I’m having a lot of fun playing around with some geometry drawing software this week (nyeeerd!), so I figured I’d use it again to make another “fun” 3-D problem for the weekend challenge. This is a bit tougher than you’d find on the SAT, but the underlying concepts, as always, are important for the SAT.

The prize this week inspired by my current situation: If there should ever come a time in your life where you’re trying to move from Brooklyn to the Bronx, you’ll be smart enough not to bother trying to move yourself in your Toyota Yaris and instead just hire movers. And when you do, they won’t break your stuff.

Point B is in the center of the top face of the cube in the figure above, and point A is one of the cube’s vertices. If the distance between points A and B is d, then what is the cube’s volume in terms of d?

Put your answers in the comments, and I’ll post the solution Monday. Good luck!

UPDATE: Nice work, JD. You were wrong at first, but you corrected yourself before I did. Solution below the cut.

The most difficult 3-D problems you’ll come across on the SAT will require you to work with right triangles, usually on the same plane as one of the solid’s sides, and then on a plane that cuts through the solid. In this question, you’re going to want to make a right triangle whose hypotenuse is AB (A.K.A. d).

To do so, we’re going to have to first figure out the distance from B to the cube’s corner. Let’s assume the cube has sides of length c:

Since we’re so good with right triangles, we know that the diagonal of the square on the right is c2, and since B is the midpoint of that segment, the distance from the corner to B must be c2/2. Now we know the lengths of both legs of the right triangle we actually care about, the one with hypotenuse d. We can use it to solve for d in terms of c. (“But wait,” you say, “aren’t we looking for the volume in terms of d?” Yes, we are. We’ll get there. Promise.)

$c^2&space;+&space;\left&space;(\frac{c\sqrt{2}}{2}\right&space;)^2&space;=&space;d^2$
$c^2&space;+&space;\frac{2c^2}{4}&space;=&space;d^2$
$c^2&space;+&space;\frac{c^2}{2}&space;=&space;d^2$
$\frac{3c^2}{2}&space;=&space;d^2$
$\frac{c\sqrt3}{\sqrt2}&space;=&space;d$

Now that we’ve got d in terms of c, we can put c in terms of d, and cube it to get the volume. Wahoo!

$c&space;=&space;\frac{d\sqrt&space;2}{\sqrt&space;3}$
$c^3&space;=&space;\left&space;(&space;\frac{d\sqrt&space;2}{\sqrt&space;3}\right&space;)^3$
$c^3=\frac{2d^3\sqrt&space;2}{3\sqrt&space;3}$

Some people like to stop here, and others like to rationalize the denominator (as JD did in his response). The SAT has been known to swing both ways.

$c^3=\frac{2d^3\sqrt&space;2}{3\sqrt&space;3}\times\frac{\sqrt3}{\sqrt3}$
$c^3=\frac{2d^3\sqrt6}{9}$

Donezo.

Note: you can check your work here by plugging in. Say c = 2, so the volume of the cube is 8. That would make d = 2√3/√2. Plug that in for d in our solution. Does it give you 8? Yup.