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(D) Let the sides of the triangle be $a=1$,$b$,and $c$. Let the altitudes corresponding to these sides be $h_a$,$h_b$,and $h_c$. The area $\Delta = \f

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$11$

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(D) Let the sides of the triangle be $a=1$,$b$,and $c$. Let the altitudes corresponding to these sides be $h_a$,$h_b$,and $h_c$. The area $\Delta = \frac{1}{2} a h_a = \frac{1}{2} b h_b = \frac{1}{2} c h_c$.Given $h_a = 1 \cdot h_a$ and $h_b = 2 h_a$ (or vice versa). Since $h_a = \frac{2\Delta}{1}$ and $h_b = \frac{2\Delta}{b}$,the condition that the longest altitude is twice the shortest implies $b=2$ or $c=2$.For a triangle with sides $1, 2, c$,the triangle inequality requires $2-1 The sides are $1, 2, 2$. This is an isosceles triangle.The semi-perimeter $s = \frac{1+2+2}{2} = \frac{5}{2}$.The area $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{\frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}} = \frac{\sqrt{15}}{4}$.The inradius $r = \frac{\Delta}{s} = \frac{\sqrt{15}/4}{5/2} = \frac{\sqrt{15}}{10}$.The circumradius $R = \frac{abc}{4\Delta} = \frac{1 \cdot 2 \cdot 2}{4 \cdot \sqrt{15}/4} = \frac{4}{\sqrt{15}}$.Then $\frac{R}{r} = \frac{4/\sqrt{15}}{\sqrt{15}/10} = \frac{40}{15} = \frac{8}{3}$.Thus $m=8$ and $n=3$. Since $m, n$ are coprime,$m+n = 8+3 = 11$.

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