1. If in a triangle ABC the exradii r₁, r₂, r₃, are such that r_{1 = 2r2 = 3r3} then a;b;c is

Dear

we know

    r1 = Δ /(s-a)

    r2   = Δ /(s-b)

    r3    =Δ/(s-c)

so     r_{1 = 2r2 = 3r3}

      Δ /(s-a) =2Δ /(s-b) =3Δ /(s-c)

     s-b =2s-2a                and     s-c = 3s-3a

     3b +c -4a=0                and  2a -b-2c=0

solve  a/5 = b/6 =c/2

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