Two Sangaku with Equal Incircles

Here is a 1897 sangaku from Chiba perfecture. The problem also appeared earlier in a printed form in a 1781 book.

a system of three equations

The point $D$ on the side $BC$ of $\Delta ABC$ is such that the incircles of triangles $ACD$ and $ABD$ have equal radii. Find the length of $AD$ in terms of the sides.



  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


  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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Copyright © 1996-2018 Alexander Bogomolny

a system of three equations

The point $D$ on the side $BC$ of $\Delta ABC$ is such that the incircles of triangles $ACD$ and $ABD$ have equal radii. Find the length of $AD$ in terms of the sides.

(This solution has been kindly sent to me by Professor J. Marshall Unger, Department of East Asian Languages and Literatures, The Ohio State University.)

In the figure above, let $r$ be the inradius of $\Delta ABC$ and $s$ its semiperimeter, $s = (a + b + c)/2.$ $\Delta ABD$ and $\Delta ACD$ have semiperimeters $s_{1}$ and $s_{2},$ respectively, but the same inradius $k.$ Using $x$ for $AD,$ observe that $s_{1} + s_{2} = s + x$ (we will use this fact twice). Adding areas, $rs = ks_{1} + ks_{2}.$ Hence $x = (rs/k) - s.$ By similar triangles,



Hence for $x,$ we have two equations:

$\displaystyle\frac{s(s-b)}{s_{1}-x} -s=x=\frac{s(s-c)}{s_{2}-x}-s.$

Expand each equation, add, and solve for $x$:

$\begin{align} s(s - b) - ss_{1} + sx &= xs_{1} - x^{2};\\ s(s - c) - ss_{2} + sx &= xs_{2} - x^{2};\\ s(2s - b - c) - s(s_{1} + s_{2}) + 2sx &= x(s_{1} + s_{2}) - 2x^{2};\\ sa - s(s + x) + 2sx &= x(s + x) - 2x^{2};\\ 2x^{2} &= (s + x)^{2} - 2sx - sa\\ x^{2} &= s^{2} - sa. \end{align}$

It follows that


$x = \sqrt{s(s - a)}.$

As a corollary, we now easily obtain a solution to another sangaku [Fukagawa & Pedoe, #2.2.3]:

$ABC$ is a right-angled triangle at $A,$ the point $D$ lies on $BC,$ and is such that the inradii of triangles $ABD$ and $ACD$ are equal. Find the common radius in terms of $a,$ $b,$ $c.$

From (1) and (2), the unknown radius $k$ is expressible in terms of $r$ and $x$. For a right triangle, $r = (b + c - a) / 2 = s - a$. Further, $sr = Area(ABC) = bc/2,$ giving $x = \sqrt{bc/2}.$ Finally,

$\displaystyle k = \frac{bc}{a + b + c + \sqrt{2bc}}.$

(P. Yiu has observed (Missouri J. Math. Sci., 15 (2003) 21-32) that if the cevian $AD$ has, as discussed above, the property that the inradii of triangles $ABD$ and $ACD$ are equal, then so are the radii of their excircles opposite vertex $A.\,$ There are two more solutions, of which one exploits P. Yiu's observation.)

|Up| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny