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]
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\(\mathbf{6b}\) \((321)\)

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,
\[\begin{aligned}\overrightarrow{AA^\prime}&\approx{}-9.556561139893\overrightarrow{AI_B},\\\overrightarrow{BB^\prime}&\approx{}-0.052320075462\overrightarrow{BI_B},\\\overrightarrow{CC^\prime}&\approx{}-0.717214264202\overrightarrow{CI_B}.\end{aligned}\]
\[\begin{alignedat}{4}A^\prime&\approx{}&3.389140284973&{}:{}&9.556561139893&{}:{}&-11.945701424867&,\\B^\prime&\approx{}&-0.039240056597&{}:{}&1.104640150925&{}:{}&-0.065400094328&,\\C^\prime&\approx{}&-0.537910698151&{}:{}&0.717214264202&{}:{}&0.820696433950&.\end{alignedat}\]
6b (321)

Hiroyasu Kamo