# 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," Loci (December 2013)