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Let $ABC$ be a triangle with sides $$a,b,c.$$ Find a point $P$ inside the triangle such that $$a(PA)^2+b(PB)^2+c(PC)^2$$ is minimum

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  • $\begingroup$ a simple calculation gives $P=(aA+bB+cC)/(a+b+c)$ - but it doesn't reveal the geometric meaning of $P$ :-) $\endgroup$ – user8268 May 9 '14 at 13:47
  • $\begingroup$ It seems to me that P is the intersection of the bisectrices $\endgroup$ – user126154 May 9 '14 at 14:05
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Let $I$ be the incenter of $ABC$. $I$ is the barycenter of the weighted points $(A;a)$,$(B;b)$ and $(C;c)$. This means that $$ a\overrightarrow{IA}+b\overrightarrow{IB}+c\overrightarrow{IC}=\overrightarrow{0}\tag{1} $$ Now $$\eqalign{ a(PA)^2+b(PB)^2+c(PC)^2&=a(\overrightarrow{IA}-\overrightarrow{IP})^2 +a(\overrightarrow{IB}-\overrightarrow{IP})^2+ a(\overrightarrow{IC}-\overrightarrow{IP})^2\cr &=a (IA)^2+b(IB)^2+c(IC)^2\cr &\phantom{=}-2\underbrace{(a\overrightarrow{IA}+b\overrightarrow{IB}+c\overrightarrow{IC})}_0 \overrightarrow{IP}+(a+b+c)(IP)^2\cr &=a (IA)^2+b(IB)^2+c(IC)^2+(a+b+c)(IP)^2 } $$ Thus, the minimum is attained if and only if $IP=0$, that is, when $P$ is the incenter of the triangle.

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  • $\begingroup$ Why is summation of a×IA term zero??? $\endgroup$ – avz2611 May 9 '14 at 14:23
  • $\begingroup$ Because $I$ is the barycenter of $(A;a)$, $(B;b)$, $(C;c)$ see $(1)$. $\endgroup$ – Omran Kouba May 9 '14 at 14:43

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