Consider the function $f\colon\mathbb{R}^{2}\to\mathbb{R}$ given by

$$f(u_{1},u_{2})=-\frac{1}{2}\log\left(1+\operatorname{e}^{-2u_{1}}+\operatorname{e}^{-4u_{1}-2u_{2}}+\operatorname{e}^{-2u_{1}-4u_{2}}\right).$$

It is a concave function of Legendre type whose stability set is the polytope $\Delta=\operatorname{conv}((0,0),(1,0),(2,1),(1,2))$. The restriction of its Legendre-Fenchel dual to $\Delta^{\circ}$ is also a concave function of Legendre type.

For $c\in\mathbb{R}$, consider the affine map

$$A_{c}\colon\mathbb{R}\to\mathbb{R}^{2},\quad u\longmapsto(-u,u+c).$$

We write $A_{c}=H+(0,c)$ for a linear function $H$. The dual of $H$ is the function $H^{\vee}\colon\mathbb{R}^{2}\to\mathbb{R}$, $(x_{1},x_{2})\mapsto x_{2}-x_{1}$. Then $\operatorname{stab}(A_{c}^{*}f)^{\circ}=H^{\vee}(\Delta^{\circ})$ is the open interval $(-1,1)$. By Proposition 3.55, there is a map $\imath_{A_{c},f}$ embedding $(-1,1)$ into $\Delta^{\circ}$ in such a way that $\imath_{A_{c},f}\circ\nabla(A_{c}^{*}f)=(\nabla f)\circ A_{c}$. For $u\in\mathbb{R}$,

	$\displaystyle\nabla(A_{c}^{*}f)(u)$	$\displaystyle=\frac{\operatorname{e}^{-2u-4c}-\operatorname{e}^{2u}-\operatorname{e}^{2u-2c}}{1+\operatorname{e}^{2u}+\operatorname{e}^{2u-2c}+\operatorname{e}^{-2u-4c}}\in(-1,1),$
	$\displaystyle(\nabla f)\circ A_{c}(u)$	$\displaystyle=\frac{\left(\operatorname{e}^{2u}+2\operatorname{e}^{2u-2c}+\operatorname{e}^{-2u-4c},\operatorname{e}^{2u-2c}+2\operatorname{e}^{-2u-4c}\right)}{1+\operatorname{e}^{2u}+\operatorname{e}^{2u-2c}+\operatorname{e}^{-2u-4c}}\in\Delta^{\circ}.$

From this, we compute $\imath_{A_{c},f}(x)=\left(x_{1},x_{2}\right)$ with

$$\left\{\begin{aligned} x_{1}&=\frac{-\operatorname{e}^{-2c}}{2(1+\operatorname{e}^{-2c})}x+\frac{2+3\operatorname{e}^{-2c}}{2(1+\operatorname{e}^{-2c})}\left(\frac{x^{2}}{\sqrt{\rho_{c}^{2}+(1-\rho_{c}^{2})x^{2}}+\rho_{c}}+\frac{\rho_{c}}{1+\rho_{c}}\right),\\ x_{2}&=\frac{2+\operatorname{e}^{-2c}}{2(1+\operatorname{e}^{-2c})}x+\frac{2+3\operatorname{e}^{-2c}}{2(1+\operatorname{e}^{-2c})}\left(\frac{x^{2}}{\sqrt{\rho_{c}^{2}+(1-\rho_{c}^{2})x^{2}}+\rho_{c}}+\frac{\rho_{c}}{1+\rho_{c}}\right),\end{aligned}\right.$$

where we have set $\rho_{c}=2\operatorname{e}^{-2c}\sqrt{1+\operatorname{e}^{-2c}}$ for short. In particular, the image of the map $\imath_{A_{c},f}$ is an arc of conic: namely the intersection of $\Delta^{\circ}$ with the conic of equation

$$(x_{2}-x_{1})^{2}=(1-\rho_{c}^{2})L_{c}(x_{1},x_{2})^{2}+2\rho_{c}L_{c}(x_{1},x_{2}),$$

with $L_{c}(x_{1},x_{2})=\frac{2+\operatorname{e}^{-2c}}{2+3\operatorname{e}^{-2c}}x_{1}+\frac{\operatorname{e}^{-2c}}{2+3\operatorname{e}^{-2c}}x_{2}-\frac{\rho_{c}}{1+\rho_{c}}$. Varying $c\in\mathbb{R}$, these arcs of conics form a foliation of $\Delta^{\circ}$, they all pass through the vertex $(1,2)$ as $x\to 1$, and their other end as $x\to-1$ parameterizes the relative interior of the edge $\operatorname{conv}((1,0),(2,1))$, see Figure 2.