Derousseau's Generalization of the Malfatti circles

The Smallest Pythagorean Triangle

\(C=90\degree\).   \(a:b:c=3:4:5\).


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

\(\mathbf{5b}\) \((303)\)

Exactly,
\[\begin{aligned}\overrightarrow{AA^\prime}&={}-\frac{\sqrt{10}-3\sqrt{5}-3\sqrt{2}+5}{2}\overrightarrow{AI_B},&\overrightarrow{BB^\prime}&={}-\frac{\sqrt{10}-3\sqrt{5}+3\sqrt{2}-5}{12}\overrightarrow{BI_B},&\overrightarrow{CC^\prime}&={}\frac{\sqrt{10}+3\sqrt{5}-3\sqrt{2}-5}{6}\overrightarrow{CI_B}.\end{aligned}\]
\[\begin{alignedat}{4}A^\prime&={}&\frac{\sqrt{10}-3\sqrt{5}-3\sqrt{2}+13}{8}&{}:{}&\frac{\sqrt{10}-3\sqrt{5}-3\sqrt{2}+5}{2}&{}:{}&-\frac{5\left(\sqrt{10}-3\sqrt{5}-3\sqrt{2}+5\right)}{8}&,\\B^\prime&={}&-\frac{\sqrt{10}-3\sqrt{5}+3\sqrt{2}-5}{16}&{}:{}&\frac{\sqrt{10}-3\sqrt{5}+3\sqrt{2}+1}{6}&{}:{}&-\frac{5\left(\sqrt{10}-3\sqrt{5}+3\sqrt{2}-5\right)}{48}&,\\C^\prime&={}&\frac{\sqrt{10}+3\sqrt{5}-3\sqrt{2}-5}{8}&{}:{}&-\frac{\sqrt{10}+3\sqrt{5}-3\sqrt{2}-5}{6}&{}:{}&\frac{\sqrt{10}+3\sqrt{5}-3\sqrt{2}+19}{24}&.\end{alignedat}\]
Approximately,
\[\begin{aligned}\overrightarrow{AA^\prime}&\approx{}1.394283479725\overrightarrow{AI_B},\\\overrightarrow{BB^\prime}&\approx{}0.358607132101\overrightarrow{BI_B},\\\overrightarrow{CC^\prime}&\approx{}0.104640150925\overrightarrow{CI_B}.\end{aligned}\]
\[\begin{alignedat}{4}A^\prime&\approx{}&0.651429130069&{}:{}&-1.394283479725&{}:{}&1.742854349656&,\\B^\prime&\approx{}&0.268955349076&{}:{}&0.282785735798&{}:{}&0.448258915126&,\\C^\prime&\approx{}&0.078480113194&{}:{}&-0.104640150925&{}:{}&1.026160037731&.\end{alignedat}\]
5b (303)

Hiroyasu Kamo