What is the size of the largest regular pentagon which can fit inside a square with a side-length of 1?
Playing with Geometers' Sketchpad one can see that the maximum comes when the line from one of the vertices of the pentagon through the center of the pentagon coincides with a diagonal of the square. Call this point D of pentagon ABCDE.
In that orientation, the two pentagonal vertices not adjacent to D (that is, A and B) form the hypotenuse of an isosceles right triangle that uses portions of the square as legs. Vertices E and C are on the other two sides of the square. Call the center of the pentagon O. Unfortunately it does not lie at the center of the square.
Segment BA is 9° from being perpendicular to A's side of the square, tilted toward point B. Symmetrically, B is 9° from being perpendicular to B's side of the square, tilted toward A.
Point C, on the opposite side of the square from A, is 27° from being perpendicular to its side of the square, tilted toward B.
If x is the distance from O to any one of the vertices A, B, etc., the unit distance that is the side of the square then equals x*(cos(9°) + cos(27°)), so
x = 1/(cos(9°) + cos(27°))
Each edge of the pentagon equals 2*x*sin(36°) which is therefore
2*sin(36°)/(cos(9°) + cos(27°)) ~= 0.625737860160924
The numerical value agrees with Geometers' Sketchpad measurements.
This makes the area approximately 0.673649260962263, which also agrees with GSP measurements.
Posted by Charlie
on 2013-11-28 13:11:40