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In $\triangle A B C$ the mid points of the sides $A B, B C$ and $C A$ are respectively $(l, 0,0),(0, m, 0)$ and $(0,0, n)$. Then, $\frac{A B^2+B C^2+C A^2}{l^2+m^2+n^2}$ is equal to
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1814 Upvotes
Verified Answer
The correct answer is:
$8$
From the figure,
$$
\begin{gathered}
x_1+x_2=2 l, y_1+y_2=0, z_1+z_2=0, \\
x_2+x_3=0, y_2+y_3=2 m, z_2+z_3=0 \\
\text { and } \quad x_1+x_3=0, y_1+y_3=0, z_1+z_3=2 n
\end{gathered}
$$
On solving, we get
$$
\begin{aligned}
x_1 & =l, x_2=l, x_3=-l, \\
& y_1=-m, y_2=m, y_3=m \\
\text { and } \quad z_1 & =n, z_2=-n, z_3=n
\end{aligned}
$$
$\therefore$ Coordinates are $A(l,-m, n), B(l, m,-n)$ and $C(-l, m, n)$
$$
\begin{aligned}
\therefore & \frac{A B^2+B C^2+C A^2}{l^2+m^2+n^2} \\
= & \frac{\left(4 m^2+4 n^2\right)+\left(4 l^2+4 n^2\right)+\left(4 l^2+4 m^2\right)}{l^2+m^2+n^2} \\
= & 8
\end{aligned}
$$
$$
\begin{gathered}
x_1+x_2=2 l, y_1+y_2=0, z_1+z_2=0, \\
x_2+x_3=0, y_2+y_3=2 m, z_2+z_3=0 \\
\text { and } \quad x_1+x_3=0, y_1+y_3=0, z_1+z_3=2 n
\end{gathered}
$$
On solving, we get
$$
\begin{aligned}
x_1 & =l, x_2=l, x_3=-l, \\
& y_1=-m, y_2=m, y_3=m \\
\text { and } \quad z_1 & =n, z_2=-n, z_3=n
\end{aligned}
$$
$\therefore$ Coordinates are $A(l,-m, n), B(l, m,-n)$ and $C(-l, m, n)$
$$
\begin{aligned}
\therefore & \frac{A B^2+B C^2+C A^2}{l^2+m^2+n^2} \\
= & \frac{\left(4 m^2+4 n^2\right)+\left(4 l^2+4 n^2\right)+\left(4 l^2+4 m^2\right)}{l^2+m^2+n^2} \\
= & 8
\end{aligned}
$$
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