HMMT 二月 2026 · COMB 赛 · 第 1 题

1 . Let S be the set of all vertices of this lattice. Compute the number of nondegenerate triangles with vertices in S that contain the center of the hexagon strictly in their interior. Proposed by: Jessica Wan, Karn Chutinan Answer: 6992 Solution: ©2026 HMMT Q R P O ′ P ′ R ′ Q By looking at hexagonal rings of lattice vertices around the center, the cardinality of S is 1 + 6 + 12 + 18 + 24 = 61 . ∗ Since triangles with the center O as a vertex cannot be counted, remove O from S to yield a set S containing 60 vertices. ∗ ′ ′ ′ Consider any 3 points P , Q , R in S such that no two are collinear with O . Let P , Q , R be their respective reflections about O . Relabel the points such that that P , Q , R lie on the same side of some ′ ′ ′ line passing through O , in that order. Of the triangles with vertices in { P, P , Q, Q , R, R } , the only ′ ′ ′ ones that strictly contain O in their interior are △ P Q R and △ P QR . Thus, the total number of triangles strictly containing O is twice the number of ways to pick the ′ ′ ′ unordered pairs { P, P } , { Q, Q } , and { R, R } such that no two of P , Q , and R are collinear with O . (Note that it makes no difference what order the pairs are in, nor does it matter which point in the ′ pair is P and which is P , etc.) We call such a set of points nondegenerate . ( ) 30 ′ ′ ′ There are ways to select the opposing pairs { P, P } , { Q, Q } , and { R, R } from the 30 pairs of 3 opposite points up to ordering. It remains to subtract out the number of such selections which are not nondegenerate. We count these in the following ways: ′ ′ ′ • Two of { P, P } , { Q, Q } , { R, R } lie on a line connecting the midpoints of opposite edges of the hexagon. There are 3 such lines, each of which has 2 pairs of opposing points; the last pair can ( ) 2 then be selected as any of the other 30 − 2 = 28 . This gives a count of 3 · · 28 = 84 . 2 ′ ′ ′ • Exactly two of { P, P } , { Q, Q } , { R, R } lie on the line connecting opposite vertices of the hexagon. There are 3 such lines, each of which has 4 pairs of opposing points; the last pair can then be ( ) 4 selected as any of the other 30 − 4 = 26 . This gives a count of 3 · · 26 = 468 . 2 ′ ′ ′ • All three of { P, P } , { Q, Q } , { R, R } lie on the line connecting opposite vertices of the hexagon. ( ) 4 Like above, this gives a count of 3 · = 12 . 3 This gives us 564 total degenerate cases. Our answer is therefore (( ) ) 30 2 − 564 = 6992 . 3 ©2026 HMMT