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# Japanese theorem of cyclic polygons.
Take a convex **cyclic polygon**, that is, a convex polygon whose vertices all lie on a circle. Now take any two triangulations, and draw in all the incircles for each small triangle in the triangulations, like so:
![[0 inbox/---files/---Japanese theorem of cyclic polygons 2023-05-14 10.44.27.excalidraw.svg]]
%%[[0 inbox/---files/---Japanese theorem of cyclic polygons 2023-05-14 10.44.27.excalidraw.md|🖋 Edit in Excalidraw]], and the [[0 inbox/---files/---Japanese theorem of cyclic polygons 2023-05-14 10.44.27.excalidraw.dark.svg|dark exported image]]%%
Then we have the following curious theorem, sometimes called the Japanese theorem of cyclic polygons:
>**Theorem.**
>The sum of the radii of the incircles of the triangles in any triangulation of a fixed convex cyclic polygon is constant.
That is, in above diagram, the sum of the radii of the circles on the left is the same as the sum of the radii of the circles on the right!
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Why is this true? Suppose this is true for **convex cyclic quadrilaterals**, that the sum of the radii of the two circles is the same in the two triangulations, then by the fact that [[0 inbox/Flip-graphs-of-convex-polygons|any two triangulations can be obtained from one another by flipping the diagonals]], we see that these flips preserve the sum of the radii. So any two triangulation (of the same convex cyclic polygon) have the same radii sum.
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So it remains to show it is true for convex cyclic quadrilaterals.[^cite]
Let us take a convex cyclic quadrilateral and consider both triangulations and their incircles. Label the segments as follows:
![[0 inbox/---files/---Japanese theorem of cyclic polygons 2023-05-14 19.42.06.excalidraw.svg]]
%%[[0 inbox/---files/---Japanese theorem of cyclic polygons 2023-05-14 19.42.06.excalidraw.md|🖋 Edit in Excalidraw]], and the [[0 inbox/---files/---Japanese theorem of cyclic polygons 2023-05-14 19.42.06.excalidraw.dark.svg|dark exported image]]%%
We will show the middle red-blue segment is the same in both triangulations. To do this, consider the quantity $T+B - L - R$ in each triangulation:
![[0 inbox/---files/---Japanese theorem of cyclic polygons 2023-05-14 20.05.39.excalidraw.svg]]
%%[[0 inbox/---files/---Japanese theorem of cyclic polygons 2023-05-14 20.05.39.excalidraw.md|🖋 Edit in Excalidraw]], and the [[0 inbox/---files/---Japanese theorem of cyclic polygons 2023-05-14 20.05.39.excalidraw.dark.svg|dark exported image]]%%
For the first triangulation, we get
$$
\begin{align*}
&T+B-L-R \\
&= (a+b+d+e)-(b+c+e+f)\\
&=(a+d)-(c+f)\\
&=(a-f)+(d-c)
\end{align*}$$
which is twice the red-blue segment for this triangulation.
For the second triangulation, we get
$$
\begin{align*}
&T+B-L-R \\
&= (p+q+s+t) - (r+s+u+p)\\
&=(q+t) - (r+u)\\
&=(q-r)+(t-u)
\end{align*}$$
which is twice the red-blue segment for the second triangulation.
Hence, the red-blue segments in both triangulation have the same length.
Next, we label the angles. Note the angles that share the same arcs will be the same. And we label the radii of the incircles as $r_1,r_2,r_3,r_4$:
![[0 inbox/---files/---Japanese theorem of cyclic polygons 2023-05-14 20.17.15.excalidraw.svg]]
%%[[0 inbox/---files/---Japanese theorem of cyclic polygons 2023-05-14 20.17.15.excalidraw.md|🖋 Edit in Excalidraw]], and the [[0 inbox/---files/---Japanese theorem of cyclic polygons 2023-05-14 20.17.15.excalidraw.dark.svg|dark exported image]]%%
Now using the lengths and angles defined, we have
$$
\begin{align*}
\tan\alpha &= \frac{r_{1}}{c} = \frac{r_{4}}{u}\\
\tan\beta &= \frac{r_{2}}{d} = \frac{r_{4}}{q}\\
\tan\gamma &= \frac{r_{2}}{f} = \frac{r_{3}}{r}\\
\tan\delta &= \frac{r_{1}}{a} = \frac{r_{3}}{t}
\end{align*}$$
So cross multiplying yields
$$
\begin{align*}
r_{1}u - r_{4}c=0\\
r_{2}q-r_{4}d=0\\
r_{2}r-r_{3}f=0\\
r_{1}t-r_{3}a=0
\end{align*}$$
Summing them alternately gives
$$
(r_{1}u - r_{4}c)-(r_{2}q-r_{4}d)+(r_{2}r-r_{3}f)-(r_{1}t-r_{3}a)=0$$
Or,
$$
r_{1}(u-t) + r_{4}(d-c) + r_{2}(r-q) + r_{3}(a-f)=0$$
Or,
$$
r_{4}(d-c) + r_{3}(a-f)=r_{1}(t-u) + r_{2}(q-r)$$
And since $d-c=a-f=t-u=q-r$, we have finally
$$
r_{1}+r_{2}=r_{3}+r_{4}$$as claimed!
---
[^cite]: Here we follow the approach by Mangho Ahuja, Wataru Uegaki, and Kayo Matsushita.
#geometry