Reuleaux Triangle in a Square

Let's make the unit square the square defined by the points (0,0), (0,1), (1,0) and (1,1). We can see that since the Reuleaux triangle was a constant width of 1, the line AB must have a length of 1. This means A has coordinates $(\cos\theta,0)$ and B has coordinates $(0,\sin\theta)$ . As these two points move along the bottom and left sides of the square, the third corner E traces out an elliptical path which we need to determine the equation of.

I will do everything in terms of the parameter $\theta$ .

C is the midpoint of AB, and so C has coordinates: $C=\left(\frac{\cos\theta}{2},\frac{\sin\theta}{2}\right)$ $\triangle ABE$ , the triangle created by connecting the points of the Reuleaux triangle with straight lines, is an equilateral triangle with side lengths of 1. This means CE, the perpendicular height where the base is AB, has a length of $\frac{\sqrt3}{2}$ . D is a point directly across from C and directly below E, we can easily prove by alternate angles and angle sum of a triangle that $\angle CED=\theta$ . This is all shown in the digram.

In triangle ABE: $\begin{aligned}\sin\theta&=\frac{a}{\sqrt3/2}&\cos\theta&=\frac{b}{\sqrt3/2}\\ a&=\frac{\sqrt3\sin\theta}{2}&b&=\frac{\sqrt3\cos\theta}{2}\end{aligned}$ Now we can add these components of CE to the coordinates of C to determine the coordinates of E: $E=\left(\frac{\cos\theta}{2}+a,\frac{\sin\theta}{2}+b\right)\\ E=\left(\frac{\sqrt3\sin\theta+\cos\theta}{2},\frac{\sqrt3\cos\theta+\sin\theta}{2}\right)$ Now we can eliminate the parameter and get an equation for x in terms of y: $\begin{aligned}\sqrt3y-x&=\frac{3\cos\theta+\sqrt3\sin\theta}{2}-\frac{\sqrt3\sin\theta+\cos\theta}{2}&\sqrt3x-y&=\frac{3\sin\theta+\sqrt3\cos\theta}{2}-\frac{\sqrt3\cos\theta+\sin\theta}{2}\\ &=\frac{2\cos\theta}{2}&&=\frac{2\sin\theta}{2}\\ &=\cos\theta&&=\sin\theta\end{aligned}$ $\begin{aligned}(\sqrt3y-x)^2+(\sqrt3x-y)^2&=\cos^2x+\sin^2x\\ 3y^2-2\sqrt3xy+x^2+3x^2-2\sqrt3xy+y^2&=1\\ 4y^2-4\sqrt3xy+4x^2-1&=0\\ y^2-\sqrt3xy+x^2-\frac{1}{4}&=0\\ y&=\frac{\sqrt3x\pm\sqrt{3x^2-4\left(x^2-\frac{1}{4}\right)}}{2}\\ y&=\frac{\sqrt3x\pm\sqrt{1-x^2}}{2}\end{aligned}$ The graph is an ellipse, but since we only want the top right corner of the ellipse, we only consider the positive part of the $\pm$ , so $y=\frac{\sqrt3x+\sqrt{1-x^2}}{2}$ Before we can make an integral, we need to find the terminals. These are the points where the ellipse touches the edges of the square, so the upper bound is obviously 1. For the lower bound, we need to look at where y=1 and find the corresponding x coordinate. $y=1\\ \frac{\sqrt3\cos\theta+\sin\theta}{2}=1\\ \cos\left(\theta-\frac{\pi}{6}\right)=1\\ \theta=\frac{\pi}{6}\\ x=\frac{\sqrt3\sin\frac{\pi}{6}+\cos\frac{\pi}{6}}{2}\\ x=\frac{\frac{\sqrt3}{2}+\frac{\sqrt3}{2}}{2}\\ x=\frac{\sqrt3}{2}$ Now we can make an integral to determine the area. Since the area is in the top right corner of the square, I'm going to move it to the bottom right corner so that it is touching the x axis by integrating 1 minus the function instead. $\begin{aligned}&\int_\frac{\sqrt3}{2}^1 1-\frac{\sqrt3x+\sqrt{1-x^2}}{2}dx\\ =&\int_\frac{\sqrt3}{2}^1 1-\frac{\sqrt3x}{2}dx-\int_\frac{\sqrt3}{2}^1\frac{\sqrt{1-x^2}}{2}dx\\ =&\left[x-\frac{\sqrt3x^2}{4}\right]_\frac{\sqrt3}{2}^1-\int_\frac{\pi}{3}^\frac{\pi}{2}\frac{\sqrt{1-\sin^2u}}{2}\cos u du\qquad\small\text{(letting x=sinu)}\\ =&\left(1-\frac{\sqrt3}{4}\right)-\left(\frac{\sqrt3}{2}-\frac{3\sqrt3}{16}\right)-\int_\frac{\pi}{3}^\frac{\pi}{2}\frac{\cos^2u}{2}du\\ =&1-\frac{9\sqrt3}{16}-\int_\frac{\pi}{3}^\frac{\pi}{2}\frac{\cos2u+1}{4}du\\ =&1-\frac{9\sqrt3}{16}-\left[\frac{\sin2u}{8}+\frac{u}{4}\right]_\frac{\pi}{3}^\frac{\pi}{2}\\ =&1-\frac{9\sqrt3}{16}-\left(\frac{\sin\pi}{8}+\frac{\pi}{8}\right)+\left(\frac{\sin\frac{2\pi}{3}}{8}+\frac{\pi}{12}\right)\\ =&1-\frac{9\sqrt3}{16}-\frac{\pi}{8}+\frac{\sqrt3}{16}+\frac{\pi}{12}\\ =&1-\frac{\sqrt3}{2}-\frac{\pi}{24}\end{aligned}$ This is the area not covered by the Reuleaux triangle in one of the corners, so we now need to multiply it by 4 and subtract it from the area of the square: $\begin{aligned}A&=1^2-4\left(1-\frac{\sqrt3}{2}-\frac{\pi}{24}\right)\\ &=1-4+2\sqrt3+\frac{\pi}{6}\\ &=-3+2\sqrt3+\frac{\pi}{6}\end{aligned}$ So $a=-3$ , $b=2$ and $c=\frac{1}{6}$ , so the answer is $-3\times2\times\frac{1}{6}=\boxed{-1}$