IMO 1974 SL 2

Problem

I 2 (POL 1) Prove that the squares with sides 1/1, 1/2, 1/3, . . . may be put into the square with side 3/2 in such a way that no two of them have any interior point in common.

Solution

We denote by qi the square with side 1 i . Let us divide the big square into rectangles ri by parallel lines, where the size of ri is 3 2 \times 1 2i for i = 2, 3, . . . and 2 \times 1 for i = 1 (this can be done because 1 + \infty i=2 2i = 2). In rectangle r1, one can put the squares q1, q2, q3, as is done on the figure. Also, since 2i + \cdot \cdot \cdot + 2i+1−1 < 2i \cdot 1 2i = 1 < 3 2, in each ri, i \geq2, one can put q2i, . . . , q2i+1−1. This completes the proof. q1 q2 q3 q4 q5q6q7 q8, . . . , q15 Remark. It can be shown that the squares q1, q2 cannot fit in any square of side less than 3 2.