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China Southeastern Mathematical Olympiad

Geometry Difficulty 5.4 AIME, harder Prove it China

Given $n$ points $P_1, P_2, \dots, P_n$ on a plane, let $M$ be any point on segment $AB$ on the plane. Denote by $|P_iM|$ the distance between $P_i$ and $M$ , $i = 1, 2, 3, \dots, n$ . Prove that
$\sum_{i=1}^n |P_iM| \le \max\left\{\sum_{i=1}^n |P_iA|, \sum_{i=1}^n |P_iB|\right\}.$
(posed by Jin Mengwei)

Solution

Let $O$ be the origin. Then we have $\overrightarrow{OM} = t \overrightarrow{OA} + (1-t) \overrightarrow{OB}$ , $t \in (0, 1)$ .
$\begin{aligned} |P_iM| &= |\overrightarrow{OM} - \overrightarrow{OP_i}| \\ &= |t \overrightarrow{OA} + (1-t) \overrightarrow{OB} - t \overrightarrow{OP_i} - (1-t) \overrightarrow{OP_i}| \\ &\le t |\overrightarrow{OA} - \overrightarrow{OP_i}| + (1-t) |\overrightarrow{OB} - \overrightarrow{OP_i}| \\ &= t |P_iA| + (1-t) |P_iB|. \end{aligned}$
Hence,
$\begin{aligned} \sum_{i=1}^{n} |P_iM| &\le t \sum_{i=1}^{n} |P_iA| + (1-t) \sum_{i=1}^{n} |P_iB| \\ &\le \max\left\{\sum_{i=1}^{n} |P_iA|, \sum_{i=1}^{n} |P_iB|\right\}. \quad \square \end{aligned}$

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