Japanese Theorem

The Japanese Theorem is a theorem which holds for cyclic polygons.

Statement

For any triangulated cyclic polygon, the sum of the inradii of the triangles is constant.

Japanese Theorem for cyclic quadrilaterals

Statement

The Japanese theorem for cyclic quadrilaterals states that for a cyclic quadrilateral $ABCD$ and incenters $M_1$ , $M_2$ , $M_3$ , $M_4$ of triangles $\triangle ABD$ , $\triangle ABC$ , $\triangle BCD$ , $\triangle ACD$ the quadrilateral $M_1M_2M_3M_4$ is a rectangle.

Proof

From $\triangle ABC$ , we can see that $\angle BM_2C = 90^{\circ} + \frac{1}{2} \angle CAB$ and similarly, from $\triangle BCD$ we have $\angle BM_3C = 90^{\circ} + \frac{1}{2} \angle CDB$ Since $ABCD$ is cyclic, therefore $\angle CDB = \angle CAB$ , which means that $\angle BM_2C = \angle BM_3C$ From this, it follows that $BM_2M_3C$ is cyclic. This means that $\angle BCM_3 + \angle BM_2M_3 = 180^{\circ}$ By symmetry, we can also derive $\angle BAM_1 + \angle BM_2M_1 = 180^{\circ}$ Adding these equations, we get $\angle BAM_1 + \angle BCM_3 + \angle BM_2M_1 + \angle BM_2M_3 = 360^{\circ}$ $\implies \angle BM_2M_1 + \angle BM_2M_3 = 360^{\circ} - \angle BAM_1 - \angle BCM_3 = 360^{\circ} - \frac{1}{2} \left(\angle CAB + \angle CDB \right)$ Which implies $M_1M_2M_3 = 90^{\circ}$

And other angles similarly. $\square$

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Japanese Theorem

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Statement