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1976 IMO Problems/Problem 3

Problem

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 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 \begin{align*} 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 \\ \end{align*} 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.

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
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