Two Sangaku with Equal Incircles

Here is a 1897 sangaku from Chiba perfecture. The problem also appeared earlier in a printed form in a 1781 book.

The point D on the side BC of triangle ABC is such that the incircles of triangles ADC and ABD have equal radii. Find the length of AD in terms of the sides.

Solution

References

1. H. Fukagawa, D. Pedoe, Japanese Temple Geometry Problems, The Charles Babbage Research Center, Winnipeg, 1989

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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. 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
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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

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Copyright © 1996-2012 Alexander Bogomolny

The point D on the side BC of triangle ABC is such that the incircles of triangles ADC and ABD have equal radii. Find the length of AD in terms of the sides.

(This solution has been kindly sent to me by Professor J. Marshall Unger, Department of East Asian Languages and Literatures, The Ohio State University.)

In the figure above, let r be the inradius of ΔABC and s its semiperimeter, s = (a + b + c)/2. ΔABP and ΔACP have semiperimeters s₁ and s₂, respectively, but the same inradius k. Using x for AD, observe that s₁ + s₂ = s + x (we will use this fact twice). Adding areas, rs = ks₁ + ks₂. Hence x = (rs/k) - s. By similar triangles,

(1)	s - b = r =  s - c s1 - x k s2 - x

Hence for x, we have two equations:

s(s - b)	- s = x =	s(s - c)	- s
s1 - x		s2 - x

Expand each equation, add, and solve for x:

	s(s - b) - ss1 + sx	= xs1 - x²
	s(s - c) - ss2 + sx	= xs2 - x²
	s(2s - b - c) - s(s1 + s2) + 2sx	= x(s1 + s2) - 2x²
	sa - s(s + x) + 2sx	= x(s + x) - 2x²
	2x²	= (s + x)² - 2sx - sa
	x²	= s² - sa.

By the quadratic formula,

As a corollary, we now easily obtain a solution to another sangaku [Fukagawa & Pedoe, #2.2.3]:

ABC is a right-angled triangle at A, the point D lies on BC, and is such that the inradii of triangles ABD and ADC are equal. Find the common radius in terms of a, b, c.

From (1) and (2), the unknown radius k is expressible in terms of r and x. For a right triangle, r = (b + c - a) / 2 = s - a. Further, sr = Area(ABC) = bc/2, giving x = √bc/2. Finally,

k = bc / (a + b + c + √2bc).

(P. Yiu observes that if the cevian AD has, as discussed above, the property that the inradii of triangles ABD and ACD are equal, then so are the radii of their excircles opposite vertex A.)

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Copyright © 1996-2012 Alexander Bogomolny