Four Incircles in an Equilateral Triangle, a Sangaku

Elsewhere we described and solved the following sangaku:

Angela Drei's proof of the four incircles sangaku

In an equilateral triangle \(ABC\) the lines \(AB'C'\), \(BC'A'\), \(CA'B'\) are drawn making equal angles with \(AB\), \(BC\), \(CA\), respectively, forming the triangle \(A'B'C'\), and so that the radius of the incircle of triangle \(A'B'C'\) is equal to the radius of the incircle of triangle \(AC'B\). Find \(A'B'\) in terms of \(AB\).

Angela Drei, an Italian math teacher, has supplied (August 4, 2012) an additional proof that also confirms an observation about three pairs of parallel lines made on the original page.

Angela's proof also proceeds in steps.

Proposition 1

The triangle \(O_{1}O_{2}O_{3}\) is equilateral.

Proof: \(\Delta ABC'\) is obtained from \(ACB'\) by a \(120^{\circ}\) counter-clockwise rotation about the center of \(ABC\) and similarly \(CA'B\) is obtained by \(ABC'\), \(O_{1}\) is the center of the incircle of triangle \(AB'C\), by the previous rotation we obtain the centers \(O_{2}\) and \(O_{3}\), then the triangle \(O_{1}O_{2}O_{3}\) is equilateral.

Proposition 2

\(L\) (point of tangency) \(\in O_{1}O_{2}\). (Similarly, \(R\in O_{1}O_{3}\), where \(R\) is another point of tangency.)

Proof: \(OT\) and \(O_{3}M\) are perpendicular to \(CB'\), \(OO_{3}\) and \(MT\) are parallel because the distances \(OT\) and \(O_{3}M\) are equal, so \(\angle MKO_{3}=\angle KO_{3}O=30^{\circ}\), \(L\) is the point of intersection of \(MT\) and \(O_{1}O_{2}\), but \(O_{3}O\) is perpendicular to \(O_{1}O_{2}\) then \(ML\) is perpendicular to \(O_{1}O_{2}\), showing that \(L\) is the point of tangency.

Proposition 3

The side of \(O_{1}O_{2}O_{3}\) is four times \(r\), the radius of the incircles.

Proof: in right triangles \(O_{1}LK\) and \(O_{3}MK\), \(\angle MKO_{3}=\angle O_{1}KL\), as vertically opposite, and \(O_{3}M=O1L=r\), so that the triangles are congruent and \(K\) is the midpoint of \(O_{1}O_{3}\), \(\angle O_{1}KL=30^{\circ}\) as was already shown, and \(\angle KO_{1}L=\angle MO_{3}K=60^{\circ}\). Further, \(\angle A'O_{3}M=30^{\circ}\); triangles \(A'RK\) and \(A'RO_{3}\) are congruent. Indeed, they have \(A'R\) as common side, and \(A'R\) is perpendicular to \(KO_{3}\), making \(O_{1}O_{3}=4r\).

Now let \(A'B'=l\); \(l=A'K+KL+LB'\) and, as a consequence of the previous claims, we deduce \(A'K=\frac{2}{\sqrt{3}}KR=\frac{2}{3}\sqrt{3}r\), \(LB'=\frac{1}{2}\left(\frac{2}{\sqrt{3}}r\right)=\frac{1}{3}\sqrt{3}r\), and \(KL=\frac{\sqrt{3}}{2}2r=\sqrt{3}r\), so that \(l=2\sqrt{3}r\).

We consider the areas of the four triangles that form \(\triangle ABC\); the sum of their areas equals that of \(\triangle ABC\), but each can be calculated separately. The area of \(\triangle AB'C\) is \(xr+\frac{\sqrt{3}}{3}r^{2}\), where \(x\) is the side of \(\triangle ABC\). The area of \(\triangle A'B'C'\) is \(\frac{3}{2}lr\) and also \(l^{2}\frac{\sqrt{3}}{4}\), making \(r=l\frac{\sqrt{3}}{6}\). The area of \(\triangle ABC=x^{2}\frac{\sqrt{3}}{4}\). Summing up leads to the equation

\(3xr + \sqrt{3}r^{2} + \frac{3}{2}lr=\frac{x^{2}\sqrt{3}}{4}.\)

Replacing \(r=l\frac{\sqrt3}}{6} we obtain

\(3xl\frac{\sqrt{3}}{6} + l^{2}\frac{\sqrt{3}}{12} + \frac{\sqrt{3}}{4}l^{}=\frac{x^{2}\sqrt{3}}{4}.\)

Or, after simplification \(6xl+4l^{2}-3x^{2}=0\), with the only positive root \(l=x\frac{\sqrt{21}-3}{4}\).

(There is another proof by George Zettler and a third one by J. John Samuel.)

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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 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
    • 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
  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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