### The Generalized Japanese Theorem

When playing with the two applets illustrating the Japanese theorem (for quadrilaterals and for \(n\)-gons), there is a strong temptation to move one vertex past another and make a polygon with crossed edges (as in Figure 10, in which we slide \(p_8\) past \(p_1 \)). Doing so yields an error message from the applet. What would happen in that case? Can we triangulate such a polygon? Can we define the total inradius, and if so, is it still invariant?

Figure 10

In this section we generalize the Japanese theorem to nonconvex polygons. In order to do so we must reexamine our definitions. We begin with a collection of points on the circle, \( p_1, \ldots, p_n \) (not necessarily distinct). We obtain a polygon by drawing a line segment from \( p_i \) to \( p_{i+1} \) for \(1 \leq i < n \) and \( p_n \) to \(p_1 . \) In particular, from here onward, when we write "cyclic polygon" we assume that we have a convex or nonconvex polygon inscribed in a circle, and we also assume implicitly that the vertices are labeled.

To define triangulation for a nonconvex polygon we must borrow some terminology from graph theory. Recall that a graph is *planar* if there exists an embedding of the graph in the plane such that no edges cross. A nonconvex polygon is *triangulated by diagonals* if there is a planar embedding of the \(n\)-gon and \(n - 2\) diagonals such that the polygon is an exterior boundary and all diagonals are nonintersecting. As with convex polygons, an easy way to obtain a triangulation is to pick one vertex and draw diagonals to all the non-neighboring vertices.

As before, we can inscribe a circle in each triangular region and measure the inradius. Now, however, we assign a sign to the inradius. If a triangle has vertices \(p_i, p_j, p_k \) with \(i < j < k\) and \(p_i, p_j, p_k \) are situated around the circle in a counterclockwise fashion, then we say that the triangle is *positively oriented*, and we take the signed inradius to be the inradius. Otherwise the triangle is *negatively oriented*, and the signed inradius is the negative of the inradius. For example, in the figure above the triangle with vertices \(p_1, p_5, \) and \(p_8\) is negatively oriented, and the dark red circle inscribed in it has a negative signed inradius. We will use \( \tilde{r} \) to denote the signed inradius of a triangle.

**Generalized Japanese theorem.** *Triangulate a cyclic polygon using diagonals. The sum of the signed inradii of the triangles is independent of the choice of triangulation.*

Use the following applet to see the invariance of the total signed inradius, which we denote \( \tilde{r_P} .\)

The proof of this version of the Japanese theorem requires a further generalization of Carnot's theorem.