Cubes and spheres... Spheres and cubes... Luke's gone mad.

Calculus Level pending

Luke inscribed a sphere in a cube of sides $x$ . Then he inscribed another cube in the sphere. Assume he could do this process infinitely. Let the sum of the volumes of all the spheres be $V_{s_{1}}$ , the sum of the volumes of all the cubes be $V_{c_{1}}$ and the quotient $\dfrac{V_{s_{1}}}{V_{c_{1}}}$ be $Q_{1}$ . If Luke started the process by a sphere of radius $x$ instead of a cube of sides $x$ , the sum of the volumes of all the spheres would be $V_{s_{2}}$ and the sum of the volumes of all the cubes would be ${V_{c_{2}}}$ . Let $\dfrac{V_{s_{2}}}{V_{c_{2}}}$ be $Q_{2}$ . If the quotient $\dfrac{Q_{1}}{Q_{2}}$ can be written as $k^{-\tfrac{3}{2}}$ , find $k$ .

The answer is 3.

1 solution

Milly Choochoo
Jun 10, 2014

For starters, because in each of the cases (first starts with the cube, second the sphere) we are constantly iterating similar sphere/cube or cube/sphere pairs and adding them, the ratio of their volumes will be the same no matter how many iterations. Here is the proof for that.

We have a cube of volume $Q$ , and so the inscribed sphere in the cube will have a volume $kQ$ , where $k$ is some constant.

$\frac{Q}{kQ} = \frac{1}{k}$

If we now inscribe another cube/sphere pair, it will be geometrically similar to the first pair by some constant $a$ .

$\frac{Q + aQ}{kQ + akQ} = \frac{Q(1+a)}{kQ(1+a)} = \frac{1}{k}$

And if we do it any $n$ number of times, we'll still get:

$\frac{Q(1+a+a^2+\dots+a^{n-1})}{kQ(1+a+a^2+\dots+a^{n-1})} = \frac{1}{k}$

With this in mind, we'll only get the first cube/sphere pair in each case, and then arrive at our answer.

Case 1: Start off with a cube of side length $x$ , and so the volume is obviously $x^3$ . The inscribed sphere will have a radius of $\frac{x}{2}$ , and so the volume will be $\frac{\pi}{6}x^3$ . Therefore,

$Q_1 = \frac{V_{s1}}{V_{c1}} = \frac{\pi}{6}$ .

Case 2: Start off with a sphere of radius $x$ , and so the volume is $\frac{4\pi}{3}x^3$ . Now for the inscribed cube, we know for certain that the corners must be touching the sphere. Thus, the length of the diagonal of the cube must be equal to the diameter of the sphere - that is, if $s$ represents the cube's side length,

$\sqrt{s^2+s^2+s^2} = \sqrt{3}s = 2x \implies s = \frac{2}{\sqrt{3}}x$

And so the volume of the cube is then $\frac{8}{3\sqrt{3}}$ . Now we have

$Q_2 = \frac{V_{s2}}{V_{c2}} = \frac{\sqrt{3}\pi}{2}$

Finally,

$\frac{Q_1}{Q_2} = \frac{1}{3\sqrt{3}}$

Thus, $\boxed{k=3}$

I put that first proof of the needlessness of the constant iterative sphere/cube pair adding because I felt that if I just put, "Well we all know that we only need to consider one, just by logic", some people would call me out and tell me to prove it.

Milly Choochoo - 7 years ago

Cubes and spheres... Spheres and cubes... Luke's gone mad.

The answer is 3.

1 solution

0 pending reports