HMMT 二月 2012 · 冲刺赛 · 第 23 题

HMMT February 2012 — Guts Round — Problem 23

[ 13 ] Points X and Y are inside a unit square. The score of a vertex of the square is the minimum distance from that vertex to X or Y . What is the minimum possible sum of the scores of the vertices of the square?

解析

[ 13 ] Points X and Y are inside a unit square. The score of a vertex of the square is the minimum distance from that vertex to X or Y . What is the minimum possible sum of the scores of the vertices of the square? √ √ 6+ 2 Answer: Let the square be ABCD . First, suppose that all four vertices are closer to X than 2 Y . Then, by the triangle inequality, the sum of the scores is AX + BX + CX + DX ≥ AB + CD = 2. Similarly, suppose exactly two vertices are closer to X than Y . Here, we have two distinct cases: the vertices closer to X are either adjacent or opposite. Again, by the Triangle Inequality, it follows that the sum of the scores of the vertices is at least 2. On the other hand, suppose that A is closer to X and B, C, D are closer to Y . We wish to compute the minimum value of AX + BY + CY + DY , but note that we can make X = A to simply minimize BY + CY + DY . We now want Y to be the Fermat point of triangle BCD , so that ∡ BY C = ◦ ◦ ∡ CY D = ∡ DY B = 120 . Note that by symmetry, we must have ∡ BCY = ∡ DCY = 45 , so ◦ ∡ CBY = ∡ CDY = 15 . ◦ ◦ sin 45 sin 15 And now we use the law of sines: BY = DY = and CY = . Now, we have BY + CY + ◦ ◦ sin 120 sin 120 √ √ 2+ 6 DY = , which is less than 2, so this is our answer. 2