Difference between revisions of "1976 IMO Problems/Problem 3"
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== Problem == | == Problem == | ||
| − | A box whose shape is a parallelepiped can be completely filled with cubes of side <math>1.</math> If we put in it the maximum possible number of cubes, each | + | A box whose shape is a parallelepiped can be completely filled with cubes of side <math>1.</math> If we put in it the maximum possible number of cubes, each of volume <math>2</math>, with the sides parallel to those of the box, then exactly <math>40</math> percent from the volume of the box is occupied. Determine the possible dimensions of the box. |
== Solution == | == Solution == | ||
| − | {{solution}} | + | The first statement tells us that the box has integer dimensions. Let the dimensions be <math>a</math>, <math>b</math>, and <math>c</math>, where they're each greater than 1. The second statement tells us that if <math>\lfloor \dfrac{a}{\sqrt[3]{2}} \rfloor *\lfloor \dfrac{b}{\sqrt[3]{2}} \rfloor *\lfloor \dfrac{c}{\sqrt[3]{2}} \rfloor</math> boxes with side length <math>\sqrt[3]{2}</math> are put into the box, that takes up <math>\dfrac{2abc}{5}</math> units of area. Thus <math>\lfloor \dfrac{a}{\sqrt[3]{2}} \rfloor *\lfloor \dfrac{b}{\sqrt[3]{2}} \rfloor *\lfloor \dfrac{c}{\sqrt[3]{2}} \rfloor=\dfrac{abc}{5}</math>. Now the range of values that the LHS can take up is <math>(\dfrac{a}{\sqrt[3]{2}}-1) (\dfrac{b}{\sqrt[3]{2}}-1) (\dfrac{c}{\sqrt[3]{2}}-1)</math> to <math>\dfrac{abc}{2}</math>, exclusive, since the fractions are not integers. Therefore we must find all <math>a</math>, <math>b</math>, and <math>c</math> such that |
| + | |||
| + | <cmath>(\dfrac{a}{\sqrt[3]{2}}-1) (\dfrac{b}{\sqrt[3]{2}}-1) (\dfrac{c}{\sqrt[3]{2}}-1)<\dfrac{abc}{5}<\dfrac{abc}{2}.</cmath> | ||
| + | |||
| + | The <math>\dfrac{abc}{2}</math> is redundant, so we can eliminate that. We simplify the LHS: | ||
| + | |||
| + | <cmath>\dfrac{(a-\sqrt[3]{2})(b-\sqrt[3]{2})(c-\sqrt[3]{2})}{2}<\dfrac{abc}{5}</cmath> | ||
| + | |||
| + | <cmath>\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}</cmath> | ||
| + | |||
| + | <cmath>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</cmath> | ||
| + | |||
| + | {{incomplete|solution}} | ||
| + | |||
== See also == | == See also == | ||
{{IMO box|year=1976|num-b=2|num-a=4}} | {{IMO box|year=1976|num-b=2|num-a=4}} | ||
Revision as of 11:30, 10 July 2008
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
The first statement tells us that the box has integer dimensions. Let the dimensions be 
, 
, and 
, where they're each greater than 1. The second statement tells us that if ![$\lfloor \dfrac{a}{\sqrt[3]{2}} \rfloor *\lfloor \dfrac{b}{\sqrt[3]{2}} \rfloor *\lfloor \dfrac{c}{\sqrt[3]{2}} \rfloor$](http://latex.artofproblemsolving.com/8/d/1/8d171264e6aa4a63948e443546e60a1750e1595c.png)
boxes with side length ![$\sqrt[3]{2}$](http://latex.artofproblemsolving.com/9/f/d/9fdfe572d1673f00f286d5ed753942d18f401cbb.png)
are put into the box, that takes up 
units of area. Thus ![$\lfloor \dfrac{a}{\sqrt[3]{2}} \rfloor *\lfloor \dfrac{b}{\sqrt[3]{2}} \rfloor *\lfloor \dfrac{c}{\sqrt[3]{2}} \rfloor=\dfrac{abc}{5}$](http://latex.artofproblemsolving.com/a/e/4/ae4a6dd209b4cd207d7338c058ff3063c1c07cb3.png)
. Now the range of values that the LHS can take up is ![$(\dfrac{a}{\sqrt[3]{2}}-1) (\dfrac{b}{\sqrt[3]{2}}-1) (\dfrac{c}{\sqrt[3]{2}}-1)$](http://latex.artofproblemsolving.com/8/0/5/8051b545f6b3ad3ef4f8c4f5e4968784afe4e45e.png)
to 
, exclusive, since the fractions are not integers. Therefore we must find all , , and such that
![\[(\dfrac{a}{\sqrt[3]{2}}-1) (\dfrac{b}{\sqrt[3]{2}}-1) (\dfrac{c}{\sqrt[3]{2}}-1)<\dfrac{abc}{5}<\dfrac{abc}{2}.\]](http://latex.artofproblemsolving.com/0/d/0/0d01652edc1f5a97153dbf77509e42ac6d01f0ad.png)
The is redundant, so we can eliminate that. We simplify the LHS:
![\[\dfrac{(a-\sqrt[3]{2})(b-\sqrt[3]{2})(c-\sqrt[3]{2})}{2}<\dfrac{abc}{5}\]](http://latex.artofproblemsolving.com/f/e/3/fe39b62d05c96d1e2991b0592c62480746dcd734.png)
![\[\dfrac{abc-\sqrt[3]{2}ab-\sqrt[3]{2}bc-\sqrt[3]{2}ac+\sqrt[3]{4}a+\sqrt[3]{4}b+\sqrt[3]{4}c-2}{2}<\dfrac{abc}{5}\]](http://latex.artofproblemsolving.com/7/f/c/7fce78d1980843b80f0747f125471b1297fa7f4e.png)
![\[3abc-5\sqrt[3]{2}ab -5\sqrt[3]{2}bc -5\sqrt[3]{2}ac+5\sqrt[3]{4}a+5\sqrt[3]{4}b+5\sqrt[3]{4}c-10<0\]](http://latex.artofproblemsolving.com/7/3/8/738b2ea2611b615e0a8eedc0f7993557e28144aa.png)
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 | ||
