HMMT 十一月 2024 · 冲刺赛 · 第 8 题

HMMT November 2024 — Guts Round — Problem 8

[7] Derek is bored in math class and is drawing a flower. He first draws 8 points A , A , . . . , A equally 1 2 8 spaced around an enormous circle. He then draws 8 arcs outside the circle where the i th arc for i = 1 , 2 , . . . , 8 has endpoints A , A with A = A , such that all of the arcs have radius 1 and any two i i +1 9 1 consecutive arcs are tangent. Compute the perimeter of Derek’s 8 -petaled flower (not including the central circle). A A 3 2 A A 4 1 A A 5 8 A A 6 7

解析

[7] Derek is bored in math class and is drawing a flower. He first draws 8 points A , A , . . . , A equally 1 2 8 spaced around an enormous circle. He then draws 8 arcs outside the circle where the i th arc for i = 1 , 2 , . . . , 8 has endpoints A , A with A = A , such that all of the arcs have radius 1 and any i i +1 9 1 two consecutive arcs are tangent. Compute the perimeter of Derek’s 8-petaled flower. A A 3 2 A A 4 1 A A 5 8 A A 6 7 Proposed by: David Dong, Evan Chang, Henrick Rabinovitz, Jackson Dryg, Krishna Pothapragada, Srinivas Arun Answer: 10 π Solution: X 2 X 1 O 2 O A A O 3 3 2 1 X 3 X 8 A A 4 1 O O 4 8 A A 5 8 X 4 X 7 O O A A 5 7 6 7 O 6 X 5 X 6 Draw the centers O , . . . , O of the arcs, and connect these centers to form a regular octagon as shown. 1 8 For each i = 1 , . . . , 8, extend line A O to hit each arc again at X . i i i The blue arcs (arc A X for each i ) are semicircles with total length 8 π . Each red arc (arc X A i i i i +1 for each i ) has central angle equal to an exterior angle of the octagon. These exterior angles total 360 degrees, so the red arcs have total length 2 π . Hence the perimeter is 10 π .