1976 IMO Problems/Problem 3
Problem
A box whose shape is a parallelepiped can be completely filled with cubes of side If we put in it the maximum possible number of cubes, each of volume , with the sides parallel to those of the box, then exactly percent from the volume of the box is occupied. Determine the possible dimensions of the box.
Solution
We name a,b,c the sides of the parallelepiped, which are positive integers. We also put
\[x = \left\lfloor\frac{a}{\sqrt[3]{2}}\right\rfloor \ \ \ \ y = \left\lfloor\frac{b}{\sqrt[3]{2}}\right\rfloor \ \ \ \ z = \left\lfloor\frac{c}{\sqrt[3]{2}}\right\rfloor \ \ \ \\] (Error compiling LaTeX. ! LaTeX Error: \begin{equation*} on input line 20 ended by \end{document}.)
It is clear that is the maximal number of cubes with sides of length that can be put into the parallelepiped with sides parallels to the sides of the box. Hence the corresponding volume is . We need , hence We give the values of and for . The same table is valid for and . By simple inspection we obtain two solutions of : and . We now show that they are the only solutions.
We can assume . So necessarily . Note that the definition of implies hence If then and since . So we have only left the cases and . But for we have and so necessarily and . It follows
Note that the definitions of imply
Moreover we have from (2) and from that
If then and we would have , which contradicts .
On the other hand, if then and since as . So we have only left the cases . But for we have and for we have and so necessarily and ()
So we arrive finally at and . If then and
since . On the other hand, for there are the only two possible values and which yield the known solutions.
See also
1976 IMO (Problems) • Resources) | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
All IMO Problems and Solutions |