# The Japanese Theorem for Nonconvex Polygons - Carnot's Theorem for Cyclic Polygons

Author(s):
David Richeson

### Carnot's Theorem for Cyclic Polygons

In order to prove the Japanese theorem we need to generalize Carnot's theorem to cyclic polygons.

Carnot's theorem for cyclic polygons. Suppose $P$ is a cyclic $n$-gon triangulated by diagonals. Let $R$ be the circumradius of $P$, let $d_1, d_2, \cdots, d_n$ be the signed distances from the sides of $P$ to the circumcenter, and let $r_1, r_2, \cdots, r_{n-2}$ be the inradii of the triangles in the triangulation. Then

$(n-2)R + \sum_{k=1}^{n-2} r_k = \sum_{i=1}^n d_i .$

Proof. This is a proof by induction. The base case, $n=3$, is simply Carnot's theorem for triangles. Now suppose the theorem holds for some $n \geq 3$. Let $P$ be a cyclic $(n+1)$-gon that is triangulated by diagonals. Because the triangulation has $n - 1$ triangles, the pigeonhole principle tells us that there is a triangle that has two edges in common with $P$. Without loss of generality, the signed distances to these two shared edges are $d_n$ and $d_{n+1}$, and the inradius of this triangle is $r_{n-1}$. Remove this triangle to obtain a cyclic $n$-gon $P'$ and let $d_n^{\prime}$ be the signed distance to the new edge. By the induction hypothesis

$\sum_{k=1}^{n-2} r_k = (2-n)R + d_n^{\prime} + \sum_{i=1}^{n-1} d_i .$

Now consider the removed triangle. The key observation is that the signed distances from the circumcenter to the sides of the triangle are $-d_n^{\prime}$, $d_n$, and $d_{n+1}$. (For example, in Figure 6(a) the dashed edge is positive when viewed as a side of triangle $p_1p_4p_5$, but negative when viewed as a side of polygon $p_1p_2p_3p_4$, and similarly in Figure 6(b) the dashed edge is negative when viewed as a side of triangle $p_1p_4p_5$, but positive when viewed as a side of polygon $p_1p_2p_3p_4 .$)

Figure 6

By Carnot's Theorem for triangles $r_{n-1} + R = -d_n^{\prime} +d_n +d_{n+1}$. Consequently,

$\sum_{k=1}^n r_k = (\sum_{k=1}^{n-2} r_k) + r_{n-1}$

$= ((2-n)R + d_n^{\prime} + \sum_{i=1}^{n-1} d_i) + (-R + (-d_n^{\prime} + d_n + d_{n+1}))$

$= (2-(n+1))R + \sum_{i=1}^{n+1} d_i .$

as was to be shown.∎

The Japanese theorem follows immediately from this version of Carnot's theorem.

Proof of the Japanese Theorem. Let $P$ be an $n$-gon inscribed in a circle of radius $R$ . Carnot's theorem for cyclic polygons says that for any triangulation of $P$ by diagonals

$r_P = \sum_{k=1}^{n-2} r_k = (2-n)R + \sum_{i=1}^n d_i ,$

where $r_k$ and $d_i$ are defined as above. But $R, n$ and the $d_i$ do not depend on the choice of triangulation, so the total inradius, $r_P$, does not either.∎

These results provide an interesting way of seeing what $r_P$ measures. Rearranging terms we obtain

$r_P = 2R - \sum_{k=1}^n (R - d_k) .$

The quantity $R - d_k$ is the amount that the perpendicular drawn to the $k$th side of $P$ differs from the radius of the circumcircle. So $r_P$ is the diameter of the circle minus the sum of these differences. Thus, larger values of $r_P$ correspond to polygons that more closely approximate the circle. Later we show that by increasing the number of sides we can find a polygon with a total inradius as close to $2R$ as we please.

David Richeson, "The Japanese Theorem for Nonconvex Polygons - Carnot's Theorem for Cyclic Polygons," Convergence (December 2013)