In the plane, Point $ O$ is the center of the equilateral triangle $ABC$ , Points $P,Q$ such that $\overrightarrow{OQ}=2\overrightarrow{PO}$. Prove that\[|PA|+|PB|+|PC|\le |QA|+|QB|+|QC|.\]
Problem
Source: China Zongqing , 17 Aug 2014
Tags: inequalities proposed, inequalities
18.08.2014 17:27
Did you mean that the length of OQ is twice the length of PO?
18.08.2014 20:36
And $O\in PQ$.
19.08.2014 13:38
Let $J,K,L$ be the midpoints of $BC,CA,AB$ respectively. Then it is clear that $AQ=2JP$, and by Ptolemy's Theorem, $AK\cdot LP+AL\cdot KP\ge KL\cdot AP$. But $\triangle AKL$ is equilateral, so $LP+KP\ge AP$. Summing up cyclically we have the desired result.
19.08.2014 14:23
Also because $AP$, $KP$ and $LP$ are sides-lengths of triangle.
15.12.2019 20:25
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