HMMT 二月 2010 · TEAM1 赛 · 第 6 题

HMMT February 2010 — TEAM1 Round — Problem 6

[ 20 ] Let S be a convex set in the plane with a finite area a . Prove that either a = 0 or S is bounded. Note: a set is bounded if it is contained in a circle of finite radius. Note: a set is convex if, whenever two points A and B are in the set, the line segment between them is also in the set.

解析

[ 20 ] Let S be a convex set in the plane with a finite area a . Prove that either a = 0 or S is bounded. Note: a set is bounded if it is contained in a circle of finite radius. Note: a set is convex if, whenever two points A and B are in the set, the line segment between them is also in the set. Solution: If all points in S lie on a straight line, then a = 0. Otherwise we may pick three points A , B , and C that are not collinear. Let ω be the incircle of 4 ABC , with I its center and r its radius. Since S is convex, S must contain ω . X A Y ω M I Z B C √ 2 a 2 Suppose S also contains a point X at a distance d from I , with d > R . We will show that d ≤ r + , 2 r which implies that the S is bounded since all points are contained within the circle centered at I of √ 2 a 2 radius r + . 2 r Let Y and Z be on ω such that XY and XZ are tangents to ω . Because S is convex, it must contain kite IY XZ , whose area we can compute in terms of d and r . IM IY Let M be the midpoint of Y Z . Since 4 IY X ∼ 4 IM Y , we know that = , that is, IM = IY IX √ 2 √ 2 4 ( IY ) r r r 1 2 2 = . Then M Y = r − = r 1 − ( ) = Y Z . 2 IX d d d 2 √ √ 1 r 2 2 2 The area of IY XZ is ( Y Z )( IX ) = rd 1 − ( ) = r d − r . This must be less than or equal to a , 2 d √ 2 2 2 2 2 4 2 2 a a 2 the area of S . This yields a ≥ r d − r or d ≤ r + . It follows that d ≤ r + , as desired. 2 2 r r