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 Pentagon in a Square (Posted on 2013-11-28)
What is the size of the largest regular pentagon which can fit inside a square with a side-length of 1?

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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

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