Proportions in Square

A 1835 Sangaku from the Miyagi prefecture {Fukagawa & Pedoe, 3.1.5] reached us with no solution. Two solutions have been posted by J. Marshall Unger from Ohio State University in his write-up of sangaku problems, (Problem #23).

Given square ABCD, with M, N the midpoints of AB, CD, inscribe a circle in kite QRSM and in triangle APQ.

Prove that the radius of the smaller circle is half that of the larger circle.

I find both solutions enticingly elegant and sufficiently simple to deserve inclusion in my collection. The problem also links to a separate problem of dividing a segment into equal parts which served as an extra inducement for putting up the current page.



  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

    Given square ABCD, with M, N the midpoints of AB, CD, inscribe a circle in kite QRSM and in triangle APQ.

    Prove that the radius of the smaller circle is half that of the larger circle.

    Solution 1

    Observe that, for any inscriptible shape, the inradius could be found as the ratio of the area and the semiperimeter. We shall juxtapose ΔAQP and the kite MQRS.

    Point P is the intersection of the diagonals in rectangle ACNM; hence P is exactly as far from AM as R which is the center of the square ABCD. It follows that triangles AMP and AMR have the same area and so are triangles AQP and MQR obtained from the former by removing the common part - ΔAMQ. We conclude that the area of the kite MQRS is twice that of ΔAQP.

    Triangles MQR and CQA are similar and, since 2·MR = AC,

    MQ + QR + RS + MS = 2(MQ + QR) = CQ + AQ.

    But, by symmetry, CQ = AT = AP + PT, while clearly PT = PQ. Therefore,

    CQ + AQ = AT + AQ = AP + PQ + AQ.

    A combination of the two identities shows that ΔAQP and the kite MQRS have equal perimeters and, hence, semiperimeters, thus proving the claim.

    Solution 2

    Note that the incircle of the kite is also the incircle of ΔDQM. Further, AN||DM, implying similarity of triangles DQM and AQP. Since DM is twice AP, the triangles are similar with the coefficient of 2, so that their inradii are in the same proportion.


    This is a well known fact (that arises in the problem of division of a segment into equal parts) that 2·AQ = QR. Thus also 2·MQ = CQ. Now, since CP = MP, 2·PQ = QM. This remark could be employed in both of the solutions above.

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

    Copyright © 1996-2018 Alexander Bogomolny