In triangle ABC,if the midpoints of the sides | vedclass

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(C) Let the vertices of $	riangle ABC$ be $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$.Given that the midpoints of $AB, BC, CA$ are $

Vedclass · Accepted Answer

$8$

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(C) Let the vertices of $	riangle ABC$ be $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$.Given that the midpoints of $AB, BC, CA$ are $(l, 0, 0), (0, m, 0), (0, 0, n)$ respectively.For midpoint of $AB$: $\frac{x_1+x_2}{2}=l, \frac{y_1+y_2}{2}=0, \frac{z_1+z_2}{2}=0 \Rightarrow x_1+x_2=2l, y_1+y_2=0, z_1+z_2=0$ ... $(i)$For midpoint of $BC$: $\frac{x_2+x_3}{2}=0, \frac{y_2+y_3}{2}=m, \frac{z_2+z_3}{2}=0 \Rightarrow x_2+x_3=0, y_2+y_3=2m, z_2+z_3=0$ ... (ii)For midpoint of $CA$: $\frac{x_3+x_1}{2}=0, \frac{y_3+y_1}{2}=0, \frac{z_3+z_1}{2}=n \Rightarrow x_3+x_1=0, y_3+y_1=0, z_3+z_1=2n$ ... (iii)Solving these equations:Adding $(i)$,(ii),and (iii): $2(x_1+x_2+x_3)=2l \Rightarrow x_1+x_2+x_3=l$. Similarly,$y_1+y_2+y_3=m$ and $z_1+z_2+z_3=n$.Subtracting (ii) from the sum: $x_1=l, y_1=-m, z_1=n$ (Wait,let's re-evaluate).From $(i)$,$x_2=2l-x_1$. Substituting into (ii): $2l-x_1+x_3=0 \Rightarrow x_1-x_3=2l$. From (iii),$x_1+x_3=0$. Adding gives $2x_1=2l \Rightarrow x_1=l, x_3=-l, x_2=l$.Similarly,$y_1=m, y_2=-m, y_3=m$ and $z_1=-n, z_2=n, z_3=n$.Thus,$A(l, m, -n), B(l, -m, n), C(-l, m, n)$.$AB^2 = (l-l)^2 + (m-(-m))^2 + (-n-n)^2 = 0 + 4m^2 + 4n^2 = 4(m^2+n^2)$.$BC^2 = (l-(-l))^2 + (-m-m)^2 + (n-n)^2 = 4l^2 + 4m^2 + 0 = 4(l^2+m^2)$.$CA^2 = (-l-l)^2 + (m-m)^2 + (n-(-n))^2 = 4l^2 + 0 + 4n^2 = 4(l^2+n^2)$.$AB^2+BC^2+CA^2 = 4(m^2+n^2+l^2+m^2+l^2+n^2) = 8(l^2+m^2+n^2)$.Therefore,$\frac{AB^2+BC^2+CA^2}{l^2+m^2+n^2} = \frac{8(l^2+m^2+n^2)}{l^2+m^2+n^2} = 8$.

In $\triangle ABC$,if the midpoints of the sides $AB, BC$ and $CA$ are respectively $(l, 0, 0), (0, m, 0)$ and $(0, 0, n)$,then $\frac{AB^2+BC^2+CA^2}{l^2+m^2+n^2}=$

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