HMMT-Nov 2015 guts, Problem 27

Let $A B C D$ 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 $A P+B P+C P+D P$.

Solution: By the triangle inequality, $A P+C P \geq A C$ and $B P+D P \geq B D$. So $P$ should be on $A C$ and $B D$; i.e. it should be the intersection of the two diagonals. Then $A P+B P+C P+D P=A C+B D$, 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 $A B C D$ is convex and this is not an issue. $\fbox{16 \sqrt{17}+8 \sqrt{5}}$.

Answer: 16 \sqrt{17}+8 \sqrt{5}