minimizing squares...

**scipa** · August 11th 2008, 04:07 PM

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.

**flyingsquirrel** · August 12th 2008, 01:48 AM

Hello,

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 ?

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