Every radius vector of the curve rho=a sin³phi/3

Every radius vector of the curve $\rho=a\sin^{3}\dfrac{\varphi}{3}$ intersects it at three points (besides the pole). Prove that the tangents at those three points form an equilateral triangle.

Step 1: The three points on one radius vector

A fixed ray $\arg=\varphi_0$ meets the curve at $\varphi_k=\varphi_0+2\pi k$, $k=0,1,2$ (three genuinely different points, since shifting $\varphi$ by $2\pi$ shifts $\varphi/3$ by only $2\pi/3$, so $\sin^3(\varphi/3)$ takes three different values).

Step 2: Tangent–radius angle

With $\rho'=a\sin^{2}\dfrac{\varphi}{3}\cos\dfrac{\varphi}{3}$, the formula $\tan\psi=\rho/\rho'$ gives

\[ \tan\psi = \frac{\sin\dfrac\varphi3}{\cos\dfrac\varphi3} = \tan\frac\varphi3 \;\Longrightarrow\; \psi = \frac\varphi3+n\pi. \tag{1} \]

Step 3: Inclination of the tangent at each point

The tangent's inclination to the polar axis is $\alpha=\varphi+\psi$; at $\varphi_k=\varphi_0+2\pi k$, \[\alpha_{k} = \varphi_{0}+2\pi k + \frac{\varphi_{0}+2\pi k}{3} = \frac{4\varphi_{0}}{3} + \frac{8\pi k}{3}. \tag{2}\]

Since a line's inclination is only meaningful modulo $\pi$, and $\dfrac{8\pi}{3}=2\pi+\dfrac{2\pi}{3}\equiv\dfrac{2\pi}{3}-\pi=-\dfrac{\pi}{3}\pmod\pi$, successive values of $\alpha_k$ ($k=0,1,2$) differ by exactly

\[ \pm\frac{\pi}{3} \pmod{\pi}. \tag{3} \]

Step 4: Equilateral triangle

Three lines whose inclinations are pairwise separated by $\pi/3$ meet each other at angle $\pi/3=60°$ pairwise (the acute angle between two lines of inclinations differing by $\pi/3$ is exactly $\pi/3$). Three distinct, pairwise non-parallel lines that meet at $60°$ to each other form an equilateral triangle. $\blacksquare$

Step 5: Verification

\[\varphi_0=1.0,\ a=1\text{: numerically computed tangent inclinations at } \varphi_0,\varphi_0+2\pi,\varphi_0+4\pi \text{ differ by} \pm\pi/3\ \checkmark\]