[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.

Solution

References

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

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 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. Circles and Semicircles in Rectangle
21. Circles in a Circular Segment
22. Circles Lined on the Legs of a Right Triangle
23. Equal Incircles Theorem
24. Equilateral Triangle, Straight Line and Tangent Circles
25. Equilateral Triangles and Incircles in a Square
26. Five Incircles in a Square
27. Four Hinged Squares
28. Four Incircles in Equilateral Triangle
29. Gion Shrine Problem
30. Harmonic Mean Sangaku
31. Heron's Problem
32. In the Wasan Spirit
33. Incenters in Cyclic Quadrilateral
34. Japanese Art and Mathematics
35. Malfatti's Problem
36. Maximal Properties of the Pythagorean Relation
37. Neuberg Sangaku
38. Out of Pentagon Sangaku
39. Peacock Tail Sangaku
40. Pentagon Proportions Sangaku
41. Pythagoras and Vecten Break Japan's Isolation
42. Radius of a Circle by Paper Folding
43. Review of Sacred Mathematics
44. Sangaku à la V. Thebault
45. Sangaku and The Egyptian Triangle
46. Sangaku in a Square
47. Sangaku Iterations, Is it Wasan?
48. Sangaku with 8 Circles
49. Sangaku with Quadratic Optimization
50. Sangaku with Three Mixtilinear Circles
51. Sangaku with Versines
52. Sangakus with a Mixtilinear Circle
53. Sequences of Touching Circles
54. Square and Circle in a Gothic Cupola
55. Tangent Circles and an Isosceles Triangle
56. The Squinting Eyes Theorem
57. Steiner's Sangaku
58. Three Incircles In a Right Triangle
59. Three Squares and Two Ellipses
60. Three Tangent Circles Sangaku
61. Triangles, Squares and Areas from Temple Geometry
62. Two Arbelos, Two Chains
63. Two Circles in an Angle
64. Two Sangaku with Equal Incircles

Copyright © 1996-2010 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,

y b  =  (a - x) a

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:   lim h → 0 f(x + h) - f(x) h 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.

Copyright © 1996-2010 Alexander Bogomolny