Seconds, anyone? .....
A concentric circle, (radius 3 2 ), and square, (side length 2 ), have a region of overlap with an area of
b a b + c a π ,
where a is coprime with both b and c . Find a + b + c .
The answer is 16.
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1 solution
Nice solution, Trevor. Thanks for posting it. :)
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Thank you, nice problem. Btw, is there a calc solution to this? I'm guessing you integrate y = x from 0 to 1-1/sqrt3 then multiply by 8. Then integrate x^2+y^2=4/3 from 1 / 3 to 1 then multiply by four. Finally you add them
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Glad you asked. :) The calculus solution would be
4 ∗ ( ∫ 0 3 1 d x + ∫ 3 1 1 3 4 − x 2 d x ) .
You would need to use a substitution of x = 3 2 sin ( θ ) to solve the second integral, but the answer does come out the same as the one in your solution.
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@Brian Charlesworth – ah, thank you, I was off by a little. I've never taken the time to learn integration to the fullest, only know the basics
@Brian Charlesworth – Not able to solve in seconds. It took 1 - 2 minute.
Looking at the circle, we can observe that the distance from the center to point of intersection is 3 2 and the distance o the mid point is 1. This forms a 30-60-90 triangle with the 30 degree angle being the one closest to the center.
Summing the area of these for triangles, we get 4 × ( 3 1 ) = 3 4 3
Looking at the sum of central triangle areas, the remaining degree of measure is 120, thus it is 1/3 the area of the circle. 3 π ( 3 2 ) 2 = 9 4 π .
Thus summing our results, we get 3 4 3 + 9 4 π . thus a+b+c= 1 6