Thread: AMC problem, finding the area inside a circle but outside a square

Hi everyone. I'm a little stuck on how to solve this problem.

2010 AMC 10B Problems - AoPSWiki

Basically, there's a circle and a square with the same center. I can't just take the area of the circle minus the area of the square, since some of the square lies outside of the circle. How do I take that into account?

The answer, by the way, should be "B."

Originally Posted by Lord Darkin

Hi everyone. I'm a little stuck on how to solve this problem.

2010 AMC 10B Problems - AoPSWiki

Basically, there's a circle and a square with the same center. I can't just take the area of the circle minus the area of the square, since some of the square lies outside of the circle. How do I take that into account?

The answer, by the way, should be "B."

I presume you are talking about problem 16? This will help you out.

-Dan

Yes, that's the problem (16) and thanks for the link.

So I played around using those formulas, and also analyzed the circle/square situation further. Apparently the triangle that is below the "circular segment" as depicted by wikipedia is an equilateral triangle, with sides of root3 over 3. Therefore theta = 60 degrees.

By using the formula of wikipedia, and multiplying it by 4 to take into account the 4 circular regions out of the square, I ended up with



Where A should be the answer to the question. But I get...



Something isn't right here ....

Edit: I should have said this before, but pi/3 for the first term represents the area of the entire circle.

Area of a sector =

Area of the triangle =

Area required =

Thanks, I forgot sin60 = 1/sqrt3