**Đây là một trong năm bài thi giỏi toán Canada năm 2011 vừa mới thi cách đây hai tuần mà cộng đồng toán học trên mạng chưa có bài giải. Bốn bài kia thì họ đã đăng bài giải. Mỗi bài họ đăng ít nhất năm cách giải khác nhau. Vậy đây là bài toán khó:**

**Amy has divided a square up into finitely many white and red rectangles, each with sides parallel to the sides of the square. Within each white rectangle, she writes down its width divided by its height. Within each red rectangle, she writes down its height divided by its width. Finally, she calculates x, the sum of these numbers. If the total area of the white rectangles equals the total area of the red rectangles, what is the smallest possible value of x?**

**Solution**

Apply the method of move and cut, if you will. We can always move the scattered rectangles that have the same color into one-half of a square. The total area of these rectangles will fill up one-half of the square. If not, cut and fill in.

Now within each half of the square we prove that the scattered rectangles now combined together have their sums of the numbers greater than the combined number if we cut them into rectangles with the vertical divisors (or vertical lines) which is 2.0 (the width is twice the height for the first half of the square). If any of the rectangle is not cut this way the sum of the numbers doubled. So the minimum is 2.0 for the top half of the square.

Doing the same for the other bottom half of the square that now has all rectangles with the same color filled up. This time we cut them with horizontal lines. With the same argument, we get the minimum number to be the height divided by the width which is 0.5.

The sum, therefore, is the addition of two minimum numbers which comes out to be the minimum that we are looking for, or 2.0 + 0.5 = 2.5.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2230157&sid=883d067f51a5af44c42ecf013d7953be#p2230157