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

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

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$8$

Vedclass · Answer

(C) Let the vertices of the triangle be $A(x_1, y_1, z_1)$, $B(x_2, y_2, z_2)$, and $C(x_3, y_3, z_3)$.Given the midpoints of $AB, BC$, and $CA$ are $(l, 0, 0), (0, m, 0)$, and $(0, 0, n)$ respectively.Using the midpoint formula:$\frac{x_1+x_2}{2} = l, \frac{y_1+y_2}{2} = 0, \frac{z_1+z_2}{2} = 0 \implies x_1+x_2=2l, y_1+y_2=0, z_1+z_2=0$$\frac{x_2+x_3}{2} = 0, \frac{y_2+y_3}{2} = m, \frac{z_2+z_3}{2} = 0 \implies x_2+x_3=0, y_2+y_3=2m, z_2+z_3=0$$\frac{x_3+x_1}{2} = 0, \frac{y_3+y_1}{2} = 0, \frac{z_3+z_1}{2} = n \implies x_3+x_1=0, y_3+y_1=0, z_3+z_1=2n$Solving these systems of equations:For $x$: $x_1+x_2=2l, x_2+x_3=0, x_3+x_1=0 \implies x_1=l, x_2=l, x_3=-l$For $y$: $y_1+y_2=0, y_2+y_3=2m, y_3+y_1=0 \implies y_1=-m, y_2=m, y_3=m$For $z$: $z_1+z_2=0, z_2+z_3=0, z_3+z_1=2n \implies z_1=n, z_2=-n, z_3=n$Thus, $A(l, -m, n), B(l, m, -n), C(-l, m, n)$.Now, calculate the squared side lengths:$AB^2 = (l-l)^2 + (m-(-m))^2 + (-n-n)^2 = 0 + (2m)^2 + (-2n)^2 = 4m^2 + 4n^2$$BC^2 = (-l-l)^2 + (m-m)^2 + (n-(-n))^2 = (-2l)^2 + 0 + (2n)^2 = 4l^2 + 4n^2$$CA^2 = (l-(-l))^2 + (-m-m)^2 + (n-n)^2 = (2l)^2 + (-2m)^2 + 0 = 4l^2 + 4m^2$Summing these: $AB^2+BC^2+CA^2 = (4m^2+4n^2) + (4l^2+4n^2) + (4l^2+4m^2) = 8(l^2+m^2+n^2)$.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 $\triangle ABC$, the midpoints of the sides $AB, BC$, and $CA$ are respectively $(l, 0, 0), (0, m, 0)$, and $(0, 0, n)$. Then, $\frac{AB^2+BC^2+CA^2}{l^2+m^2+n^2}$ is equal to

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