Here is an old geometry/measurement chestnut for you.

[Whateverist]

Draw a quarter circle and label the center point C and the points on either end of its arc, A and B.

Let the length of the quarter circle's radius = 1 unit 

Now draw the circle with diameter AB, and label another point on this circle D such that point D is not inside the original quarter circle CAB.

Find the area of circle ABD outside of quarter circle CAB. Place answer inside hide tags and justify your result.

[ignoramus]

42?

[Whateverist]

42?

Wait, does your approach rely on string theory?

[polymath257]

Draw a quarter circle and label the center point C and the points on either end of its arc, A and B.

Let the length of the quarter circle's radius = 1 unit 

Now draw the circle with diameter AB, and label another point on this circle D such that point D is not inside the original quarter circle CAB.

Find the area of circle ABD outside of quarter circle CAB.  Place answer inside hide tags and justify your result.

An old problem of Hipparchus.

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The area of the half-circle ABD is the same as the area of the quarter circle. Subtract the area outside of the triangle and the quarter circle from both.
The answer is the same as the area of the triangle OAB where O is the center of the quarter circle. In other words, 1/2.

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[SteelCurtain]

I have no idea how a circle can be defined by three points. I think this is what you're getting at.

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π/4

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[ignoramus]

(where's Rob when I need him!)

(Rob, it's a joke man, relax)

[Whateverist]

Draw a quarter circle and label the center point C and the points on either end of its arc, A and B.

Let the length of the quarter circle's radius = 1 unit 

Now draw the circle with diameter AB, and label another point on this circle D such that point D is not inside the original quarter circle CAB.

Find the area of circle ABD outside of quarter circle CAB.  Place answer inside hide tags and justify your result.

An old problem of Hipparchus.

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The area of the half-circle ABD is the same as the area of the quarter circle. Subtract the area outside of the triangle and the quarter circle from both.
The answer is the same as the area of the triangle OAB where O is the center of the quarter circle. In other words, 1/2.

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Thanks for the Hipparchus reference.  I didn't know that and will have to look into who that was.  I don't recall where I encountered it but for whatever reason I woke up too early this morning thinking about it and had to recreate it.

I love the elegance of the solution though and how unlikely it seems when you consider the shape whose area we must find in itself.  Very satisfying I think.

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I have no idea how a circle can be defined by three points. I think this is what you're getting at.

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π/4

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That is basically the figure we're interested in but the circle which includes the point A and B whose center is at C looks a little misshapen, almost as though you were trying to get point C on the circle as well.  It shouldn't be on the circle since it is the circle's center point.  So the shaded region should include only the crescent shape inside the half circle but outside the quarter circle.  Is that clear?  

Sorry I didn't provide an image.  I certainly provided it when I gave it to advanced middle schoolers but I was keen to see if I could give specific enough directions for the sophisticates we get around here.

I'll leave it there if that clarifies the picture for you so you can enjoy polishing it off.

[The Valkyrie]

Still looking for the chestnut.

[Whateverist]

Still looking for the chestnut.

Is that a nipple reference?  Glad to know you're an officianado (sp?) too.  My wife isn't into boobs but she knows my tastes in girl butt and will point them out for me.

[Fireball]

My take-

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Not convinced that point "D" is required.