Four circles of unit radius are drawn with centers (0,1), (-1,0), (0,1), and (0,-1). A circle with radius 2 is drawn with the origin as its center. What is the area of all points which are contained in an odd number of these 5 circles? (Express your answer in the form "a pi + b" or "a pi - b", where a and b are integers.)
4pi - 8
The Answer = 4pi-8.
confirmed.
no actually the answer is:
Solution:
We want the area of the regions labeled "One Circle" and "Three Circles" (these are the odd numbers covered). We can get these by taking the area of the big circle with a radius of 2 (which is ) and then subtracting the area of "Two Circles". Now "Two Circles" has 1 part in each of the 4 little radius-1 circles. Looking at one of them, we can see the the areas labeled "1" fits perfectly in the areas labeled "2". Thus each part in the 4 circles has an area of 2. Since we have four of these, the total "Two Circles" area is 8. Subtracting this from , we have .
I still don't get it. So what's the answer. Your solution is very confusing.
whats the answer then
DFS
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img38.imageshack.us/img38/4959/circles2.png
put an h t t p thing before the above urL and you will find an image which corresponds to my solution
since the large circle encloses all the small ones, a point in an odd number of the 5 circles is enclosed in an even number of the 4 small circles.
Since no point lies in more than two of the small circles, we want all the points in the small-circle intersections, and all the points outside all the small circles.
Since the figure is symmetric about the origin, we can just work with the first octant, and multiply the result by 8.
Draw a detailed diagram of one quadrant, with the two small arcs and the diagonal, and the large arc. Just work with one octant for calculations.
The circles centered at (1,0) and (0,1) intersect at (1,1), with the two arcs subtending an angle of pi/2.
The arcs enclose an area (a sector minus a triangle) of (pi-2)/4.
The area of the large circle outside the small circle is thus pi(2^2)/8 - pi/4 = pi/4
Adding up the intersection and the outside areas, we get pi/4 + (pi-2)/4 = (pi-1)/2
For the whole figure then, the desired area is 4pi - 4