Try this problem from Duke Math Meet 2009 Problem 7 based on Area of Ellipse. This problem was asked in the individual round.
Let $ R_I , R_{II} , R_{III} , R_{IV} $ be areas of the elliptical region $ \frac{(x-10)^2}{10} + \frac {(y-31)^2}{31} \le 2009 $ that lie in the first, second, third, and fourth quadrants, respectively. Find $ R_I - R_{II} + R_{III} - R_{IV} $
Discussion:
Special Note: The answer to this problem is given as 1240. This is the wrong answer. It approximates the region $ R_I - R_{II} + R_{III} - R_{IV} $ as a rectangle. However, we provide a solution using symmetry using that assumption and that works fine. Computing area of the ellipse is a tricky business.
First, we draw an approximate picture of the ellipse. The centre is at (10, 31). Here is a figure of it.
To find $ R_{I} - R_{II} $ reflect region $ R_{II} $ about y-axis. Look at the figure. The shaded region is $ R_{I} - R_{II} $. Its width along the line through centre (y=31) is 20 by symmetry (as the centre is 10 unit away from x-axis so $ R_I $ is 20 unit 'thicker' than $ R_{II} $ . You may convince yourself about this by solving for x setting y = 31 ).
Now to find $ R_{III} - R_{IV} $ we reflect $ R_{III} $ about y-axis again. The strip (shaded in blue) is negative of $ R_{III} - R_{IV} $
Note that the blue region 'begins' 31 unit 'below' the minor axis of the ellipse. So if we go 31 unit 'above' the minor axis and take the portion of the red strip, by symmetry it will be equal to the blue strip. We have shaded it in black and red.
So if we want to find $ R_{I} - R_{II} - { - (R_{III} -R_{IV}) } $ we remove the black stripe from the red strip and get the final region whose area is $ R_I - R_{II} + R_{III} - R_{IV} $.
Here is where we apply approximation. Width of the strip is 20 and its height is (31+31) = 62. Hence if we approximate the area as a rectangle, then answer is $ 1240 (62 \times 20) $.
But note that the strip is not ACTUALLY a rectangle. So this is only an approximate answer.
