The figure shows a region consisting of all points inside a square that are closer to the center than to the sided of the square. Find the area of the region.
The area of the given region
Without loss of generality we may assume that the centre of square is at origin. Then the square is symmetric about both axes. Divide the square into four right angled triangles of equal area using the diagonals. Owing to symmetry to find the total area it suffices to find the area of the shaded region consisting of all points in the triangle that are closer to the center than to that side. We can then multiply this area by to find the total area of the region.
Find the equation of the curve whose points are equidistant from the center and side. This curve(purple) shall be the top-boundary of the area. Consider an arbitrary point on the curve. Then the distance of the point from the side is and its distance from the center is .
So the distances are equal if
Take square of both sides.
Subtract from both sides.
Add to both sides.
Divide by and simplify.
Thus the crescent upper boundary of the region belongs to a parabola that opens down.
The area is equal to the area of the black triangle plus the blue crescent shaped area
To find these areas, find coordinate of the horizontal line separating them.Consider the top right corner point of R. We may note that this point is of the type for some . This can be seen in two ways. Notice that the diagonal of the square is part of the line , and the point lies on the line. Or notice that the top right corner point is part of top boundary curve as well as right boundary curve. Suppose the point is and its distance from origin is . Then by construction the distance from top side of square to point is same as its distance from origin. Similarly the distance from right side of square to point is also same as its distance from origin . Thus . So we have
Let be the corner point. Then,
Add on both sides.
Divide both sides by , and simplify.
Notice that the last answer is obtained by multiplying to both denominator and numerator.Due to symmetry the top left corner shall be
Thus area of the crescent shaped section (blue) is