A better solution to a difficult sangaku problem

J. Marshall Unger
Department of East Asian Languages & Literatures
The Ohio State University

PROBLEM. In ΔABC, AB = BC. If one chooses D on AB and J on CD such that AJ ⊥ CD and the incircles of ΔACJ, ΔADJ, and ΔBCD all have radius r, then r = AJ/4.

Chapter 6, problem 3, sangaku from Fukagawa & Rothman


The trick to solving the problem expeditiously is to clarify what is given and to prove the converse proposition first.

Notice that the only connection between AB = BC and the geometry of ΔACD is that the incircle of ΔABC (dashes) is tangent to AC at its midpoint. The two incircles of ΔACJ and ΔADJ cannot be tangent to the shared side AJ at the same point, as shown, unless J is the point of tangency of the incircle of ΔACD (dots) with side CD [Honsberger, see also Three Incircles in a Triangle]. Hence AC = AD. And the two incircles cannot be congruent unless, in addition, either ∠CAJ = ∠DAJ or AJ ⊥ CD, in which case, ΔACJ = ΔADJ and the other condition follows. Therefore, given a fixed isosceles ABC, if the congruence of all three small incircles implies r = AJ/4, as the problem asserts, then the converse must also be true. Indeed, it is easier to prove.

In the figure above, let a, b, d, and h be the lengths of CD, AD = AC, DN, and AJ, respectively. Let r1 and r2 be the radii of the incircles of ΔACD and ΔBCD, respectively. We use two lemmas, which we will prove later:

1.a - b= 2d
2.2 r1r2= ad

Given h = 4r, square the equation for the inradius of a right triangle: b = h + a/2 - 2r = 2r + a/2 to get b² = 4r² + 2ar + a²/4, and equate this with the Pythagorean result b² = (a/2)² + h² = a²/4 + 16r². This yields a = 6r, which, with h = 4r, implies b = 5r. Hence, by Lemma 1, 2d = r.

Now area( ΔACD) = ha/2 = 12r² = r1s, where s is the semiperimter of ΔACD or 8r. Thus r1 = 3r/2 = 3d. Using Lemma 2, we find r2 = a/6 = r.

Chapter 6, problem 3, sangaku from Fukagawa & Rothman.

From this, it is clear that solving the original problem comes down to proving, without knowing the value of h, that ΔACJ and ΔADJ are 3:4:5 right triangles.

It's easy to show that ΔDTU is similar to ΔDO2U. Hence DU is the mean proportional between DT and O2U = DT + DN. In terms of lengths, we therefore have

 a/2 - r + d= [r² + (a/2 - r)²] / (a/2 - r)
 (a/2 - r)² + d (a/2 - r)= r² + (a/2 - r
 d (a/2 - r)= r²
 ad/2 - dr= r²
 r1r - dr= r²
 r1 - d= r.

(Lemma 2 takes the form 2r1r = ad since we are assuming r2 = r.) Since JS = r, O1S = d. Thus JO1 = r + d = 3d = r1.

Hence, by Lemma 2, 2rr1 = 2(2d)(3d) = 12d² = ad, so a = 12d = 6r. Since a - b = 2d (Lemma 1), b = 10d = 5r. And because ADJ is a right triangle, a/2 = 3r and b = 5r imply h = 4r.

The lemmas follow from the general case of any triangle ABC with incircle I. Draw a cevian CD and the incircles O1 and O2 of the resulting triangles ACD and BCD. Label the points of tangency of these incircles as shown. Then EH = GK = DN.

Chapter 6, problem 3, sangaku from Fukagawa & Rothman.

PROOF: By equal tangents, CE = CF (1), and

AH = AK BM = BN CH = CJ DN = DL.

Subtracting the equations in second row from those in the first, we get

EH = GK (2) FM = GN (3) CM - CH = JL (4) JL = DK - DN (5).

Equating the left and right sides of (4) and (5),

 CM + DN= DK + CH
 (CF + FM) + DN= DK + (CE + EH)
 (by 1) FM + DN= DK + EH
 (by 3) GN + DN= DK + EH
 (GN - DG) + DN= (DK - DG) + EH
 2DN= GK + EH
 (by 2) DN= GK (6).

Linking equations (2) and (6), EH = GK = DN.

Many more inferences can be drawn from this figure, but we need just two.

Chapter 6, problem 3, sangaku from Fukagawa & Rothman.

Lemma 1 follows immediately if E and J are the midpoints of AC and CD, respectively. Then CD/2 - AC/2 = EH = DN. In the problem, as explained earlier, J must be the midpoint of CD. Requiring that ABC be isosceles forces E to be the midpoint of AC.

Lemma 2 in the general case is r1r2 = DJ·DL. The dashed lines in the figure show that

(r1 + r2)² + (DJ - DL)² = (O1O2)² = (r1 - r2)² + KN²,

but KN = DK + DN = DJ + DL by equal tangents. The rest is just algebra.

In introducing the solution of Kitagawa Moko (1764 -1833), Fukagawa and Rothman remark, "It is somewhat complicated and we hope that readers can find a simpler one." I do not know whether the solution presented above is what Fukagawa and Rothman had in mind, but Kitagawa's solution is certainly complicated. It proves Lemma 1 relying in part on the given that ΔABC is isosceles; finds expressions for r1, a, and d in terms of r and h; plugs these directly into Lemma 2; produces a third-degree equation in r and h through more painstaking algebra; rejects the roots h = ± r2 because they are inconsistent with the figure; and so concludes that h = 4r. In addition, Kitagawa's proof gives no hint that the lemmas hold even for scalene ABC, the true significance of requiring that ABC be isosceles, that the converse is true, or that 3:4:5 right triangles play a role.


  1. H. Fukagawa, A. Rothman, Sacred Mathematics: Japanese Temple Geometry, Princeton University Press, 2008
  2. R. Honsberger, Episodes in nineteenth and twentieth-century Euclidean geometry, MAA, 1995


  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

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