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A rectangle $OACB$ with two axes as two sides,the origin $O$ as a vertex is drawn in which the length $OA$ is four times the width $OB$.A circle is drawn passing through the points BC and touching $OA$ at its mid-point,thus dividing the rectangle into three parts. Find the ratio of the areas of these three parts.

My work:
I found out the co-ordinates of the points $A(0,4a),B(a,0),C(a,4a)$.I found out the co-ordinates of the point where the circle touches the line $OA$ to be $M(0,2a)$.
Now, I found out the centre of the circle just to verify whether $BC$ is the diameter to be $O~'(\frac52a,2a)$. I can understand the the ratio of $AMC$ to $OMB$ is $1$ by symmetry. But, I cannot prove it. Also I have no idea how to find out the area of $BMC$. I have no clue but to apply definite integral over the equation of the circle. Help me with this. It would be better if the solution does not include definite integral.

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  • $\begingroup$ This would be useful to avoid calculus. $\endgroup$ – John Habert Feb 7 '14 at 16:17
  • $\begingroup$ Yes, that works great. Thanks a lot! $\endgroup$ – Hawk Feb 7 '14 at 16:20
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Take $a=2$ so that all points including the center of the circle become integral. That simplifies the computation and has no effect on the resulting ratios. You can thgen draw the situation like this:

Illustration

Notice the green triangle I drew there. You should be able to compute its area as well as all of its edge lengths. Using the Wikipedia article on circular segments you should also be able to compute the area between the circle and that triangle. After that, it's only a bit of symmetry as well as sums and differences.

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