilborn's book, Am Math Monthly 108, 2001, pp 182-188

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