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Question Number 124189 by ajfour last updated on 01/Dec/20

Commented by ajfour last updated on 01/Dec/20

$$Findradiiofthethreecircles$$$$intermsofa,b,c.$$

Commented by ajfour last updated on 01/Dec/20

$$thanksSir!$$

Commented by mr W last updated on 01/Dec/20

$$so-calledMalfatticircles$$

Commented by mr W last updated on 01/Dec/20

Commented by MJS_new last updated on 01/Dec/20

$$\mathrm{given}a,b,c\mathrm{we}\mathrm{can}\mathrm{put}\mathrm{the}\mathrm{triangle}\mathrm{as}\mathrm{follows}:$$$$A=\left(\begin{array}{c}0\\ 0\end{array}\right)B=\left(\begin{array}{c}c\\ 0\end{array}\right)C=\left(\begin{array}{c}\frac{-a^2+b^2+c^2}{2c}\\ \frac{\delta }{2c}\end{array}\right)$$$$\mathrm{the}\mathrm{incenter}\mathrm{is}[M\mathrm{for}\mathrm{mid}-\mathrm{point}\mathrm{to}\mathrm{avoid}$$$$\mathrm{another}C]$$$$M_I=\left(\begin{array}{c}\frac{-a+b+c}{2}\\ r_I\end{array}\right)\mathrm{with}{r}_I=\frac{\delta }{2(a+b+c)}$$$$\mathrm{the}\mathrm{distances}\mathrm{of}\mathrm{this}\mathrm{to}\mathrm{the}\mathrm{vertices}\mathrm{are}$$$$d=\mid M_{I}A\mid =\sqrt{\frac{(-a+b+c)bc}{a+b+c}}$$$$e=\mid M_{I}B\mid =\sqrt{\frac{a(a-b+c)c}{a+b+c}}$$$$f=\mid M_{I}C\mid =\sqrt{\frac{a(a+b-c)c}{a+b+c}}$$$$\delta =\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}$$$$\mathrm{thought}\mathrm{this}\mathrm{might}\mathrm{help}$$

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