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):

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,

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).

Sangaku

  1. Sangaku: Reflections on the Phenomenon
  2. Critique of My View and a Response
  3. 1 + 27 = 12 + 16 Sangaku
  4. 3-4-5 Triangle by a Kid
  5. 7 = 2 + 5 Sangaku
  6. A 49th Degree Challenge
  7. A Geometric Mean Sangaku
  8. A Hard but Important Sangaku
  9. A Restored Sangaku Problem
  10. A Sangaku: Two Unrelated Circles
  11. A Sangaku by a Teen
  12. A Sangaku Follow-Up on an Archimedes' Lemma
  13. A Sangaku with an Egyptian Attachment
  14. A Sangaku with Many Circles and Some
  15. A Sushi Morsel
  16. An Old Japanese Theorem
  17. Archimedes Twins in the Edo Period
  18. Arithmetic Mean Sangaku
  19. Bottema Shatters Japan's Seclusion
  20. Chain of Circles on a Chord
  21. Circles and Semicircles in Rectangle
  22. Circles in a Circular Segment
  23. Circles Lined on the Legs of a Right Triangle
  24. Equal Incircles Theorem
  25. Equilateral Triangle, Straight Line and Tangent Circles
  26. Equilateral Triangles and Incircles in a Square
  27. Five Incircles in a Square
  28. Four Hinged Squares
  29. Four Incircles in Equilateral Triangle
  30. Gion Shrine Problem
  31. Harmonic Mean Sangaku
  32. Heron's Problem
  33. In the Wasan Spirit
  34. Incenters in Cyclic Quadrilateral
  35. Japanese Art and Mathematics
  36. Malfatti's Problem
  37. Maximal Properties of the Pythagorean Relation
  38. Neuberg Sangaku
  39. Out of Pentagon Sangaku
  40. Peacock Tail Sangaku
  41. Pentagon Proportions Sangaku
  42. Proportions in Square
  43. Pythagoras and Vecten Break Japan's Isolation
  44. Radius of a Circle by Paper Folding
  45. Review of Sacred Mathematics
  46. Sangaku à la V. Thebault
  47. Sangaku and The Egyptian Triangle
  48. Sangaku in a Square
  49. Sangaku Iterations, Is it Wasan?
  50. Sangaku with 8 Circles
  51. Sangaku with Angle between a Tangent and a Chord
  52. Sangaku with Quadratic Optimization
  53. Sangaku with Three Mixtilinear Circles
  54. Sangaku with Versines
  55. Sangakus with a Mixtilinear Circle
  56. Sequences of Touching Circles
  57. Square and Circle in a Gothic Cupola
  58. Steiner's Sangaku
  59. Tangent Circles and an Isosceles Triangle
  60. The Squinting Eyes Theorem
  61. Three Incircles In a Right Triangle
  62. Three Squares and Two Ellipses
  63. Three Tangent Circles Sangaku
  64. Triangles, Squares and Areas from Temple Geometry
  65. Two Arbelos, Two Chains
  66. Two Circles in an Angle
  67. Two Sangaku with Equal Incircles
  68. Another Sangaku in Square
  69. Sangaku via Peru
  70. FJG Capitan's Sangaku

References

  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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