# minimizing squares...

• August 11th 2008, 05:07 PM
scipa
minimizing squares...
Given a triangle $\triangle ABC$ and a straight line $l$, find the point $P$ on $l$ such that $(PA)^2 + (PB)^2 + (PC)^2$ is the smallest.
• August 12th 2008, 02:48 AM
flyingsquirrel
Hello,
Quote:

Originally Posted by scipa
Given a triangle $\triangle ABC$ and a straight line $l$, find the point $P$ on $l$ such that $(PA)^2 + (PB)^2 + (PC)^2$ is the smallest.

Let $A_h,\,B_h$ and $C_h$ be the orthogonal projection of $A,\,B$ and $C$ on $l$, respectively. Using Pythagorean theorem :

$\begin{cases}
PA^2=PA_h^2+AA_h^2\\
PB^2=PB_h^2+BB_h^2\\
PC^2=PC_h^2+CC_h^2\\
\end{cases}$

hence

$PA^2+PB^2+PC^2=\underbrace{AA_h^2+BB_h^2+CC_h^2}_{ \text{that's a constant}}+PA_h^2+PB_h^2+PC_h^2
$

As $AA_h^2+BB_h^2+CC_h^2$ doesn't depend on $P$ and $PA_h^2+PB_h^2+PC_h^2\geq 0$, minimizing $PA^2+PB^2+PC^2$ is equivalent to minimizing $PA_h^2+PB_h^2+PC_h^2$. How can this be achieved ?