# 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 $\Delta ABC$ is such that the incircles of triangles $ACD$ and $ABD$ have equal radii. Find the length of $AD$ in terms of the sides.

### References

H. Fukagawa, D. Pedoe,

*Japanese Temple Geometry Problems*, The Charles Babbage Research Center, Winnipeg, 1989Write to:

Charles Babbage Research Center

P.O. Box 272, St. Norbert Postal Station

Winnipeg, MB

Canada R3V 1L6

## 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 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

The point $D$ on the side $BC$ of $\Delta ABC$ is such that the incircles of triangles $ACD$ 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 $\Delta ABC$ and $s$ its semiperimeter, $s = (a + b + c)/2.$ $\Delta ABD$ and $\Delta ACD$ have semiperimeters $s_{1}$ and $s_{2},$ respectively, but the same inradius $k.$ Using $x$ for $AD,$ observe that $s_{1} + s_{2} = s + x$ (we will use this fact twice). Adding areas, $rs = ks_{1} + ks_{2}.$ Hence $x = (rs/k) - s.$ By similar triangles,

(1)

$\displaystyle\frac{s-b}{s_{1}-x}=\frac{s-c}{s_{2}-x}.$

Hence for $x,$ we have two equations:

$\displaystyle\frac{s(s-b)}{s_{1}-x}=x=\frac{s(s-c)}{s_{2}-x}-s.$

Expand each equation, add, and solve for $x$:

$\begin{align} s(s - b) - ss_{1} + sx &= xs_{1} - x^{2};\\ s(s - c) - ss_{2} + sx &= xs_{2} - x^{2};\\ s(2s - b - c) - s(s_{1} + s_{2}) + 2sx &= x(s_{1} + s_{2}) - 2x^{2};\\ sa - s(s + x) + 2sx &= x(s + x) - 2x^{2};\\ 2x^{2} &= (s + x)^{2} - 2sx - sa\\ x^{2} &= s^{2} - sa. \end{align}$

It follows that

(2)

$x = \sqrt{s(s - a)}.$

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 $ACD$ 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 $x = \sqrt{bc/2}.$ Finally,

$\displaystyle k = \frac{bc}{a + b + c + \sqrt{2bc}}.$

(P. Yiu has observed (Missouri J. Math. Sci., 15 (2003) 21-32) 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 $A.\,$ There are two more solutions, of which one exploits P. Yiu's observation.)

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

Copyright © 1996-2018 Alexander Bogomolny

67585382