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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.