## 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.

3 December 2015, Created with GeoGebra

### 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 = 90^{o.
}

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

PX/RW = TX/TW = QX/SW,

so that

PX/QX = RW/SW.

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,

p^{2} + (a + d)^{2} = (R - a)^{2} and
p^{2} + (b - d)^{2} = (R - b)^{2}.

These are solved for a and b to get

a = (R^{2} - p^{2} - d^{2}) / 2(R + d),
b = (R^{2} - p^{2} - d^{2}) / 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 D_{1} and D_{2} 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

Circle A is transformed into a straight line parallel to L through the image of M, which is D_{2}. Similarly, B is mapped onto D_{1}.

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. D_{2}. 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

## Sangaku

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### References

- H. Fukagawa, D. Pedoe,
*Japanese Temple Geometry Problems*, The Charles Babbage Research Center, Winnipeg, 1989, #4.2.4 - R. Honsberger,
*From Erdös To Kiev*, MAA, 1996.

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