Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.7 AIME Prove it China

Given four fixed points $A(-3, 0)$ , $B(1, -1)$ , $C(0, 3)$ , $D(-1, 3)$ and a variable point $P$ in a plane rectangular coordinates system, the minimum of $|PA| + |PB| + |PC| + |PD|$ is ______.

Solution

As shown in the figure, assuming that $AC$ and $BD$ meet at point $F$ , we have
$|PA| + |PC| \ge |AC| = |FA| + |FC|$
and
$|PB| + |PD| \ge |BD| = |FB| + |FD|.$
When $P$ coincides with $F$ , $|PA| + |PB| + |PC| + |PD|$ reaches the minimum.
That is, $|AC| + |BD| = 3\sqrt{2} + 2\sqrt{5}$ .
So $3\sqrt{2} + 2\sqrt{5}$ is the required answer.

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