# The Japanese Theorem for Nonconvex Polygons - A Further Generalization of Carnot's Theorem

Author(s):
David Richeson

### A Further Generalization of Carnot's Theorem

When we introduced Carnot's theorem we gave a technique for computing the signed distance of a side of a polygon to the circumcenter. We must now generalize this to oriented triangles and more generally to nonconvex polygons.

Suppose we have a cyclic polygon with vertices $p_1, \ldots, p_n .$ We now define $d_i$, the signed distance from the $i$i th side of the polygon to the center of the circle. If the side is not a diameter and is not degenerate, then it cuts off an arc of the circle shorter than a semicircle.The endpoints are neighbors in the cyclic ordering of the vertices; that is, $p_{i+1}$ follows $p_i$ for $1 \leq i < n ,$ and $p_1$ follows $p_n$. So they induce an orientation on the circle. If the orientation is counterclockwise, then let $d_i$ be the distance from the segment to the center. Otherwise, let $d_i$ be the negative of the distance. If the segment has no length, then we take $d_i = 0 .$

In Figure 11 we see three different cyclic polygons with the signs of the $d_i$ labeled. Notice that the positively oriented triangle in Figure 11(a) corresponds to the original definition.

Figure 11

Carnot's theorem for oriented triangles. Let $T$ be an oriented triangle with circumradius $R$ and inradius $r .$ Suppose $a, b,$ and $c$ are the signed distances from the circumcenter of $T$ to the sides of $T$ and that $\tilde{r}$ is the signed inradius. If $T$ is positively oriented, then $R + r = R + \tilde{r} = a + b + c .$ If $T$ is negatively oriented, then $- R - r = - R + \tilde{r} = a + b + c .$

Proof. If $T$ is positively oriented, then this is simply the usual Carnot's theorem. Suppose $T$ is negatively oriented. The signed distances to the sides are the negatives of what they would have been had the triangle been positively oriented. Thus, by Carnot's theorem $R + r = - a - b - c ,$ and by definition $\tilde{r} = - r .$∎

Generalized Carnot's theorem for cyclic polygons. Suppose $P$ is a cyclic $n$-gon that is triangulated by diagonals. Let $d_1, \ldots, d_n$ be the signed distances from the sides of $P$ to the circumcenter and let $d_1, \ldots, d_n$ be the signed inradii of the triangles in the triangulation. Suppose there are $p$ triangles that are positively oriented and $q$ that are negatively oriented. Then

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

Proof. This is a proof by induction on the number of vertices. We may assume they are all distinct; if not, . The base case, $n = 3$ is simply Carnot's theorem for oriented triangles. Now suppose the theorem holds for all $3 \leq i \leq n$ for some $n .$ Let $P$ be a cyclic $(n+1)$-gon that is triangulated by diagonals. Furthermore, suppose it has $p$ triangles that are positively oriented and $q$ that are negatively oriented. By the pigeonhole principle there is a triangle that has two edges in common with $P .$ Without loss of generality, we may assume that this triangle has vertices $p_1, p_n$ and $p_{n+1} ,$ that the signed distances to the two edges shared with $P$  are $d_n ,$ and $d_{n + 1} ,$ and that the signed inradius of this triangle is $\tilde r_{n-1} .$ Remove this triangle to obtain a cyclic $n$-gon $P^{\prime} .$ The key fact is that the sign of the signed distance to the newly created side (which has endpoints $p_1$ and $p_n$ is different if viewed as an edge of $P^{\prime}$ and as an edge of the triangle. (For example, in Figure 12 the sign is positive if viewed as a side of $P^{\prime}$ and negative if viewed as a side of the triangle.) Let $d_n^{\prime}$ be the signed distance to this side of $P^{\prime}$ and $-d_n^{\prime}$ be the signed distance to this side of the triangle.

Figure 12

Case 1: the removed triangle is positively oriented. By the induction hypothesis

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

Now consider the removed triangle. By Carnot's theorem for oriented triangles, $R + \tilde{r}_{n-1} = - d_n^{\prime} + d_n + d_{n+1} .$ Consequently,

$\sum_{k=1}^{n-1} \tilde{r_k} = \left( \sum_{k=1}^{n-2} \tilde{r_k} \right) + \tilde{r}_{n-1}$

$= \left( R ( q - p + 1) + d_n^{\prime} + \sum_{i=1}^{n-1} d_i \right) + ( - R + (- d_n^{\prime} + d_n + d_{n+1}))$

$= R(q-p) + \sum_{i=1}^{n+1} d_i ,$

as was to be shown.

Case 2: the triangle is negatively oriented. This case proceeds similarly, except that the induction hypothesis gives

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

and for the removed triangle $- R + \tilde{r}_{n-1} = -d_n^{\prime} + d_n + d_{n+1} .$

Case 3: the triangle is degenerate. In this case two or three of the vertices $p_1, p_n ,$ and $p_{n+1}$ coincide. Because $\tilde{r}_{n-1} = 0 ,$ our induction hypothesis gives us

$\sum_{k=1}^{n-1} \tilde{r}_k = \sum_{k=1}^{n-2} \tilde{r}_k = R (q - p ) + d_n^{\prime} + \sum_{i=1}^{n-1} d_i .$

We now consider three subcases. (a) Suppose $p_1 = p_{n+1} .$ Then $d_{n+1} = 0$ and $d_n = d_n^{\prime} .$ So $d_n^{\prime} = d_n + d_{n+1} .$ Substituting this into the formula yields the desired conclusion. (b) The case $p_n = p_{n+1}$ is similar. (c) Suppose $p_1 = p_n .$ Then $d_n^{\prime} = 0$ and $d_n + d_{n+1} = 0 .$ Thus, the result follows.∎

David Richeson, "The Japanese Theorem for Nonconvex Polygons - A Further Generalization of Carnot's Theorem," Convergence (December 2013)