Let ABCD be a square of side length 3. P is a point on the plane such that each of angle
APB, BPC, CPD and DPA is at least 60. If each possible position of P is
painted red, find the area of the red region.
The red area is the overlap of four circles, each of which is circumscribed about an equilateral triangle internal to the square with its base as one side of the square.
This area looks like a "puffed up square" as the intersections of the four circles form a square with side length equal to 3/sqrt(3)=sqrt(3). The radius of curvature for the arcs defining the area is also sqrt(3) and each of the four arcs subtends 60°.
The central square therefore has area 3, to which we have to add the areas of the four segments of circles with the edges of the square as base (chord).
The area of a full circle would be pi*3, but 60° contains only 1/6 of that or pi/2 square units. The triangle that needs to be subtracted to get the segment area has base sqrt(3) and height 3/2 for an area of 3*sqrt(3)/4. Therefore the area of each circle segment is pi/2 - 3*sqrt(3)/4. Four of these then add to 2*pi - 3*sqrt(3).
To that add the area 3 of the central square and get 3 + 2*pi - 3*sqrt(3) ~= 4.08703288447294.
Posted by Charlie
on 2014-02-16 12:30:22