Math Help - Problem with infinite geometric series and circles tangent to each other + a square

1. Problem with infinite geometric series and circles tangent to each other + a square

Hey I've got a problem I'm trying to do to brush up on my skills (exam season...):

Consider a circle inscribed in a square with each side of the square having a length of 1 unit. Now at each the four corners, additional circles are constructed that are externally tangent to the original circle and tangent to the square. This is continued forever; what is the total area of all circles?

I think I can do the part of computing the by making the formula for the series fairly easy. Obviously, the first (and biggest) circle has a radius of 0.5.

My problem is: how do I find the size/radius/area of the next triangle(s) - the ones in the corner?

Thanks, any help is greatly appreciated!

2. Hint: Use Pythagoras to evaluate the length of the diagonal of the square.

3. That was actually the first thing I tried, but I couldn't get the next circle's size from that:

1^2 + 1^2 = c^2
2 = c^2
\sqrt{2} = c

But from there, how can I get that circle's size since I can't get it's radius?
Oh, and what am I doing wrong with my latex? do I need latex tags? :/

4. Re: Problem with infinite geometric series and circles tangent to each other + a squa

Following on from Prove It's hint,
you could use similar triangles to express the ratio
of the radii of the large and small circles.

Hence, in the attached .pdf, we want the ratio

$\frac{sB}{BC}$

and we can painlessly find this using the ratio

$\frac{AB}{AC}$

since the ratios of corresponding linear measurements are equal.

Continue on with Pythagoras' Theorem to evaluate these lengths,
giving the ratio of the radii,
which is also the ratio of the small circle's radius to the "next smallest" circle's radius.

Then the sum of all the areas can be expressed as a geometric series.

My calculations give

$Area=\pi R^2\left[1+4\left[\left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right)^2+\left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right)^4+.....\right]\right]$

$=\pi R^2\left[1+4\left(\frac{3-2\sqrt{2}}{4\sqrt{2}}\right)\right]=\pi R^2\left[1+\frac{3-2\sqrt{2}}{\sqrt{2}}\right]=1.12 \pi R^2$

where R is the radius of the main circle.