HMMT 二月 2019 · 冲刺赛 · 第 25 题
HMMT February 2019 — Guts Round — Problem 25
题目详情
- [ 15 ] A 5 by 5 grid of unit squares is partitioned into 5 pairwise incongruent rectangles with sides lying on the gridlines. Find the maximum possible value of the product of their areas.
解析
- [ 15 ] A 5 by 5 grid of unit squares is partitioned into 5 pairwise incongruent rectangles with sides lying on the gridlines. Find the maximum possible value of the product of their areas. Proposed by: Yuan Yao Answer: 2304 The greatest possible value for the product is 3 · 4 · 4 · 6 · 8 = 2304 , achieved when the rectangles are 3 × 1 , 1 × 4 , 2 × 2 , 2 × 3 , 4 × 2 . To see that this is possible, orient these rectangles so that the first number is the horizontal dimension and the second number is the vertical dimension. Then, place the bottom-left corners of these rectangles at (2 , 4) , (4 , 0) , (2 , 2) , (0 , 2) , (0 , 0) respectively on the grid. We will now prove that no larger product can be achieved. Suppose that there is at least one rectangle 4 2 of area at most 2 . Then the product is at most 2 · 5 . 75 = 2 · 33 . 0625 < 2 · 1100 = 2200 by AM-GM. Now 4 suppose that there is at least one rectangle of area at least 9 . Then the product is at most 9 · 4 = 2304 by AM-GM. (Neither of these is tight, since you cannot have non-integer areas, nor can you have four rectangles all of area 4 . ) Now consider the last possibility that is not covered by any of the above: that there are no rectangles of size at most 2 and no rectangles of area at least 9 . There can be at most one rectangle of area 3 , 5 , 6 , 8 each, at most two rectangles of area 4 , and no rectangles of area 7 . The only way to achieve a sum of 25 with these constraints is 3 , 4 , 4 , 6 , 8 , which produces a product of 2304 . We have shown through the earlier cases that a larger product cannot be achieved, so this is indeed the maximum.