Sangaku with Quadratic Optimization

[Fukagawa and Rothman, p. 117, problem 42] refer to Fujita Kagen's book Zoku Shinpeki Sanpo for the following problem

Take a point on the hypotenuse AB of the right triangle ABC. From that point draw perpendiculars to the legs of the triangles. The perpendiculars form a rectangle. Find the location of the point on the hypotenuse that maximizes the area of the rectangle.

According to Fukagawa and Rothman, the problem was originally proposed in 1806 by Hotta Sensuke, a student of Fujita school, and written on a tablet hung in the Gikyosha shrine of Niikappugun, Hokkaido.



  1. H. Fukagawa, A. Rothman, Sacred Mathematics: Japanese Temple Geometry, Princeton University Press, 2008


  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

Using the notations from the diagram, the task is to minimize the are S = xy.

The configuration features several pairs of similar triangles that may be used to derive a relationship between x and y. In particular,


(a - x)


from which S = xy = (b/a)(ax - x²) and the problem reduces to finding maximum of the quadratic function f(x) = ax - x². As the leading coefficient of the function is -1, the graph of the function is a parabola with its "horns" extended downwards, thus insuring existence of a maximum.

I quote from [Fukagawa and Rothman, p. 139]:

Here is a traditional solution from the manuscript Solutions to Problems of Zoku Shinpeki Sanpo by Kitagawa Moko (1763-1833). In this case the traditional solution is pretty much what any calculus student would do (but see chapter 9.)

... Setting the derivative df/dx to 0 gives x = a/2 and, consequently, y = b/2.

As I am sure most of the calculus students will indeed find the maximum of function f(x) = ax - x² by equating the derivative to 0, I think it is worthwhile to quote from chapter 9 as mentioned by Fukagawa and Rothman above:

The question of differentiation in some sense is more difficult than that of integration because, although traditional Japanese mathematicians wrote volumes on integration techniques, they were virtually silent about how they took derivatives. One thing is certain: the Japanese did not have the concept of differentiation as we know it. In no traditional manuscript do we the fundamental formula for the derivative:

h → 0
f(x + h) - f(x)


Without this concept it is difficult or impossible to develop a formal theory of differentiation. Perhaps for this reason in the wasan differentiation was confined to finding the maximum and minimum of functions.

How did the traditional geometers do this? For quadratic equations, parabolas, one knows without calculus that the maximum or minimum is at the vertex of the parabola, and so for functions of the form y = ax² + bx + c, one can write down the answer xmin/max = -b/2a and ymin/max = y(xmin/max). By induction, one can do this for a few polynomials ...

My interpretation of the above is that, faced with the task of finding the maximum of the function obtained in the sangaku at hand, the traditional Japanese mathematicians would not, perhaps for the wrong reasons, calculate and subsequently equate to zero the derivative of quadratic functions. Unlike a majority of the present day calculus students, they would simply point out to a solution which is especially easy to surmise (and remember) for the function f(x) = ax - x² = x (a - x). Indeed, the parabola has symmetry around its axis which is known to pass through its vertex. The maximum (minimum) is, therefore, midway between the roots of the parabola which, in the case of the quadratic function f(x) = x (a - x), are obviously x = 0 and x = a, giving the maximum at x = (0 + a)/2 = a/2. Almost as easily one can obtain the same result by completing the square: ax - x² = -(x - a/2)² + a²/4.

Related material

A Sample of Optimization Problems II

  • Mathematicians Like to Optimize
  • Building a Bridge
  • Building Bridges
  • Optimization Problem in Acute Angle
  • Geometric Optimization from the Asian Pacific Mathematical Olympiad
  • Cassini's Ovals and Geometric Optimization
  • Heron's Problem
  • Optimization in Parallelepiped
  • Matrices and Determinants as Optimization Tools: an Example
  • An Inequality between AM, QM and GM
  • |Contact| |Front page| |Contents| |Geometry| |Up|

    Copyright © 1996-2018 Alexander Bogomolny