IMO 1976 SL 6

Origin: NET

Problem

A rectangular box can be filled completely with unit cubes. If one places cubes with volume 2 in the box such that their edges are parallel to the edges of the box, one can fill exactly 40% of the box. Determine all possible (interior) sizes of the box.

Solution

Suppose a1 \leqa2 \leqa3 are the dimensions of the box. If we set bi = [ai/ 3\sqrt 2], the condition of the problem is equivalent to a1 b1 \cdot a2 b2 \cdot a3 b3 = 5. We list some values of a, b = [a/ 3\sqrt 2] and a/b: a 2 3 b 1 2 a/b 2 1.5 1.33 1.67 1.5 1.4 1.33 1.29 1.43 We note that if a > 2, then a/b \leq5/3, and if a > 5, then a/b \leq3/2. If a1 > 2, then a1 b1 \cdot a2 b2 \cdot a3 b3 < (5/3)3 < 5, a contradiction. Hence a1 = 2. If also a2 = 2, then a3/b3 = 5/4 \leq 3\sqrt 2, which is impossible. Also, if a2 \geq6, then a2 b2 \cdot a3 b3 \leq(1.5)2 < 2.5, again a contradiction. We thus have the following cases: (i) a1 = 2, a2 = 3, then a3/b3 = 5/3, which holds only if a3 = 5; (ii) a1 = 2, a2 = 4, then a3/b3 = 15/8, which is impossible; (iii) a1 = 2, a2 = 5, then a3/b3 = 3/2, which holds only if a3 = 6. The only possible sizes of the box are therefore (2, 3, 5) and (2, 5, 6).