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(A) Given $\frac{a+b}{7} = \frac{b+c}{8} = \frac{c+a}{9} = \lambda$.Then $a+b = 7\lambda$,$b+c = 8\lambda$,and $c+a = 9\lambda$.Adding these equations

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$\frac{5}{2}$

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(A) Given $\frac{a+b}{7} = \frac{b+c}{8} = \frac{c+a}{9} = \lambda$.Then $a+b = 7\lambda$,$b+c = 8\lambda$,and $c+a = 9\lambda$.Adding these equations,$2(a+b+c) = 24\lambda$,so $a+b+c = 12\lambda$.Subtracting the given equations from $a+b+c = 12\lambda$,we get $c = 5\lambda$,$a = 4\lambda$,and $b = 3\lambda$.Since $a^2 + b^2 = (4\lambda)^2 + (3\lambda)^2 = 25\lambda^2 = c^2$,the triangle is a right-angled triangle with $\angle C = 90^{\circ}$.For a right-angled triangle,the circumradius $R = \frac{c}{2} = \frac{5\lambda}{2}$ and the inradius $r = \frac{a+b-c}{2} = \frac{4\lambda+3\lambda-5\lambda}{2} = \lambda$.Therefore,$\frac{R}{r} = \frac{5\lambda/2}{\lambda} = \frac{5}{2}$.

Let $a, b$ and $c$ be the lengths of the sides of a triangle $ABC$ such that $\frac{a+b}{7} = \frac{b+c}{8} = \frac{c+a}{9}$. If $r$ and $R$ are the inradius and circumradius of the triangle $ABC$,respectively,then the value of $\frac{R}{r}$ is equal to

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