A Sangaku Follow-Up on an Archimedes' Lemma

The following Sangaku problem sports a publication trail:

X is a point on a chord L of a given circle C, and circles A and B are drawn on opposite sides of L so as to touch L at X and be tangent to C. Prove that the ratio of the radii of A and B is independent of the position of X on L.

The problem has been included in Fukagawa and Pedoe's Japanese Temple Geometry Problems, offered by Crux Mathematicorum, 1987, 194, and appeared in R. Honsberger's collection From Erdös To Kiev with two solutions from the Crux. The first two solutions below, have been adapted from Honsberger's book.

Solution 1

This solution is by Dan Sokolowsky, Williamsburg, VA.

Start by drawing PXQ perpendicular to L at X. Since L is a common tangent to A and B, PX and QX are diameters of A and B. If M and N are their points of tangency with C, angles at M and N subtended by PX and QX are right:

∠XMP = ∠XNS = 90o.

Extending MX and MP to meet C in R and S, we get a diameter RS (since RS subtends right angles in C.)

Now, the centers U and V of A and C are in line with the point of tangency M, and it is clear that triangles MUX and MVS are isosceles. Since they share a common base angle at M, the base angles at X and S are also equal, and the lines PXQ and RS are parallel. Thus RS is perpendicular to L.

Similarly, extending NX and NQ we'll get a diameter perpendicular to L. Therefore, R lies on NX and S on NQ.

The next step is to extend MR and NS beyond M and N to meet in T. In ΔRST, RN and SM are altitudes, so that X is the orthocenter. TX is then the third altitude and is perpendicular to RS. But L is a line through X perpendicular to RS, which leads us to conclude that T lies on L.

The final step is to consider two pairs of similar triangles (TXP, TWR) and (TXQ, TWS):


so that


Since the right side here is independent of X, so is the left side. The conclusion follows.

I have only one remark concerning this proof. We may have started with constructing RS, a diameter perpendicular to PXQ. Then PR passes through M (and QS through N) by Lemma 1 from Archimedes' Book of Lemmas. (The situation has been explored elsewhere.)

Solution 2

This solution is by Sam Baethge, San Antonio, TX. It's a straightforward application of a theorem attributed to another famous Greek mathematician.

Let p and d be the horizontal and vertical components of VX. And let a, b and R be the radii of A, B, and C, respectively. By the Pythagorean theorem,

p2 + (a + d)2 = (R - a)2 and p2 + (b - d)2 = (R - b)2.

These are solved for a and b to get

a = (R2 - p2 - d2) / 2(R + d), b = (R2 - p2 - d2) / 2(R - d),

from which a/b = (R - d)/(R + d) independent of X.

This unexciting solution proved extremely useful in constructing the diagram and, ultimately, writing the applet illustration.

Solution 3

This simple solution that is due to Michel Cabart employs inversion which I am not sure has been available to the Japanese during the period of seclusion.

Trace the tangents D1 and D2 to circle C in R and S. Now transform the figure by the inversion with center X, and power equal to the power of X relative to C. The latter equals k = XM·XS = XN·XR. In particular this means that M and X are inverse images of each other, and so are N and R.

Circle A is transformed into a straight line parallel to L through the image of M, which is D2. Similarly, B is mapped onto D1.

The inverse of P stays on the perpendicular to L through X, i.e., PQ, but also lies on the image of A, i.e. D2. Hence the distance from X to the image of P is R + d. The distance from P itself to X is of course 2a. Thus 2a·(R + d) = k. Similarly, for the inverse of Q, 2b·(R - d) = k. Taking the ratio, we have the desired a/b = (R - d)/(R + d).


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  12. A Sangaku Follow-Up on an Archimedes' Lemma
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  1. H. Fukagawa, D. Pedoe, Japanese Temple Geometry Problems, The Charles Babbage Research Center, Winnipeg, 1989, #4.2.4
  2. R. Honsberger, From Erdös To Kiev, MAA, 1996.

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