HMMT 二月 2026 · 冲刺赛 · 第 3 题

HMMT February 2026 — Guts Round — Problem 3

[5] A rectangle with length 4 and height 2 is placed in the bottom left corner of a square with side length

解析

[5] A rectangle with length 4 and height 2 is placed in the bottom left corner of a square with side length 10 . Regions A and B are separated by a line segment drawn from the rectangle’s upper right corner to some point on the boundary of the square. Given that regions A and B have the same perimeter, compute the positive difference between the areas of regions A and B . A 10 B 2 4 Proposed by: Sebastian Attlan Answer: 4 Solution: Let the line segment splitting A and B have length ℓ , and let it split the perimeter of the square that is not shared by the rectangle into lengths x and y , which are on the borders of A and B , respectively. Then x + y = 8 + 10 + 10 + 6 = 34 . The perimeter of A is 4 + x + ℓ , while the perimeter of B is 2 + y + ℓ . Then 4 + x + ℓ = 2 + y + ℓ, so y = x + 2 . Combining this with the condition x + y = 34 gives us ( x, y ) = (16 , 18) . Knowing this, the length of segments in the figure can be calculated as below: 8 2 8 A 10 B 2 4 6 Hence, the area of A (a trapezoid) is 4 + 8 · 8 = 48 2 while the area of B (a rectangle missing a triangle in its top left corner) is 4 · 8 6 · 10 − = 44 , 2 ©2026 HMMT so their difference is 48 − 44 = 4 .