HMMT 十一月 2015 · 冲刺赛 · 第 33 题

HMMT November 2015 — Guts Round — Problem 33

[ 17 ] How many ways are there to place four points in the plane such that the set of pairwise distances between the points consists of exactly 2 elements? (Two configurations are the same if one can be obtained from the other via rotation and scaling.) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2015, November 14, 2015 — GUTS ROUND Organization Team Team ID#

解析

[ 17 ] How many ways are there to place four points in the plane such that the set of pairwise distances between the points consists of exactly 2 elements? (Two configurations are the same if one can be obtained from the other via rotation and scaling.) Proposed by: Alexander Katz Answer: 6 Let A, B, C, D be the four points. There are 6 pairwise distances, so at least three of them must be equal. Case 1: There is no equilateral triangle. Then WLOG we have AB = BC = CD = 1. • Subcase 1.1: AD = 1 as well. Then AC = BD 6 = 1, so ABCD is a square. • Subcase 1.2: AD 6 = 1. Then AC = BD = AD , so A, B, C, D are four points of a regular pentagon. Case 2: There is an equilateral triangle, say ABC , of side length 1. • Subcase 2.1: There are no more pairs of distance 1. Then D must be the center of the triangle. • Subcase 2.2: There is one more pair of distance 1, say AD . Then D can be either of the two intersections of the unit circle centered at A with the perpendicular bisector of BC . This gives us 2 kites. ◦ • Subcase 2.3: Both AD = BD = 1. Then ABCD is a rhombus with a 60 angle. This gives us 6 configurations total.