A Sangaku with an Egyptian Attachment:
What Is This About?
A Mathematical Droodle

 

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Solution

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Copyright © 1996-2018 Alexander Bogomolny

The applet purports to suggest the following sangaku [Temple Geometry, p. 3]:

  Assume points A and B are fixed on a line AB and two circles are drawn touching AB at A and B and tangent to each other. A circle Q is tangent to AB and the two circles externally. Prove that, as the two circles change, circle Q remains tangent to a fixed circle through A and B. Moreover, the radius of the latter is 5/8·AB.

 

(This is a Sangaku from the Miyagi prefecture whose tablet has disappeared long ago.)

We already met this configuration and now how to compute the radius and location of circle Q. If the radii of the circles on A and B are R1 and R2, the radius of circle Q r and the horizontal distances between the left circle and Q and Q and the right circle are x and y then

(1) x² = 4R1r,
y² = 4R2r,
(x + y)² = 4R1R2.

Let for some configuration of the circles, a circle with center O, radius R and passing through A and B also touches circle, Q as shown. This gives us two right triangles with sides

  h = OM, (x + y)/2, R and
r + h, (x - y)/2, R - r.

The Pythagorean theorem implies two identities:

  h² + (x + y)²/4 = R² and
(r + h)² + (x - y)²/4 = (R - r)².

Substituting h² from the first into the second gives

(2) 2Rr = xy - 2rh.

However, from (1), xy = 2r(x + y), which transforms (2) into

(2) 2Rr = 2r(x + y) - 2rh,

or

  R + h = x + y = 2·(x + y)/2.

It's not hard to see that the only right triangle in which twice one leg equals the sum of the hypotenuse and the other leg are equal is the famous 3-4-5 triangle, often referred to as the Egyptian triangle. Indeed, let a, b, c be the legs and the hypotenuse of a right triangle that satisfy say

(3) a + c = 2b.

Then since a² + b² = c², we have

  a² + b² = (2b - a)²,

from which

  3b = 4a,

and then (3) implies

  3c = 5a and 4c = 5b.

If we define t = a/3, then

  a = 3t,
b = 4t,
c = 5t.

This tells us that R and H and hence O do not depend on the values of x and y but only on their sum (x + y) and are invariant for a fixed pair of points A and B.

Applying inversion Michel Cabart found a shorter solution to the problem:

 

Let's transform the figure by an inversion with center A letting point B invariant. Then

  • line (AB) is globally invariant,
  • circle (A) becomes line (A') parallel to (AB),
  • circle (B) becomes circle (B') tangent to (AB) in B and to line (A'),
  • circle (Q) becomes circle (Q') tangent to (B') and to lines (A') and (AB), thus equal to (B').

The tangent (BF) to circle (Q') satisfies: tan (FBE/2) = t = 1/2. Its position is thus fixed. It is the inverse of a fixed circle in the original figure passing by A and B (red) and tangent to circle (Q). In this circle we have: AH × AC = AB² (as H is the inverse of C) or 2R x sin FBE = AB; but sin FBE = 2t/(1+t2) = 4/5. Thus R = (5/8)AB.

References

  1. H. Fukagawa, D. Pedoe, Japanese Temple Geometry Problems, The Charles Babbage Research Center, Winnipeg, 1989

    Write to:

    Charles Babbage Research Center
    P.O. Box 272, St. Norbert Postal Station
    Winnipeg, MB
    Canada R3V 1L6

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

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

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