HMMT 二月 2026 · 冲刺赛 · 第 32 题
HMMT February 2026 — Guts Round — Problem 32
题目详情
- [16] Kelvin the frog starts at the center of a regular hexagon ABCDEF with side length 100 , facing ◦ towards A . He hops forward an integer distance between 0 and 200 units, inclusive, then turns 120 clockwise. He repeats this process two more times (possibly jumping different distances), ending up within hexagon ABCDEF (possibly on its boundary). Compute the number of distinct paths he could have taken. ©2026 HMMT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT February 2026, February 14, 2026 — GUTS ROUND Organization Team Team ID#
解析
- [16] Kelvin the frog starts at the center of a regular hexagon ABCDEF with side length 100 , facing ◦ towards A . He hops forward an integer distance between 0 and 200 units, inclusive, then turns 120 clockwise. He repeats this process two more times (possibly jumping different distances), ending up within hexagon ABCDEF (possibly on its boundary). Compute the number of distinct paths he could have taken. Proposed by: Derek Liu 4 4 Answer: 101 − 100 = 4060401 Solution: Kelvin the frog essentially jumps along a unit triangular grid; any point within 100 steps of the origin along this grid could be his ending spot. We can view the triangular grid as the 3 -dimensional lattice points projected onto the plane x + y + z = 0 (and scaled accordingly). The hexagon is then the projection of the region satisfying | x − y | ≤ 100 , | y − z | ≤ 100 , and | z − x | ≤ 100 (in other words, the region in which no two coordinates differ by more than 100 ). If Kelvin’s jumps have lengths a , b , and c in that order, then his ending position will be the projection of ( a, b, c ) . To count the number of such points, we casework on max( a, b, c ) . 3 3 • If max( a, b, c ) ≤ 100 , then there are 101 possible points, as any point in the region [0 , 100] has no two coordinates differ by more than 100 . • If max( a, b, c ) = m > 100 , then each of a , b , and c must be at least m − 100 . Since their maximum 3 must still be exactly m , the points that satisfy this are all points in [ m − 100 , m ] minus all 3 points in [ m − 100 , m − 1] . There are 100 possible values for m ( 101 through 200 ), so there are 3 3 100(101 − 100 ) points in this case. ©2026 HMMT The answer is thus 3 3 3 4 4 101 + 100(101 − 100 ) = 101 − 100 = 4060401 .