HMMT 二月 2004 · 冲刺赛 · 第 24 题

HMMT February 2004 — Guts Round — Problem 24

[9] We say a point is contained in a square if it is in its interior or on its boundary. Three unit squares are given in the plane such that there is a point contained in all three. Furthermore, three points A , B , C , are given, each contained in at least one of the squares. Find the maximum area of triangle ABC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 28, 2004 — GUTS ROUND 3 2

解析

We say a point is contained in a square if it is in its interior or on its boundary. Three unit squares are given in the plane such that there is a point contained in all three. Furthermore, three points A , B , C , are given, each contained in at least one of the squares. Find the maximum area of triangle ABC . √ Solution: 3 3 / 2 Let X be a point contained in all three squares. The distance from X to any point √ in any of the three squares is at most 2, the length of the diagonal of the squares. √ Therefore, triangle ABC is contained in a circle of radius 2, so its circumradius is at √ most 2. The triangle with greatest area that satisfies this property is the equilateral √ triangle in a circle of radius 2. (This can be proved, for example, by considering that the maximum altitude to any given side is obtained by putting the opposite vertex at the midpoint of its arc, and it follows that all the vertices are equidistant.) The equilateral triangle is also attainable, since making X the circumcenter and positioning the squares such that AX , BX , and CX are diagonals (of the three squares) and ABC √ is equilateral, leads to such a triangle. This triangle has area 3 3 / 2, which may be calculated, for example, using the sine formula for area applied to ABX , ACX , and √ 2 ◦ BCX , to get 3 / 2( 2) sin 120 . (See diagram, next page.) 3 2