HMMT 十一月 2015 · 冲刺赛 · 第 27 题
HMMT November 2015 — Guts Round — Problem 27
题目详情
英文原题
- [ 13 ] Let ABCD be a quadrilateral with A = (3 , 4) , B = (9 , − 40) , C = ( − 5 , − 12) , D = ( − 7 , 24). Let Pbe 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 yx +2 z x − 3 z +1
解析
英文解析
- [ 13 ] Let ABCD be a quadrilateral with A = (3 , 4) , B = (9 , − 40) , C = ( − 5 , − 12) , D = ( − 7 , 24). Let Pbe 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 yx +2 z x − 3 z +1