Another Sangaku in Square

This is problem #3.2.5 from Fukagawa & Pedoe's book. It is simple enough, but surprisingly the configuration exhibits several curious properties. In particular, there are several occurrences of the Egyptian 3-4-5 triangle that have been missed on the page specifically devoted to the appearance of that triangle in several sangaku problems. The sangaku is said to have been written in 1838, in the Iwate prefecture, on a tablet which has not survived.

$ABCD$ is a square and the semicircle on $AB$ as diameter, center $O_1,$ lies within it. $CG,$ where $G$ lies on $AD,$ touches this semicircle, and $O_{2}(r_{2})$ is the incircle of $\Delta CDG.$ An external common tangent of the semicircle and $O_{2}(r_{2})$ meets $CD$ in $R,$ $BC$ in $N,$ and intersects $CG$ in $M.$ The circle $O_{3}(r_{3})$ is the incircle of $\Delta CMN.$

sangaku in square

Prove that $r_{2}:r_{3}=3:2.$


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$ABCD$ is a square and the semicircle on $AB$ as diameter, center $O_1,$ lies within it. $CG,$ where $G$ lies on $AD,$ touches this semicircle, and $O_{2}(r_{2})$ is the incircle of $\Delta CDG.$ An external common tangent of the semicircle and $O_{2}(r_{2})$ meets $CD$ in $R,$ $BC$ in $N,$ and intersects $CG$ in $M.$ The circle $O_{3}(r_{3})$ is the incircle of $\Delta CMN.$

sangaku in square

Prove that $r_{2}:r_{3}=3:2.$

Since the configuration contains several Egyptian triangles, it is worth while to first establish the following


The inradius $r$ of an Egyptian triangle with side lengths $3w,$ $4w,$ $5w,$ equals $w.$

The proof is straightforward. As we know, in a right triangle with legs $a$ and $b$ and hypotenuse $c,$ the inradius equals $(a+b-c)/2.$ Thus, for the Egyptian triangle, we have $r=(3w+4w-5w)/2=w.$

Now, $\Delta CDG$ is an Egyptian triangle (this has been established earlier $(\Delta ABH,$ there.) There is a direct derivation. Assume for convenience that the side of the square equals $12.$

sangaku in square, solution, step 1

Let $K$ be the point of tangency of $CG$ and the semicircle. Then $CK=CB=12.$ Let $AG=GK=x.$ In $\Delta CDG,$ $CG^{2}=CD^{2}+DG^{2},$ so that $(12+x)^{2}=12^{2}+(12-x)^{2},$ from which $x=3.$ This exactly means that the sides of $\Delta CDG$ equal $9,$ $12,$ and $15,$ making it an Egyptian triangle. By Lemma, $r_{2}=3.$

Of course, $r_{1}=6.$ Let $EL$ be the midline of square $ABCD$ as shown.

sangaku in square, solution, step 2

Then $EL$ is tangent to both the semicircle and $(O_{2}).$ Let $U$ and $T$ be respective points of tangency. If $J$ the point of tangency of $CG$ and $(O_{2})$ then both $JK$ and $TU$ are internal common tangents of the semicircle and $(O_{2}),$ implying $JK=TU=3.$

Let $KH'$ be a second tangent from $K$ to $(O_{2}).$ Then $H'K=JK$ so that $KJO_{2}H'$ is a rhombus and, since $JK\perp JO_{2},$ it is a square. Thus $H'K\perp KJ$ (i.e., $CG)$ and $K$ lies on $H'O_{1}.$ It follows that $CG\perp H'O_{1}$ and that $H'O_{1}$ is tangent to $(O)_{2})$ at $H'.$

Let $H$ and $V$ on $NR$ be the points of tangency of $NR$ with $(O_{2})$ and $(O_{1}),$ respectively. $HV$ is parallel (and equal $H'O_{1})$ and so $HV\perp CG.$ In other words, $CG\perp RN.$ Thus triangles $CNR,$ $CMR,$ and $CMN$ are all similar to $\Delta CDG$ and, hence, are all Egyptian. As the final step we need to evaluate the "size" of $\Delta CMN.$

We have already found that $GK=AG=3.$ Also $JK=JM=3.$ Thus $GM=9,$ whereas $CG=15,$ giving $CM=6.$ From here the inradius $r_2$ of $\Delta CMN$ equals $2,$ and this solves the problem.


The configuration has several additional properties:

  1. Let $F$ be the intersection of $EL$ and $NR.$ Then $C,O_{3},F,O_{1}$ are collinear. $FL=\frac{1}{4}EL.$

    sangaku in square, solution, extra 1

  2. $FO_{1}GO_{2}$ is a square.

  3. $CO_{1}O_{2}$ is a right isosceles triangle.

  4. $G, O_{2},R$ are collinear and $DR=\frac{3}{8}CD.$

  5. $S$ the intersection of $O_{2}F$ with $CG$ is the midpoint of $O_{2}F$ and $CG.$

  6. $D,M,L$ are collinear.

  7. $BN=\frac{1}{6}BC.$


  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 © 1996-2017 Alexander Bogomolny


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