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Math Problem Help

Old 08-22-05, 08:17 AM
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Math Problem Help

Anyone know the name of this math problem or any links with info on it. I solved it, just wanted to see if there was anything else on it.

You take a 1 x 1 square. Draw 4 quarter circles with a radius of 1 with the center of each at the four vertices of the square. Compute the area of the intersection.

Last edited by Venusian; 08-22-05 at 08:47 AM.
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Old 08-22-05, 08:20 AM
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Old 08-22-05, 08:44 AM
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Old 08-22-05, 10:03 AM
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That can't be right, because it evaluates to ~2.05, and you started with a 1 x 1 square, and excluded part of it.
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Old 08-22-05, 10:21 AM
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Ugh. Damn signs. I screwed up a sign in one of my steps. Amongst another couple of mistakes . Sorry. Fixed now.

Detailed solution:
Spoiler:
The intersection area can be described as a smaller square surrounded by four areas with a circle segment on each side. The arc of each of those circle segments is 30 degrees. You can figure this out by sheer geometry.
-Since the arc is 30, you can use the Law of Cosines to get the length of that side of the small square in the center as sqrt( 2 - sqrt(3)). The area of the center square is therefore 2 - sqrt(3).
-The area of that whole sector is 30/360 * pi = pi/12. The area of the isoceles triangle from one corner to the opposite side of that square is (.5 * sin 30) = 1/4. The area, therefore, of the segment on the edge of the center square is pi/12 - 1/4.
-Then you add the area of the center square and the area of the 4 segments surrounding it and you get 2 - sqrt(3) + 4 * (pi/12 - 1/4) = 2 - sqrt(3) + pi/3 -1 = 1 + pi/3 - sqrt(3).

Last edited by Otto; 08-22-05 at 10:38 AM.
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Old 08-22-05, 10:44 AM
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Agreed. I had a solution that evaluated numerically to that 30 minutesago, but I made half a dozen mistakes "simplifying" it to that form. Finally got the form to agree with the number. Numerically, is is 0.3151 of the area of the square.
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