Reuleaux triangle

Example

Let $w > 0$, and let $A$, $B$, $C$ be the vertices of an equilateral triangle of side $w$ in the plane. The Reuleaux triangle is the intersection of the three closed discs of radius $w$ centred at $A$, $B$ and $C$. Its boundary consists of three arcs of circles of radius $w$: the arc from $B$ to $C$ centred at $A$, and the two similar arcs. Each arc has angle $\pi/3$.

Its boundary is a curve of constant width $w$. Check: take a unit vector $u$ and the two supporting lines orthogonal to $u$. The arc centred at $A$ consists of the points $A + w v$ with $v$ in an angle $V_A$ of size $\pi/3$. Its outer normal at $A + w v$ is $v$. The outer normals of the body at the vertex $A$ are the vectors $-v$ with $v \in V_A$. These six angles of size $\pi/3$ cover all directions. So for some vertex, say $A$, one of $u$ and $-u$ lies in $V_A$ and the other is an outer normal at $A$. Then one supporting line passes through $A$ and the other through a point $A \pm w u$ of the arc centred at $A$. The distance between the two lines is $w$.

The perimeter is $3 \cdot w \cdot \pi/3 = \pi w$. The area is the sum of three circular sectors of angle $\pi/3$ minus twice the triangle: $3 \cdot \frac{\pi w^2}{6} - 2 \cdot \frac{\sqrt{3}}{4} w^2 = \frac{\pi - \sqrt{3}}{2} w^2$.