I’m going to be out of town for a few days, so I won’t be able to post a solution for this until probably Monday, but hopefully I’ll be able to check in and at least congratulate the winner, who will receive a free copy of my Math Guide.

A student is drawing concentric circles, each with a diameter one centimeter longer than that of the one before. When he tires of this random task, he has drawn 21 circles. If the area enclosed by his circles increased by 300% from the time he drew his first circle until the time he was done, what is the radius of the smallest circle?

Put your answers in the comments—first correct response wins. Full contest rules here.

LATE UPDATE: David’s the winner—he got the correct answer first. I’ve been buried under a bunch of other things and have been slow to get the solution posted. Sorry about that. It’s below the cut.
The first task here is to figure out how much bigger the last circle is than the first. This requires a tiny bit of care. Note that the second circle’s diameter is 1 cm bigger than that of the first. The 3rd circle’s diameter is 2 cm bigger than that of the first. And so on. So the 21st circle’s diameter will be 20 cm larger than that of the first.

Of course, this means the radius of the largest circle is 10 bigger than the radius of the smallest circle, which is what we really care about if we’re talking about areas.

Let’s call the radius of the smallest circle r. The area of the smallest circle is πr2. The area of the largest circle is π(+ 10)2.

The next tricky part is the 300% increase. An increase of 300% does not mean that the new area is 3 times bigger than the old. It means the area increased by its original size 3 times over.

This is an important insight, and it makes this problem a lot easier. I’ll include the mathy way below, because I like that kind of thing and I think it’s important to show that there are multiple ways to solve this, but note that it’s a bit more involved if you don’t know that a 300% increase means you end up with 4 times what you started with.

If the new area is 4 times the original area, we can set up a simple equation:

$4(\pi r^2)=\pi(r+10)^2$
$4\pi r^2=\pi(r^2+20r+100)$
$4r^2=r^2+20r+100$

At this point, you’re going to have to use the quadratic formula, or do some factoring. You never need the quadratic formula on the SAT, so I’d recommend factoring, but really it’s whatever makes you happy. 🙂

You know your first terms have to be 3r and r, and you know one factor must contain subtraction to get -100 in the last term, etc. A little trial and error later, you get…

$(r-10)(3r+10)=0$

So your solutions are r = 10, or r = -10/3. Since a circle’s radius can’t be negative, the answer is r = 10.

That was a bit mathy, right? Here’s the other, slightly mathier way:

If the area enclosed increased by 300%, then we can use our percent change formula to solve.

$\textup{Percent\; Change}=\frac{\textup{Amount\;of\;Change}}{\textup{Original\;Value}}\times100\%$
$300\%=\frac{\pi(r+10)^2-\pi r^2}{\pi r^2}\times100\%$

Let’s clean that up a little bit by dividing both sides by 100% and knocking the π term out…

$3=\frac{(r+10)^2-r^2}{r^2}$

And simplify a little more…

$3=\frac{r^2+20r+100-r^2}{r^2}$
$3=\frac{20r+100}{r^2}$
$3r^2=20r+100$
$3r^2-20r-100=0$

And that’s the same place we ended up at above. You’ll still get r = 10.

Something tells me that with no response that I’m wrong. Looks like it is back to the drawing board.

Sorry, I have been so swamped that I haven’t posted the solution yet. David got it right first…the answer is 10cm.

TofuBean says:

I think that there is a quicker way that does not require factoring or the quadratic formula. Is this a legal way?
Total Area = Area of smallest circle x 4
pi(r + 10)^2 = 4pi(r^2)
Then we can cancel pi out
(r + 10)^2 = 4r^2
FInd square root of both sides
r + 10 = 2r
r = 10
[EDIT: Oh, I don’t think I took the +/- into account when square rooting. I guess I was lucky that the answer I came up with wasn’t the negative one!]