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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\(\mathbf{2a}\) \((031)\)

Exactly,
\[\begin{aligned}\overrightarrow{AA^\prime}&={}\frac{2\sqrt{10}+\sqrt{5}-2\sqrt{2}+1}{12}\overrightarrow{AI_A},&\overrightarrow{BB^\prime}&={}\frac{2\sqrt{10}+\sqrt{5}+2\sqrt{2}-1}{2}\overrightarrow{BI_A},&\overrightarrow{CC^\prime}&={}\frac{2\sqrt{10}-\sqrt{5}-2\sqrt{2}-1}{4}\overrightarrow{CI_A}.\end{aligned}\]
\[\begin{alignedat}{4}A^\prime&={}&-\frac{2\sqrt{10}+\sqrt{5}-2\sqrt{2}-7}{8}&{}:{}&\frac{2\sqrt{10}+\sqrt{5}-2\sqrt{2}+1}{18}&{}:{}&\frac{5\left(2\sqrt{10}+\sqrt{5}-2\sqrt{2}+1\right)}{72}&,\\B^\prime&={}&-\frac{2\sqrt{10}+\sqrt{5}+2\sqrt{2}-1}{4}&{}:{}&-\frac{2\sqrt{10}+\sqrt{5}+2\sqrt{2}-7}{6}&{}:{}&\frac{5\left(2\sqrt{10}+\sqrt{5}+2\sqrt{2}-1\right)}{12}&,\\C^\prime&={}&-\frac{2\sqrt{10}-\sqrt{5}-2\sqrt{2}-1}{8}&{}:{}&\frac{2\sqrt{10}-\sqrt{5}-2\sqrt{2}-1}{6}&{}:{}&-\frac{2\sqrt{10}-\sqrt{5}-2\sqrt{2}-25}{24}&.\end{alignedat}\]
Approximately,
\[\begin{aligned}\overrightarrow{AA^\prime}&\approx{}0.561016347758\overrightarrow{AI_A},\\\overrightarrow{BB^\prime}&\approx{}5.194525211291\overrightarrow{BI_A},\\\overrightarrow{CC^\prime}&\approx{}0.065015054523\overrightarrow{CI_A}.\end{aligned}\]
\[\begin{alignedat}{4}A^\prime&\approx{}&0.158475478364&{}:{}&0.374010898505&{}:{}&0.467513623131&,\\B^\prime&\approx{}&-2.597262605646&{}:{}&-0.731508403764&{}:{}&4.328771009409&,\\C^\prime&\approx{}&-0.032507527261&{}:{}&0.043343369682&{}:{}&0.989164157580&.\end{alignedat}\]
2a (031)

Hiroyasu Kamo