### Carnot's Theorem

For proofs of the quadrilateral case of the Japanese theorem or the polygonal case with all diagonals eminating from a single vertex see [AUM1, AUM2, FP, FR, Gr, Hay, Jo1, Mi, Yo]. We will not prove those separately, but instead prove the more general Japanese theorem for cyclic polygons. Our proof, which is in essence Honsberger's proof ([Ho]), relies on a classical theorem discovered by Lazare Nicolas Marguérite Carnot (1753–1823) that exhibits a relationship between a triangle, its inradius, and its circumradius.

The *signed distance* of a side of a convex cyclic polygon to the circumcenter is the distance, unless the segment joining the cirucumcenter to the side lies completely outside the polygon, in which case it is the negative of the distance. For example, for the triangles in Figures 4(a) and 4(b), all the signed distances \(a, b,\) and \(c\) are positive except \(b\) in Figure 4(b). The sign contributions of the cyclic hexagon in Figure 4(c) are indicated.

Figure 4

**Carnot's theorem.** *If \(a\), \(b\), and \(c\) are the signed distances from the circumcenter of a triangle to the sides of the triangle, \(R\) is the circumradius, and \(r\) is the inradius, then \(R + r = a + b + c\).*

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*Figure 5*

See [AN] for a short proof of Carnot's theorem for acute triangles (in this case \(a\), \(b\), and \(c\) are all positive); the given argument can be modified to cover the non-acute case.

To see the theorem in action, move the vertices of the triangle below and see the equality of the two sums.