HMMT 二月 2026 · COMB 赛 · 第 7 题
HMMT February 2026 — COMB Round — Problem 7
题目详情
- Let S be the set of vertices of a right prism whose bases are regular decagons A A . . . A and 1 2 10 B B . . . B . A plane, not passing through any vertex of S , partitions the vertices of S into two sets, 1 2 10 one of which is M . Compute the number of possible sets M that can arise out of such a partition.
解析
- Let S be the set of vertices of a right prism whose bases are regular decagons A A . . . A and 1 2 10 B B . . . B . A plane, not passing through any vertex of S , partitions the vertices of S into two sets, 1 2 10 one of which is M . Compute the number of possible sets M that can arise out of such a partition. Proposed by: Tiger Zhang Answer: 1574 Solution: We first deal with edge cases. Let A = { A , . . . , A } and B = { B , . . . , B } be the two 1 10 1 10 sets of 10 vertices that lie on a single base. • The empty set, all 20 vertices, A , and B are all possibilities for M . • If M contains all of A and between 1 and 9 vertices of B (inclusive), then M can contain any consecutive sequences of vertices from B . For each n between 1 and 9 , there are 10 such sequences of n vertices, so there are 9 · 10 = 90 such sets M . • Likewise, there are 90 possible sets containing all of B and between 1 and 9 vertices of A . • The complements of the 180 sets from the previous cases are also possibilities for M . This gives us 364 edge cases. Henceforth, we assume M contains between 1 and 9 vertices on each base. Let the plane cut A A . . . A at line ℓ and B B . . . B at line ℓ . Note that M consists of the 1 2 10 1 1 2 10 2 vertices of A lying on one side of ℓ (call this set X ), along with the vertices of B lying on the 1 corresponding side of ℓ (call this set Y ). Moreover, ℓ and ℓ are parallel, and any parallel lines ℓ 2 1 2 1 and ℓ are achievable by some plane. 2 Therefore, we may make the following simplification to the problem: we project the prism onto a single decagonal base A A . . . A , and we count the number of pairs ( X, Y ) of subsets of vertices of this 1 2 10 decagon such that X and Y can be cut by parallel lines ℓ and ℓ . 1 2 ℓ 1 Y A 1 X P A 1 A 2 10 ℓ 2 Q 1 A A 3 9 P 2 A A 4 8 A A 5 7 Q 2 A 6 ©2026 HMMT Let Γ be the circumcircle of A A . . . A . Let ℓ intersect Γ at P and P , and let ℓ intersect Γ at 1 2 10 1 1 2 2 Q and Q (with P Q < P Q ). Then X is precisely the set of vertices of A A . . . A lying on arc 1 2 1 1 2 1 1 2 10 P P , and Y is precisely the set of vertices of A A . . . A lying on arc Q Q (with the two arcs going 1 2 1 2 10 1 2 in the same direction). The key claim is as follows. Claim 1. The positive difference between the numbers of vertices of A lying on arc P Q and lying 1 1 on arc P Q is at most 1 . 2 2 Proof. Suppose for the sake of contradiction that this is not the case. Without loss of generality, assume k vertices of A lie on arc P Q , and at least k + 2 vertices of A lie on arc P Q . Then arc 1 1 2 2 2 π 2 π P Q has measure less than ( k + 1) , while arc P Q has measure greater than ( k + 1) . But since 1 1 2 2 10 10 ℓ and ℓ are parallel, these arcs must have the same measure, contradiction. 1 2 Now we are ready to finish the problem. We will count the number of possible pairs ( X, Y ) given their sizes x = | X | and y = | Y | . We casework by their parities. In both cases, we assume without loss of generality that x ≥ y and X = { A , A , . . . , A } ; the same arguments apply if y ≥ x or X is rotated. 1 2 x • If x and y have the same parity, let x − y = 2 k , the total number of vertices of A on arcs P Q 1 1 and P Q . Our key claim implies that each arc must have k of these points, so we must have 2 2 Y = { A , . . . , A } , which gives 1 possibility. k +1 x − k • If x and y have opposite parity, let x − y = 2 k − 1 , the total number of vertices of A on arcs P Q 1 1 and P Q . Our key claim implies that one of these arcs must have k − 1 points and the other 2 2 must have k , so we must have Y = { A , . . . , A } or Y = { A , . . . , A } , which gives 2 k x − k k +1 x − k +1 possibilities. There are (5)(5) + (4)(4) = 41 cases where x and y have the same parity, and (5)(4) + (4)(5) = 40 cases where x and y have opposite parities. Accounting for the fact that there are 10 possible sets X for each x , we get a total of 10(41 · 1 + 40 · 2) = 1210 possibilities. Adding back the edges cases, we get a final answer of 364 + 1210 = 1574 .