We put in it the maximum possible number of cubes

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We put in it the maximum possible number of cubes

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Source: IMO ShortList, Netherlands 2, IMO 1976, Day 1, Problem 3

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orl

#1 h Nov 12, 2005, 8:30 PM • 3 Y

Y by Adventure10, megarnie, Mango247

A box whose shape is a parallelepiped can be completely filled with cubes of side $1.$ If we put in it the maximum possible number of cubes, each of volume $2$ , with the sides parallel to those of the box, then exactly $40$ percent of the volume of the box is occupied. Determine the possible dimensions of the box.

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cval

#2 h Nov 6, 2008, 2:06 AM • 5 Y

Y by viperstrike, Adventure10, and 3 other users

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 \ \ \$
It is clear that $xyz$ is the maximal number of cubes with sides of length $\sqrt[3]{2}$ that
can be put into the parallelepiped with sides parallels to the sides of the box.
Hence the corresponding volume is $V_2=2\cdot xyz$ . We need $V_2=0.4\cdot V_1=0.4\cdot abc$ ,
hence $\frac ax\cdot \frac by\cdot \frac cz=5\ \ \ \ \ \ \ \ (1)$
We give the values of $x$ and $a/x$ for $a=1,\dots ,8$ . The same table is valid for $b,y$ and $c,z$ .
$\begin{tabular}{|c|c|c|} \hline a & x & a/x \\ \hline 1 & 0 & - \\ \hline 2 & 1 & 2 \\ \hline 3 & 2 & 3/2 \\ \hline 4 & 3 & 4/3 \\ \hline 5 & 3 & 5/3 \\ \hline 6 & 4 & 3/2 \\ \hline 7 & 5 & 7/5 \\ \hline 8 & 6 & 4/3 \\ \hline \end{tabular}$
By simple inspection we obtain two solutions of $(1)$ : $\{a,b,c\}=\{2,5,3\}$ and $\{a,b,c\}=\{2,5,6\}$ .
We now show that they are the only solutions.

We can assume $\frac ax\ge \frac by \ge \frac cz$ . So necessarily $\frac ax\ge \sqrt[3]{5}$ . Note that
the definition of $x$ implies $x< a/\sqrt[3]2 < x+1,$
hence $\sqrt[3]2< a/x < \sqrt[3]2(1+\frac 1x)$
If $a\ge 4$ then $x\ge 3$ and $\frac ax<\sqrt[3]2(1+\frac 1x)\le \sqrt[3]2(\frac 43)<\sqrt[3]5$
since $2\cdot \frac {4^3}{3^3}<5$ . So we have only left the cases $a=2$ and $a=3$ . But for $a=3$
we have $a/x=3/2<\sqrt[3]5$ and so necessarily $a=2$ and $a/x=2$ .
It follows
$\frac by \cdot \frac cz =\frac 52 \ \ \ \ \ \ (2)$

Note that the definitions of $y,z$ imply $y< b/\sqrt[3]2 < y+1,\ \ \textrm{and} \ \ z< c/\sqrt[3]2 < z+1.\ \ \ \ (3)$
Moreover we have from (2) and from $b/y\ge c/z$ that
$\frac by \ge \sqrt{5/2}\ \ \ \ \ (4)$

If $b=2$ then $b/y=2$ and we would have $c/z=5/4<\sqrt[3]2$ , which contradicts $(3)$ .

On the other hand, if $b>5$ then $y>4$ and $\frac by<\sqrt[3]2(1+\frac 1y)\le \sqrt[3]2(\frac 54)<\sqrt{5/2}$
since $2^2\cdot \frac {5^6}{4^6}<\frac{5^3}{2^3}$ as $5^3<2^7$ . So we have only left the
cases $b=3,4,5$ . But for $b=3$ we have $b/y=3/2<\sqrt{5/2}$ and for $b=4$ we have
$b/y=4/3<\sqrt{5/2}$ and so necessarily $b=5$ and $b/y=5/3$ ( $>\sqrt{5/2}$ )

So we arrive finally at $a=2,b=5$ and $c/z=3/2$ . If $c\ge 8$
then $z\ge 6$ and $\frac cz<\sqrt[3]2(1+\frac 1z)\le \sqrt[3]2(\frac 76)<\frac 32$
since $2\cdot \frac {7^3}{6^3}<\frac{3^3}{2^3}$ . On the other hand, for $c\le 7$ there are the only two possible values
$c=3$ and $c=6$ which yield the known solutions.

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#3 h Apr 27, 2022, 8:29 PM • 3 Y

Y by Mango247, Mango247, Mango247

How can you fill a parallelpiped completely with cubes of dimension 1? Isn't this only possible when the faces are all either parallel or perpendicular? (i.e. a rectangular prism). Not sure why the wording includes "parallelpiped" rather than "rectangular prism." Am I misunderstanding something here?

This post has been edited 2 times. Last edited by arvind_r, Apr 27, 2022, 8:40 PM

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jred

#4 h Nov 15, 2022, 7:59 AM

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The official wording is a rectangular box.

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