# Regular pentagon in a square

Recently I stumbled upon the following problem:

What is the area of set of points of unit square that can be covered with a regular pentagon that is contained entirely in the square and whose diagonal has length $1$?

In other words, fix a unit square $S$ and let $$\mathcal P = \{P \subset S \mid P \mathrm{\ is \ a \ regular \ pentagon \ with \ diagonal \ of \ length \ } 1 \}.$$ What is the area of $\bigcup \mathcal P$?

• JUst want to know how many elements are there in $\mathcal P$?
– Mick
Commented Aug 9, 2016 at 10:08
• $\mathcal P$ contains continuum elements Commented Aug 9, 2016 at 10:12
• Do you mean:- Inside the unit square, draw as many regular pentagons (with unit diagonals) as possible and sum their areas up?
– Mick
Commented Aug 9, 2016 at 10:23
• No, I mean: calculate the area of the figure obtained by taking the sum of these pentagons. Commented Aug 9, 2016 at 10:28

Let $\mathcal U$ be the union of all regular pentagons of diagonal 1 lying wholly in the unit square. We refer to the diagram below.
Consider a pentagon flush with the square's left side. It rotates while keeping point contact with the bottom and left sides at A and B respectively until it is flush with the bottom side. The locus of R during this rotation then forms part of the boundary of $\mathcal U$; four of these loci (one at each corner) and four straight lines of the pentagon's side length $\Phi=\frac{\sqrt5-1}{2}$ form the entire boundary of $\mathcal U$. The regions of the unit square not in $\mathcal U$ are shaded.
Despite initial appearances, the locus of R is not a straight line; we must derive and integrate an expression for it. We know that BA = 1 and BR = RA = $\Phi$. If O is the origin, OX = $x$ and OY = $y$ then a simple application of the Pythagorean theorem yields $$\left(x+\sqrt{\Phi^2-y^2}\right)^2+\left(y+\sqrt{\Phi^2-x^2}\right)^2=1$$ $$\left(x+\sqrt{\frac{3-\sqrt5}2-y^2}\right)^2+\left(y+\sqrt{\frac{3-\sqrt5}2-x^2}\right)^2=1$$ From here on we will be relying on Wolfram Alpha for our expressions. Solving for y in terms of x yields $$y=\frac{\sqrt{10}-\sqrt2}8\sqrt{-2x^2-\sqrt5+3} -\frac{\sqrt{10+2\sqrt5}}4 x$$ The indefinite integral of this (up to a constant) is $$\frac{ 2\sqrt2(\sqrt5-2)\sin^{-1}\left(x\sqrt{\frac{3+\sqrt5}2}\right) -x(2x\sqrt{5+\sqrt5}-(\sqrt5-1)\sqrt{3-\sqrt5-2x^2}) }{8\sqrt2}$$ This expression vanishes when $x=0$, so the area of one of the triangular regions not in $\mathcal U$ is its value when $x=\frac{3-\sqrt5}4$, the x-coordinate of the bottom-left vertex of the pentagon flush with the square's bottom side. After substituting and a drastic simplification we get the very nice $$\frac{\sqrt5-2}{40}\pi$$ The area of $\mathcal U$ and the answer to the original question is therefore $$1-4\left(\frac{\sqrt5-2}{40}\pi\right)=1-\frac{\sqrt5-2}{10}\pi=0.925837\dots$$ This constant is indexed as A275967 in Sloane's OEIS.