minimum value of aAP^2 + bBP^2 + cCP^2

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minimum value of aAP^2 + bBP^2 + cCP^2

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Source: 6-th Korean Mathematical Olympiad 1993, Final Round

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#1 h May 11, 2007, 5:40 AM • 6 Y

Y by Adventure10, Mango247, Mango247, Mango247, and 2 other users

Let be given a triangle $ABC$ with $BC = a, CA = b, AB = c$ . Find point $P$ in the plane for which $aAP^{2}+bBP^{2}+cCP^{2}$ is minimum, and compute this minimum.

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#2 h May 11, 2007, 6:09 AM • 4 Y

Y by Adventure10, Mango247, and 2 other users

N.T.TUAN wrote:

Let be given a triangle $ABC$ with $BC = a$ , $CA = b$ , $AB = c$ . Find point $P$ in the plane for which $a\cdot AP^{2}+b\cdot BP^{2}+c\cdot CP^{2}$ is minimum.

$(\forall )P\in \mathcal P$ (plane), $\{\begin{array}{c}a\cdot PA^{2}+b\cdot PB^{2}+c\cdot PC^{2}=2p\cdot PI^{2}+abc\\\\ a\cdot PA^{2}+b\cdot PB^{2}+c\cdot PC^{2}\ge abc\\\\ \sum a\cdot PA^{2}=abc\Longleftrightarrow P: =I\end{array}$ , where $2p=a+b+c$ and $I$ is the incenter of $\triangle ABC$ (it is well-known).

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Raúl

#3 h Aug 30, 2010, 3:08 PM • 1 Y

Y by Adventure10

Who can I proof the 1st equation :O ?

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#4 h Aug 30, 2010, 9:08 PM • 1 Y

Y by Adventure10

Raúl wrote:

Who can I proof the first identity $a\cdot PA^{2}+b\cdot PB^{2}+c\cdot PC^{2}=(a+b+c)\cdot PI^2+abc$ ?

Denote $a+b+c=2s$ , $D\in BC\cap AI$ and the length $l_a$ of the $A$ -bisector. I'll use remarkable relations : $\begin{array}{c} \frac {DB}{c}=\frac {DB}{b}=\frac {a}{b+c}\ .\\\\ \frac {IA}{b+c}=\frac {ID}{a}=\frac {l_a}{2s}\ .\\\\ l_a^2=\frac {4bcs(s-a)}{(b+c)^2}\ .\end{array}$

Apply the Stewart's relation to cevians : $\begin{array}{cc} PD/\triangle BPC\ : & (b+c)^2\cdot PD^2=\left(b\cdot PB^2+c\cdot PC^2\right)-a^2bc\ .\\\\ PI/\triangle APD\ : & 4s^2\cdot PI^2=2s\left[a\cdot PA^2+(b+c)\cdot PD^2\right]-\frac {4abcs(s-a)}{b+c}\ .\end{array}$

and eliminate $PD$ between above relations. Can obtain similarly and the relation $PA^2+PB^2+PC^2=3\cdot PG^2+\frac {a^2+b^2+c^2}{3}$ .

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