# A Trigonometric Solution to a Difficult Sangaku Problem

### Michel Cabart

Feb. 25, 2010

PROBLEM. In ΔABC, AB = BC. If one chooses D on AB and J on CD such that

SOLUTION

Let's say ∠ADC = x, AD = 1 and tan(x/2) = t. We calculate ρ, the radius of the incircle of ΔADJ by a known formula for the radius of the incircle:

ρ = AD·tan(∠DAJ)·tan(∠ADJ) / (tan(∠DAJ) + tan(∠ADJ)),

with: tan(∠ADJ) = t, tan(∠DAJ) = tan(45° - x/2) = (1 - t)/(1 + t). So that

ρ = t(1 - t)/(t² + 1) .

Similarly r2 = CD·tan(∠CDO_{2})·tan(∠DCO_{2}) / (tan(∠CDO_{2}) + tan(∠DCO_{2})),

with CD = 2 cos(x) = 2 (1 - t²)/(1 + t²). Further,

tan(∠CDO_{2} = tan(90° - x/2) = 1/t,

tan DCO_{2} = tan (90° - 3x/2) = (1 - 3t^{2})/(3t - t^{3}),

so that r_{2} = ½(1 - 3t²)/t(1 + t²).

Equating ρ and r_{2} yields the equation:

2t^{3} - 5t^{2} + 1 = (2t - 1)·(t^{2} - 2t - 1) = 0.

The second (quadratic) factor cannot be zero as it would lead to

The solution is then t = 1/2, leading to ρ = r_{2} = 1/5,

## Sangaku

- Sangaku: Reflections on the Phenomenon
- Critique of My View and a Response
- 1 + 27 = 12 + 16 Sangaku
- 3-4-5 Triangle by a Kid
- 7 = 2 + 5 Sangaku
- A 49
^{th}Degree Challenge - A Geometric Mean Sangaku
- A Hard but Important Sangaku
- A Restored Sangaku Problem
- A better solution to a difficult sangaku problem
- A Simple Solution to a Difficult Sangaku Problem
- A Trigonometric Solution to a Difficult Sangaku Problem

- A Sangaku: Two Unrelated Circles
- A Sangaku by a Teen
- A Sangaku Follow-Up on an Archimedes' Lemma
- A Sangaku with an Egyptian Attachment
- A Sangaku with Many Circles and Some
- A Sushi Morsel
- An Old Japanese Theorem
- Archimedes Twins in the Edo Period
- Arithmetic Mean Sangaku
- Bottema Shatters Japan's Seclusion
- Chain of Circles on a Chord
- Circles and Semicircles in Rectangle
- Circles in a Circular Segment
- Circles Lined on the Legs of a Right Triangle
- Equal Incircles Theorem
- Equilateral Triangle, Straight Line and Tangent Circles
- Equilateral Triangles and Incircles in a Square
- Five Incircles in a Square
- Four Hinged Squares
- Four Incircles in Equilateral Triangle
- Gion Shrine Problem
- Harmonic Mean Sangaku
- Heron's Problem
- In the Wasan Spirit
- Incenters in Cyclic Quadrilateral
- Japanese Art and Mathematics
- Malfatti's Problem
- Maximal Properties of the Pythagorean Relation
- Neuberg Sangaku
- Out of Pentagon Sangaku
- Peacock Tail Sangaku
- Pentagon Proportions Sangaku
- Proportions in Square
- Pythagoras and Vecten Break Japan's Isolation
- Radius of a Circle by Paper Folding
- Review of Sacred Mathematics
- Sangaku à la V. Thebault
- Sangaku and The Egyptian Triangle
- Sangaku in a Square
- Sangaku Iterations, Is it Wasan?
- Sangaku with 8 Circles
- Sangaku with Angle between a Tangent and a Chord
- Sangaku with Quadratic Optimization
- Sangaku with Three Mixtilinear Circles
- Sangaku with Versines
- Sangakus with a Mixtilinear Circle
- Sequences of Touching Circles
- Square and Circle in a Gothic Cupola
- Steiner's Sangaku
- Tangent Circles and an Isosceles Triangle
- The Squinting Eyes Theorem
- Three Incircles In a Right Triangle
- Three Squares and Two Ellipses
- Three Tangent Circles Sangaku
- Triangles, Squares and Areas from Temple Geometry
- Two Arbelos, Two Chains
- Two Circles in an Angle
- Two Sangaku with Equal Incircles
- Another Sangaku in Square
- Sangaku via Peru
- FJG Capitan's Sangaku

|Up| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

68251210