Derousseau's Generalization of the Malfatti circles

Angle Bisectors


Jump
[Guy]
[Lob & Richmond]
(0**)
(1**)
(2**)
(3**)

\(\mathbf{0}\)   \((000)\)

\[\begin{aligned}\overrightarrow{AA^{\prime}} &= \dfrac{\left(1+\tan\dfrac{B}{4}\right)\left(1+\tan\dfrac{C}{4}\right)}{2\left(1+\tan\dfrac{A}{4}\right)}\overrightarrow{AI}\\\overrightarrow{BB^{\prime}} &= \dfrac{\left(1+\tan\dfrac{A}{4}\right)\left(1+\tan\dfrac{C}{4}\right)}{2\left(1+\tan\dfrac{B}{4}\right)}\overrightarrow{BI}\\\overrightarrow{CC^{\prime}} &= \dfrac{\left(1+\tan\dfrac{A}{4}\right)\left(1+\tan\dfrac{B}{4}\right)}{2\left(1+\tan\dfrac{C}{4}\right)}\overrightarrow{CI}\end{aligned}\]

Hiroyasu Kamo