There is a regular hexagon whose inradius in 1 unit. A region is created inside the hexagon which is made up of all the points that are closer to the center of the hexagon than any of its sides. Find the area of this region.

(In reply to

A start by Charlie)

Take the triangle/parabola described by Charlie and put them on a coordinate grid with the vertex of the parabola at the origin. Then the directrix is at y = -1/2 and the focus is at (0, 1/2).

The equation of the parabola is then y = (1/2)*x^2. And the other two sides of the triangle are y = -sqrt(3)*x + 1/2 and y = -sqrt(3)*x + 1/2. These sides intersect the parabola at (2-sqrt(3), (7/2-2*sqrt(3)) and (-2+sqrt(3), (7/2-2*sqrt(3)) respectively.

By symmetry only one half of the parabola/triangle is needed and can be doubled for the whole triangle, or multiplied by 12 for the entire area sought. Then the area between the parabola and the right side of the triangle is given by:

Integ {0 to 2-sqrt(3)} -sqrt(3)*x + 1/2 - (1/2)*x^2 dx.

The integral eventually resolves to a value of (16-9sqrt(3))/6, which when multiplied by 12 yields an area of **32-18*sqrt(3) = 0.8231**.

As a check, the shape composed of the six parabolic arcs is circumscribed by a regular hexagon with an inradius of 1/2. That hexagon has an area of sqrt(3)/2 = 0.8660, which is slightly more than the parabolas' area of 0.8231.