HMMT 二月 2020 · 几何 · 第 3 题

HMMT February 2020 — Geometry — Problem 3

Consider the L-shaped tromino below with 3 attached unit squares. It is cut into exactly two pieces of equal area by a line segment whose endpoints lie on the perimeter of the tromino. What is the longest possible length of the line segment?

解析

Consider the L-shaped tromino below with 3 attached unit squares. It is cut into exactly two pieces of equal area by a line segment whose endpoints lie on the perimeter of the tromino. What is the longest possible length of the line segment? Proposed by: James Lin 5 Answer: 2 Solution: Let the line segment have endpoints A and B . Without loss of generality, let A lie below √ √ the lines x + y = 3 (as this will cause B to be above the line x + y = 3) and y = x (we can reflect about y = x to get the rest of the cases): B A Now, note that as A ranges from (0 , 0) to (1 . 5 , 0), B will range from (1 , 1) to (1 , 2) to (0 , 2), as indicated by the red line segments. Note that these line segments are contained in a rectangle bounded by x = 0, √ 5 2 2 y = 0, x = 1 . 5, and y = 2, and so the longest line segment in this case has length 2 + 1 . 5 = . 2 √ 3 As for the rest of the cases, as A = ( x, 0) ranges from (1 . 5 , 0) to ( 3 , 0), B will be the point (0 , ), so x √ √ 9 2 2 it suffices to maximize x + given 1 . 5 ≤ x ≤ 3. Note that the further away x is from 3, the 2 x 9 2 larger x + gets, and so the maximum is achieved when x = 1 . 5, which gives us the same length as 2 x before. 5 Thus, the maximum length is . 2