Harmonic ratios of the pentagon

Consider the pentagon enclosed in two circles, as shown here. Again, from Euclid, the line segment from the center of the pentagon (and circles) to the midpoint of any edge is perpendicular to that edge.
Consider the triangle shown here at the heart of the figure. As the equal angles of the pentagon are 360/5=72 degrees, the upper angle of the yellow triangle is 72/2=36 degrees. The angle on the right, as before, is right. So the lower angle is 180-90-36=54 degrees. This is a 36-54 right triangle, not to familiar. The harmonic ratio we seek is the ratio of the vertical hypotenuse to the lower side, evidently (to the modern eye) the cosecant of 54 degrees, or ~1.23612.
But Euclid, not having access to such trig tables, had a more precise method. In Book XIII, Proposition 11 (Heath, 1960; v.3, p. 461) he shows that the ratio of the side of the pentagon to the diameter of the circumscribed circle is "the minor", which he had carefully studied in Book X. Copying from (Heath, p. 466): if R is the radius of the circumscribed circle, and S the side of the pentagon, then:
[1] S = R/2 * sqrt(10-2*sqrt(5))
Looking at our yellow triangle, we have from the Pythagorean theorem,
[2] R2 = r2 + (S/2)2
where r is the radius of the inscribed circle. Replacing S in [2] by its value as a function of R in [1], we have,
[3] R2 = r2 + R2/16 * (10-2*sqrt(5))
or, collecting terms in R on the left,
[4] R2{1-(10-2*sqrt(5))/16} = r2
At this point we may as well evaluate (appoximately) the expression in {brackets}, finding,
[5] R2{0.6545085} = r2
or, R/r ~= 1.236068. Kepler knew this.
Rev'd 17 jan 2002 by ralph abraham