A shape-regular family and a degenerating sliver family

Each cell in the stated triangular-grid subdivision has side $a=1\mathrm{/}n$. Its altitude is $\sqrt{3}a\mathrm{/}2$, so its area is $\sqrt{3}{a}^2\mathrm{/}4$ and its incenter lies one third of the altitude from each side; hence $\rho =\sqrt{3}a\mathrm{/}6$. Its diameter is $h=a$, so $h\mathrm{/}\rho =2\sqrt{3}$, independently of $n$. By [F1] this family is shape regular.

For $0<ϵ\le 1$, the sliver triangle's incenter joined to its three vertices partitions it into three triangles with common altitude ${\rho }_{ϵ}$, so its area is ${\rho }_{ϵ}$ times half its perimeter. Its side lengths are $1,ϵ,\sqrt{1+{ϵ}^2}$, its longest side is $h_{ϵ}=\sqrt{1+{ϵ}^2}$ and its area is $ϵ\mathrm{/}2$. Its perimeter is $1+ϵ+\sqrt{1+{ϵ}^2}$, so ${\rho }_{ϵ}=\frac{ϵ}{1+ϵ+\sqrt{1+{ϵ}^2}}\mathrm{.}$ The numerator of $h_{ϵ}\mathrm{/}{\rho }_{ϵ}$ tends to $1$, while its denominator is of order $ϵ$; hence the ratio tends to infinity. The second triangle in each strip is congruent, so every cell has the same ratio. Thus no uniform shape-regularity constant exists for this fixed-square mesh family.