# Area of a region | CMI Entrance 2014 solution

This is a problem from Chennai Mathematical Institute, CMI Entrance 2014 based on area of a region. Try to solve it.

Problem: Area of a region

$\mathbf{ A= {(x, y), x^2 + y^2 \le 144 , \sin(2x+3y) le 0 } }$ . Find the area of A.

Discussion:

$\mathbf{ x^2 + y^2 \le 144 }$ is a disc of radius 12 with center (0, 0).

Now notice that $\mathbf{ \sin(2x + 3y) \le 0 \implies \sin ( 2(-x) + 3(-y) ) = \sin(-(2x+3y)) = -\sin(2x+3y) \ge 0 }$

Hence if a point (x, y) is in A then (-x, -y) is not in A.
Similarly if there is a point (x, y) not in A then we get a corresponding point (-x, -y) in A.

Therefore we have a bijection between points in A and not in A.

Thus area of A is exactly half the area of the disc = $\mathbf{ \frac{\pi (12)^2 }{2} = 72 \pi }$

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### 2 comments on “Area of a region | CMI Entrance 2014 solution”

1. […] . Find the area of A. Solution […]

2. This can in fact be generalised to any odd function f(x). f(ax+by) divides the area of the disc in half if f is odd.

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