# 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.

Solution

### References

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

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## Sangaku

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.)