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

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.