HMMT 十一月 2015 · 冲刺赛 · 第 27 题

HMMT November 2015 — Guts Round — Problem 27

[ 13 ] 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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2015, November 14, 2015 — GUTS ROUND Organization Team Team ID# y − 1 y x +2 z x − 3 z +1

解析

[ 13 ] 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 . Proposed by: Alexander Katz √ √ Answer: 16 17 + 8 5 By the triangle inequality, AP + CP ≥ AC and BP + DP ≥ 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 17 + 8 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. y − 1 y x +2 z x − 3 z +1