A better solution to a difficult sangaku problem

J. Marshall Unger
Department of East Asian Languages & Literatures
The Ohio State University

PROBLEM. In ΔABC, AB = BC. If one chooses D on AB and J on CD such that AJ ⊥ CD and the incircles of ΔACJ, ΔADJ, and ΔBCD all have radius r, then r = AJ/4.

SOLUTION

The trick to solving the problem expeditiously is to clarify what is given and to prove the converse proposition first.

Notice that the only connection between AB = BC and the geometry of ΔACD is that the incircle of ΔABC (dashes) is tangent to AC at its midpoint. The two incircles of ΔACJ and ΔADJ cannot be tangent to the shared side AJ at the same point, as shown, unless J is the point of tangency of the incircle of ΔACD (dots) with side CD [Honsberger, see also Three Incircles in a Triangle]. Hence AC = AD. And the two incircles cannot be congruent unless, in addition, either ∠CAJ = ∠DAJ or AJ ⊥ CD, in which case, ΔACJ = ΔADJ and the other condition follows. Therefore, given a fixed isosceles ABC, if the congruence of all three small incircles implies r = AJ/4, as the problem asserts, then the converse must also be true. Indeed, it is easier to prove.

In the figure above, let a, b, d, and h be the lengths of CD, AD = AC, DN, and AJ, respectively. Let r₁ and r₂ be the radii of the incircles of ΔACD and ΔBCD, respectively. We use two lemmas, which we will prove later:

Given h = 4r, square the equation for the inradius of a right triangle: b = h + a/2 - 2r = 2r + a/2 to get b² = 4r² + 2ar + a²/4, and equate this with the Pythagorean result b² = (a/2)² + h² = a²/4 + 16r². This yields a = 6r, which, with h = 4r, implies b = 5r. Hence, by Lemma 1, 2d = r.

Now area(ΔACD) = ha/2 = 12r² = r₁s, where s is the semiperimter of ΔACD or 8r. Thus r₁ = 3r/2 = 3d. Using Lemma 2, we find r₂ = a/6 = r.

From this, it is clear that solving the original problem comes down to proving, without knowing the value of h, that ΔACJ and ΔADJ are 3:4:5 right triangles.

It's easy to show that ΔDTU is similar to ΔDO₂U. Hence DU is the mean proportional between DT and O₂U = DT + DN. In terms of lengths, we therefore have

(Lemma 2 takes the form 2r₁r = ad since we are assuming r₂ = r.) Since JS = r, O₁S = d. Thus JO₁ = r + d = 3d = r₁.

Hence, by Lemma 2, 2rr₁ = 2(2d)(3d) = 12d² = ad, so a = 12d = 6r. Since a - b = 2d (Lemma 1), b = 10d = 5r. And because ADJ is a right triangle, a/2 = 3r and b = 5r imply h = 4r.

The lemmas follow from the general case of any triangle ABC with incircle I. Draw a cevian CD and the incircles O₁ and O₂ of the resulting triangles ACD and BCD. Label the points of tangency of these incircles as shown. Then EH = GK = DN.

PROOF: By equal tangents, CE = CF (1), and

Subtracting the equations in second row from those in the first, we get

Equating the left and right sides of (4) and (5),

Linking equations (2) and (6), EH = GK = DN.

Many more inferences can be drawn from this figure, but we need just two.

Lemma 1 follows immediately if E and J are the midpoints of AC and CD, respectively. Then CD/2 - AC/2 = EH = DN. In the problem, as explained earlier, J must be the midpoint of CD. Requiring that ABC be isosceles forces E to be the midpoint of AC.

Lemma 2 in the general case is r₁r₂ = DJ·DL. The dashed lines in the figure show that

(r1 + r2)² + (DJ - DL)² = (O1O2)² = (r1 - r2)² + KN²,

but KN = DK + DN = DJ + DL by equal tangents. The rest is just algebra.

In introducing the solution of Kitagawa Moko (1764 -1839), Fukagawa and Rothman remark, "It is somewhat complicated and we hope that readers can find a simpler one." I do not know whether the solution presented above is what Fukagawa and Rothman had in mind, but Kitagawa's solution is certainly complicated. It proves Lemma 1 relying in part on the given that ΔABC is isosceles; finds expressions for r₁, a, and d in terms of r and h; plugs these directly into Lemma 2; produces a third-degree equation in r and h through more painstaking algebra; rejects the roots h = ± r√2 because they are inconsistent with the figure; and so concludes that h = 4r. In addition, Kitagawa's proof gives no hint that the lemmas hold even for scalene ABC, the true significance of requiring that ABC be isosceles, that the converse is true, or that 3:4:5 right triangles play a role.

References

1. H. Fukagawa, A. Rothman, Sacred Mathematics: Japanese Temple Geometry, Princeton University Press, 2008
2. R. Honsberger, Episodes in nineteenth and twentieth-century Euclidean geometry, MAA, 1995

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
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 Three Mixtilinear Circles
50. Sangaku with Versines
51. Sangakus with a Mixtilinear Circle
52. Sequences of Touching Circles
53. Square and Circle in a Gothic Cupola
54. Tangent Circles and an Isosceles Triangle
55. The Squinting Eyes Theorem
56. Steiner's Sangaku
57. Three Incircles In a Right Triangle
58. Three Squares and Two Ellipses
59. Three Tangent Circles Sangaku
60. Triangles, Squares and Areas from Temple Geometry
61. Two Arbelos, Two Chains
62. Two Circles in an Angle
63. Two Sangaku with Equal Incircles

Copyright © 1996-2009 Alexander Bogomolny