HMMT 十一月 2020 · 冲刺赛 · 第 11 题

HMMT November 2020 — Guts Round — Problem 11

[8] Two diameters and one radius are drawn in a circle of radius 1, dividing the circle into 5 sectors. The a largest possible area of the smallest sector can be expressed as π , where a, b are relatively prime positive b integers. Compute 100 a + b . 9

解析

[8] Two diameters and one radius are drawn in a circle of radius 1, dividing the circle into 5 sectors. a The largest possible area of the smallest sector can be expressed as π , where a, b are relatively prime b positive integers. Compute 100 a + b . Proposed by: William Qian Answer: 106 Solution: Let the two diameters split the circle into four sectors of areas A , B , A , and B , where π A + B = . Without loss of generality, let A ≤ B . 2 If our radius cuts into a sector of area A , the area of the smallest sector will be of the form min( x, A − x ). A π Note that min( A − x, x ) ≤ ≤ . 2 8 If our radius cuts into a sector of area B , then the area of the smallest sector will be of the form B π A π π A π min( A, x, B − x ) ≤ min( A, ) = min( A, − ). This equals A if A ≤ and it equals − if A ≥ . 2 4 2 6 4 2 6 π π This implies that the area of the smallest sector is maximized when A = , and we get an area of . 6 6 9