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

The first statement tells us that the box has integer dimensions. Let the dimensions be $a$ , $b$ , and $c$ , where they're each greater than 1. The second statement tells us that if $\lfloor \dfrac{a}{\sqrt[3]{2}} \rfloor *\lfloor \dfrac{b}{\sqrt[3]{2}} \rfloor *\lfloor \dfrac{c}{\sqrt[3]{2}} \rfloor$ boxes with side length $\sqrt[3]{2}$ are put into the box, that takes up $\dfrac{2abc}{5}$ units of area. Thus $\lfloor \dfrac{a}{\sqrt[3]{2}} \rfloor *\lfloor \dfrac{b}{\sqrt[3]{2}} \rfloor *\lfloor \dfrac{c}{\sqrt[3]{2}} \rfloor=\dfrac{abc}{5}$ . Now the range of values that the LHS can take up is $(\dfrac{a}{\sqrt[3]{2}}-1) (\dfrac{b}{\sqrt[3]{2}}-1) (\dfrac{c}{\sqrt[3]{2}}-1)$ to $\dfrac{abc}{2}$ , exclusive, since the fractions are not integers. Therefore we must find all $a$ , $b$ , and $c$ such that

$(\dfrac{a}{\sqrt[3]{2}}-1) (\dfrac{b}{\sqrt[3]{2}}-1) (\dfrac{c}{\sqrt[3]{2}}-1)<\dfrac{abc}{5}<\dfrac{abc}{2}.$

The $\dfrac{abc}{2}$ is redundant, so we can eliminate that. We simplify the LHS:

$\dfrac{(a-\sqrt[3]{2})(b-\sqrt[3]{2})(c-\sqrt[3]{2})}{2}<\dfrac{abc}{5}$

$\dfrac{abc-\sqrt[3]{2}ab-\sqrt[3]{2}bc-\sqrt[3]{2}ac+\sqrt[3]{4}a+\sqrt[3]{4}b+\sqrt[3]{4}c-2}{2}<\dfrac{abc}{5}$

$3abc-5\sqrt[3]{2}ab -5\sqrt[3]{2}bc -5\sqrt[3]{2}ac+5\sqrt[3]{4}a+5\sqrt[3]{4}b+5\sqrt[3]{4}c-10<0$

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1976 IMO (Problems) • Resources
Preceded by Problem 2	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 4
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1976 IMO Problems/Problem 3

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