On Mathoverflow, I saw this great result on the "Japanese Theorem".
- “Japanese Theorem” on cyclic polygons: Higher-dimensional generalizations?
- Given triangulation of a cyclic polygon, the sum of the
areasinradii of the incircles of the triangles is independent of the triangulation.
How do we prove this result that make the independents of the triangulation obvious.
The inradius is related to the area by $\boxed{\text{Area} = \text{semiperimeter} \times \text{inradius}}$. Perhaps this can be used to re-create the conservation law above?
Also, it is sufficient to prove this result for a cyclic quadrilateral and compare the two triangulations.
Proof: Wikipedia says is based on Carnot's Theorem: $OO_A+OO_B+OO_C = R + r$, where $r$ is the inradius, and $R$ is the circumradius, $OO_A,OO_B,OO_C$ distances to the sides of the triangle.
In that case, I am not understanding proof of this Carnot's result, or why - if we sum over the triangles in the triangulation - this sum is independent of the triangulation.
To prove Carnot's result, you can turn it to a purely trigonometric identity.
First $$ \sum OO_A=R\sum \cos A $$ Second $r(a+b+c)=2Rr(\sum \sin A)=2 \times Area=ac\sin B=4R^2\Pi\sin A $, so $$ r=2R\frac{\Pi\sin A}{\sum\sin A}. $$
Now applying the well-known identities $$ \sum \sin A=4\Pi\cos\frac{A}{2} $$ and $$ \sum \cos A=1+4\Pi \sin \frac{A}{2} $$