Given a 1*1 square.
Find the area of the subset of points that are closer to its center than to any of the square’s sides.
Divide the square into four isosceles triangles, with the square's center as the apex of each of the triangles and the sides of the square as the bases.
As a coordinate system, choose the center of the bottom edge of the square as the origin. The right edge of the bottom triangle is then y = 1/2 - x. The point on this line that is also on the boundary of the area to be calculated has sqrt(2) * x = y.
sqrt(2) * x = 1/2 - x
x * (1 + sqrt(2)) = 1/2
x = 1 / (2 * (1 + sqrt(2)))
y = 1/2 - 1 / (2 * (1 + sqrt(2)))
The area sought is a central square plus four segments of parabolas, as the boundary is a set of four parabolas with the center of the square as a focus and each of the sides of the square as directrices. The central square then has each side equal
2 * (1/2 - (1/2 - 1 / (2 * (1 + sqrt(2)))))
= 2 * 1 / (2 * (1 + sqrt(2)))
= 1 / (1 + sqrt(2)) = sqrt(2) - 1 =~ 0.414213562373095
This is also the base of the piece of parabola included in one of the constituent triangles. The height of that piece is
1/4 - (sqrt(2) - 1) / 2 =~ 0.0428932188134524
The area of a parbola cut off by a chord is
(2/3) * b * h
In this case thats
(2/3) * (sqrt(2) - 1) * (1/4 - (sqrt(2) - 1) / 2)
which simplifies to
5/(3*sqrt(2)) - 7/6 =~ 0.0118446353109123.
The area sought is the total of the central square plus four of these parabola segments:
(sqrt(2) - 1)^2 + 4 * (5/(3*sqrt(2)) - 7/6) =~ 0.218951416497459.
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Posted by Charlie
on 2024-03-28 13:34:40 |