Chapter 12.P, Problem 1P

Each edge of a cubical box has length 1 m. The box contains nine spherical balls with the same radius r. The center of one ball is at the center of the cube and it touches the other eight balls. Each of the other eight balls touches three sides of the box. Thus the balls are tightly packed in the box (see the figure). Find r. (If you have trouble with this problem, read about the problem-solving strategy entitled Use Analogy on page 98.)

To determine

To find:

The radius $r$

Answer to Problem 1P

Solution:

The radius of each ball is

$\mathrm{r}\mathrm{ }=\sqrt{3}-\frac{3}{2}m$

Explanation of Solution

1) Concept:

We create an analogous problem in two dimensions and use the strategy in two dimension to solve the problem in three dimension.

2) Calculation:

Since three dimensional situations are often diffucult to visualize and work with, we create an analogous problem in two dimensions and use the strategy in two dimension to solve the problem in three dimension.

Let us try to find the analogous problem in two dimensions. The analogue of a cube is a square, and the analogue of a sphere is a circle.Thus, a simillar problem in two dimensions is as follows:

What is r, if five circles with the same radius r are contained in a square of side 1m so that the circles touch each other and four of the circles touch two sides of square, as shown in the figure,

By pythagorean Theorem, the diagonal of the square is $\sqrt{2}$

Also from the figure, the diagonal of the square is $4r+2x$

Therefore, $4r+2x=\sqrt{2}$

But, $x$ is the diagonal of the smallar square of side $r$.

Therefore,

$x=\sqrt{2 }r$

Therefore,

$4r+2x=\sqrt{2}$ implies that

$4r+2\sqrt{2}\mathrm{ }\mathrm{r}\mathrm{ }=\sqrt{2}$

That is, $\left(4+2\sqrt{2}\right)\mathrm{ }\mathrm{r}\mathrm{ }=\sqrt{2}$

Therefore,

$\mathrm{ }\mathrm{r}\mathrm{ }=\frac{\sqrt{2}}{\left(4+2\sqrt{2}\right)}$

Let’s use this idea to solve the original three dimensional problem.

The diagonal of the cube is $\sqrt{1^2+1^2+1^2}=\sqrt{3}$

And the diagonal of the cube is also $4r+2x$.

Therefore,$4r+2x=\sqrt{3}$

 where $x$ is the diagonal of the cube with edge $r$.

Therefore,

$x=\sqrt{r^2+r^2+r^2}=\sqrt{3} r$

Therefore,

$4r+2x=\sqrt{3}$ implies that $4r+2\sqrt{3} r=\sqrt{3}$

That is, $\left(4+2 \sqrt{3}\right)\mathrm{ }\mathrm{r}\mathrm{ }=\sqrt{3}$

Therefore,

$\mathrm{ }\mathrm{r}\mathrm{ }=\frac{\sqrt{3}}{\left(4+2\sqrt{3}\right)}$

By rationalization,

$\mathrm{r}\mathrm{ }=\frac{2\sqrt{3}-3}{2}$

Therefore, the radius of each ball is

$\mathrm{r}\mathrm{ }=\sqrt{3}-\frac{3}{2}m$

Conclusion:

The radius of each ball is

$\mathrm{r}\mathrm{ }=\sqrt{3}-\frac{3}{2}m$