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).

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.

Solutions

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 49^th Degree Challenge
   7. A Geometric Mean Sangaku
   8. A Hard but Important Sangaku
   9. A Restored Sangaku Problem
      • A better solution to a difficult sangaku problem
      • A Simple Solution to a Difficult Sangaku Problem
      • A Trigonometric Solution to a Difficult 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
       • Equal Incircle Theorem, Angela Drei's Proof
   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
       • Four Incircles in an Equilateral Triangle, a Sangaku
   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
       • An Extension of a Sangaku with Touching Circles
   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.

   Remark

   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