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Geometry Difficulty 4.9 AIME Prove it United States

Problem:
Let $ABCD$ be a quadrilateral with $A=(3,4)$ , $B=(9,-40)$ , $C=(-5,-12)$ , $D=(-7,24)$ . Let $P$ be a point in the plane (not necessarily inside the quadrilateral). Find the minimum possible value of $AP + BP + CP + DP$ .

Solution

Solution:
Answer: $16 \sqrt{17} + 8 \sqrt{5}$
By the triangle inequality, $AP + CP \geq AC$ and $BP + DP \geq BD$ . So $P$ should be on $AC$ and $BD$ ; i.e., it should be the intersection of the two diagonals. Then $AP + BP + CP + DP = AC + BD$ , which is easily computed to be $16 \sqrt{17} + 8 \sqrt{5}$ by the Pythagorean theorem.

Note that we require the intersection of the diagonals to actually exist for this proof to work, but $ABCD$ is convex and this is not an issue.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.