In a triangle ABC,if 2r_1 = 3r_2 = r_3,then a | vedclass

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(A) We know that the exradii of a triangle are given by $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.Given $2r

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$4 : 3 : 5$

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(A) We know that the exradii of a triangle are given by $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.Given $2r_1 = 3r_2 = r_3 = k$ (let).Then $r_1 = \frac{k}{2}$,$r_2 = \frac{k}{3}$,and $r_3 = k$.Thus,$\frac{1}{r_1} = \frac{2}{k}$,$\frac{1}{r_2} = \frac{3}{k}$,and $\frac{1}{r_3} = \frac{1}{k}$.We know that $\frac{1}{r_1} = \frac{s-a}{\Delta}$,$\frac{1}{r_2} = \frac{s-b}{\Delta}$,and $\frac{1}{r_3} = \frac{s-c}{\Delta}$.Summing these,$\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{3s - (a+b+c)}{\Delta} = \frac{3s - 2s}{\Delta} = \frac{s}{\Delta} = \frac{1}{r}$.Also,$\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{2}{k} + \frac{3}{k} + \frac{1}{k} = \frac{6}{k} = \frac{1}{r}$,so $r = \frac{k}{6}$.Using $s-a = \frac{\Delta}{r_1} = \frac{\Delta}{k/2} = \frac{2\Delta}{k}$,$s-b = \frac{3\Delta}{k}$,and $s-c = \frac{\Delta}{k}$.Since $s = \frac{6\Delta}{k}$,we have $a = s - (s-a) = \frac{6\Delta}{k} - \frac{2\Delta}{k} = \frac{4\Delta}{k}$,$b = \frac{3\Delta}{k}$,and $c = \frac{5\Delta}{k}$.Therefore,$a : b : c = 4 : 3 : 5$.

In a triangle $ABC$,if $2r_1 = 3r_2 = r_3$,then $a : b : c =$

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