HMMT 十一月 2021 · 团队赛 · 第 6 题

HMMT November 2021 — Team Round — Problem 6

[40] The taxicab distance between points ( x , y ) and ( x , y ) is | x − x | + | y − y | . A regular octagon 1 1 2 2 2 1 2 1 is positioned in the xy plane so that one of its sides has endpoints (0 , 0) and (1 , 0). Let S be the set of 2 all points inside the octagon whose taxicab distance from some octagon vertex is at most . The area 3 m of S can be written as , where m, n are positive integers and gcd( m, n ) = 1. Find 100 m + n . n 3

解析

[40] The taxicab distance between points ( x , y ) and ( x , y ) is | x − x | + | y − y | . A regular octagon 1 1 2 2 2 1 2 1 is positioned in the xy plane so that one of its sides has endpoints (0 , 0) and (1 , 0). Let S be the set of 2 all points inside the octagon whose taxicab distance from some octagon vertex is at most . The area 3 m of S can be written as , where m, n are positive integers and gcd( m, n ) = 1. Find 100 m + n . n Proposed by: David Vulakh Answer: 2309 Solution: F D E G C A B X Y In the taxicab metric, the set of points that lie at most d units away from some fixed point P form a square centered at P with vertices at a distance of d from P in directions parallel to the axes. The 2 diagram above depicts the intersection of an octagon with eight such squares for d = centered at its 3 √ 2 vertices. (Note that since 2 > · 2, the squares centered at adjacent vertices that are diagonal from 3 each other do not intersect.) The area of the entire shaded region is 4[ ABCDEF G ] = 4(2([ AF G ] + [ AY F ]) − [ EXY ]), which is easy to evaluate since AF G , AY F , and EXY are all 45-45-90-degree √ ( ) 2 2 1 2 4 1 23 √ triangles. Since AF = , GF = , and EX = , the desired area is 4 + − = . 3 3 9 9 36 9 3 2 3