Let $H$ be embedded in a cartesian plane with its center at the origin and one of its cusps positioned at $\tuple {a, 0}$. :400px We have that $\LL$ is $3$ times the length of one arc of the deltoid. From Arc Length for Parametric Equations: :$\ds \LL = 3 \int_{\theta \mathop = 0}^{\theta \mathop = 2 \pi/3} \sqrt {\paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2} \rd \theta$ where, from Equation of Deltoid: :$\begin{cases} x & = 2 b \cos \theta + b \cos 2 \theta \\ y & = 2 b \sin \theta - b \sin 2 \theta \end{cases}$ We have: {{begin-eqn}} {{eqn | l = \frac {\d x} {\d \theta}       | r = -2 b \sin \theta - 2 b \sin 2 \theta       | c =  }} {{eqn | l = \frac {\d y} {\d \theta}       | r = 2 b \cos \theta - 2 b \cos 2 \theta       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | o =        | r = \paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2       | c =  }} {{eqn | r = \paren {-2 b \sin \theta - 2 b \sin 2 \theta}^2 + \paren {2 b \cos \theta - 2 b \cos 2 \theta}^2       | c =  }} {{eqn | r = 4 b^2 \paren {\paren {-\sin \theta - \sin 2 \theta}^2 + \paren {\cos \theta - \cos 2 \theta}^2}       | c =  }} {{eqn | r = 4 b^2 \paren {\sin^2 \theta + 2 \sin \theta \sin 2 \theta + \sin^2 2 \theta + \cos^2 \theta - 2 \cos \theta \cos 2 \theta + \cos^2 2 \theta}       | c =  }} {{eqn | r = 4 b^2 \paren {2 + 2 \sin \theta \sin 2 \theta - 2 \cos \theta \cos 2 \theta}       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = 8 b^2 \paren {1 + \sin \theta \sin 2 \theta - \cos \theta \cos 2 \theta}       | c =  }} {{eqn | r = 8 b^2 \paren {1 + 2 \sin^2 \theta \cos \theta - \cos \theta \cos 2 \theta}       | c = Double Angle Formula for Sine }} {{eqn | r = 8 b^2 \paren {1 + 2 \sin^2 \theta \cos \theta - \cos \theta \paren {1 - 2 \sin^2 \theta} }       | c = Double Angle Formula for Cosine: Corollary 2 }} {{eqn | r = 8 b^2 \paren {1 - \cos \theta + 4 \sin^2 \theta \cos \theta}       | c = simplifying }} {{eqn | r = 8 b^2 \paren {1 - \cos \theta + 4 \cos \theta \paren {1 - \cos^2 \theta} }       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = 8 b^2 \paren {1 - \cos \theta + 4 \cos \theta \paren {1 + \cos \theta} \paren {1 - \cos \theta} }       | c = Difference of Two Squares }} {{eqn | r = 8 b^2 \paren {1 - \cos \theta} \paren {1 + 4 \cos \theta \paren {1 + \cos \theta} }       | c =  }} {{eqn | r = 8 b^2 \paren {1 - \cos \theta} \paren {1 + 4 \cos \theta + 4 \cos^2 \theta}       | c =  }} {{eqn | r = 8 b^2 \paren {1 - \cos \theta} \paren {1 + 2 \cos \theta}^2       | c =  }} {{eqn | r = 8 b^2 \paren {2 \sin^2 \frac \theta 2} \paren {1 + 2 \cos \theta}^2       | c = Half Angle Formula for Sine }} {{eqn | r = 16 b^2 \sin^2 \frac \theta 2 \paren {1 + 2 \cos \theta}^2       | c =  }} {{end-eqn}} Thus: :$\sqrt {\paren {\dfrac {\d x} {\d \theta} }^2 + \paren {\dfrac {\d y} {\d \theta} }^2} = 4 b \sin \dfrac \theta 2 \size {1 + 2 \cos \theta}$ In the range $0$ to $2 \pi / 3$, $1 + 2 \cos \theta$ is not less than $0$, and so: :$\ds \LL = 3 \int_0^{2 \pi / 3} 4 b \sin \dfrac \theta 2 \paren {1 + 2 \cos \theta} \rd \theta$ Put: :$u = \cos \dfrac \theta 2$ so: :$2 \dfrac {\d u} {\d \theta} = -\sin \dfrac \theta 2$ As $\theta$ increases from $0$ to $\dfrac {2 \pi} 3$, $u$ decreases from $1$ to $\dfrac 1 2$. Then: {{begin-eqn}} {{eqn | l = 1 + 2 \cos \theta       | r = 1 + 2 \paren {2 \paren {\cos \dfrac \theta 2}^2 - 1}       | c = Half Angle Formula for Cosine }} {{eqn | r = 4 u^2 - 1       | c =  }} {{end-eqn}} Substituting: :$2 \dfrac {\d u} {\d \theta} = -\sin \dfrac \theta 2$ and the limits of integration: :$u = 1$ for $\theta = 0$ :$u = \dfrac 1 2$ for $\theta = \dfrac {2 \pi} 3$ we obtain, after simplifying the sign: {{begin-eqn}} {{eqn | l = \LL       | r = 12 b \int_1^{1/2} \paren {1 - 4 u^2} 2 \rd u       | c = Integration by Substitution }} {{eqn | r = 24 b \intlimits {u - \frac 4 3 u^3} 1 {1/2}       | c =  }} {{eqn | r = 24 b \paren {\paren {\frac 1 2 - \frac 4 3 \frac 1 {2^3} } - \paren {1 - \frac 4 3} }       | c =  }} {{eqn | r = 24 b \paren {\frac 1 2 - \frac 1 6 - 1 + \frac 4 3}       | c =  }} {{eqn | r = 24 b \paren {\frac 1 3 + \frac 1 3}       | c =  }} {{eqn | r = 16 b       | c =  }} {{eqn | r = \frac {16 a} 3       | c =  }} {{end-eqn}} {{qed}}