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. Unknown error_msg)
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 atand
. 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 |