Let a, b, c be the side-lengths of a triangle | vedclass

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(C) Let the side lengths of $	riangle ABC$ be $BC = a, AC = b, AB = c$ and the lengths of the medians be $AD = l, BE = m, CF = n$.In any triangle,the

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$\left(\frac{3}{4}, 1\right)$

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(C) Let the side lengths of $	riangle ABC$ be $BC = a, AC = b, AB = c$ and the lengths of the medians be $AD = l, BE = m, CF = n$.In any triangle,the length of a median is less than the semi-sum of the adjacent sides. Thus,$l Summing these inequalities,we get $l+m+n Therefore,$K = \frac{l+m+n}{a+b+c} Also,let $G$ be the centroid of the triangle. In $	riangle BGC$,by the triangle inequality,$BG + GC > BC$. Since the centroid divides the median in the ratio $2:1$,we have $BG = \frac{2}{3}m$ and $GC = \frac{2}{3}n$. Thus,$\frac{2}{3}(m+n) > a$.Similarly,$\frac{2}{3}(n+l) > b$ and $\frac{2}{3}(l+m) > c$.Adding these three inequalities,we get $\frac{4}{3}(l+m+n) > a+b+c$,which implies $K = \frac{l+m+n}{a+b+c} > \frac{3}{4}$.Combining these results,$K \in \left(\frac{3}{4}, 1ight)$.

Let $a, b, c$ be the side-lengths of a triangle and $l, m, n$ be the lengths of its medians. Put $K = \frac{l+m+n}{a+b+c}$. Then,as $a, b, c$ vary,$K$ can assume every value in the interval

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