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 345 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.
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
There is in fact a general formula, which has been derived elsewhere. If we denote the radius of the circle centered at K as R_{1}, that of the circle with center at P as R_{2} and so on, then Archimedes' formula admits a generalization:
R_{n} = Rr/(r^{2} + r + n^{2}). 
For the case at hand, where r = 1,
R_{1} + R_{2} = OD, 
which means that KPAB 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 345 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 O_{n}, their radii as R_{n}, and the projection of O_{n} on AB as P_{n}, this in agreement with the previous usage of R_{1} and R_{2}, so that
As we know from Pappus' theorem,
O_{n}P_{n} = 2nR_{n}. 
Since in the arc, r = 1, we may simplify the expression for R_{n} to
R_{n} = R/(2 + n^{2}). 
Combining the two gives
O_{n}P_{n} = 2nR/(2 + n^{2}). 
We can also show that
DP_{n} = (n^{2}  4)R/[2(2 + n^{2})]. 
It then follows by the Pythagorean theorem that
DO_{n} = (n^{2} + 4)R/[2(2 + n^{2})], 
Thus all triangles DP_{n}O_{n} are Pythagorean. In particular ΔDP_{1}O_{1} is 345. Wait, so far there is no big deal as ΔDP_{1}O_{1} is none other than ΔDOK. But take n = 6. Then

which shows that ΔDP_{6}O_{6} is also 345. As a matter of fact, ΔOP_{3}O_{3} is also 345.
References
 L. Bankoff, Are the Twin Circles of Archimedes Really Twin?, Mathematics Magazine, Vol. 47, No. 4 (Sept., 1974), 214218
 L. Bankoff, C. W. Trigg, The Ubiquitous 345 Triangle, Mathematics Magazine, Vol. 47, No. 2 (Mar., 1974), 6170
 J. L. Heilborn, Geometry Civilized. History, Culture, and Technique, Oxford University Press, 2000
 D. Schattschneider, Reviwew of Heilborn's book, Am Math Monthly 108, 2001, pp 182188

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