HMMT 二月 2026 · 团队赛 · 第 1 题

HMMT February 2026 — Team Round — Problem 1

[20] Let P be a regular hexagon with center O . Distinct points A and B lie on the boundary of P . The segments OA and OB divide the interior of P into two polygons R and R . Let a and a be 1 2 1 2 the areas of R and R respectively, and let p and p be the perimeters of R and R respectively. 1 2 1 2 1 2 Given that a /p = a /p , prove that a = a . 1 1 2 2 1 2

解析

[20] Let P be a regular hexagon with center O . Distinct points A and B lie on the boundary of P . The segments OA and OB divide the interior of P into two polygons R and R . Let a and a be 1 2 1 2 the areas of R and R respectively, and let p and p be the perimeters of R and R respectively. 1 2 1 2 1 2 Given that a /p = a /p , prove that a = a . 1 1 2 2 1 2 Proposed by: Linus Yifeng Tang Answer: N/A Solution: l 1 r r r A O r B Let r be the inradius of the polygon. Let • l be the length of the shared boundary of R and P , 1 1 • l be the length of the shared boundary of R and P , and 2 2 • k be the length of the shared boundary of R and R , i.e., P A + P B . 1 2 rl 1 Then the area of R is a = , as we can divide R into triangles each of height r with total base 1 1 1 2 rl 2 length l . Likewise, the area of R is a = . Meanwhile, the perimeters of R and R are p = l + k 1 2 2 1 2 1 1 2 and p = l + k , respectively. 2 2 The condition a /p = a /p thus implies that 1 1 2 2 rl / 2 rl / 2 1 2 = . l + k l + k 1 2 Dividing both sides by r/ 2 and cross-multiplying, we get l ( l + k ) = l ( l + k ) , 1 2 2 1 which implies that l k = l k , so l = l . Hence, a = rl / 2 = rl / 2 = a , as desired. 1 2 1 2 1 1 2 2