Existence of $1$-Lipschitz map between triangles

Question

Consider two (Euclidean) triangles $T$ and $T'$. Let's say that $T$ majorizes $T'$ if there exists a 1-Lipschitz map that sends vertices to vertices and sides to sides (for some labeling of the vertices).

My question is, what are necessary and sufficient conditions for $T$ to majorize $T'$ ?

I know a sufficient condition. Let's say that the lengths $(l_1, l_2, l_3)$ of $T$ and $(l_1', l_2', l_3')$ of $T'$ satisfy the strong triangle inequalities if $l_i + l_j - l_k \ge l_i' + l_j' - l_k'$ for all pairwise distinct $i,j,k$. Then if $T$ and $T'$ satisfy the strong triangle inequalities, then $T$ majorizes $T'$. Is this condition necessary ?

Answer 1

i) Recall that $1$-Lipschitz map is area-decreasing. Note that condition in OP (cf. Heron's formula) is not equivalent condition :

Proof : Consider an equilateral triangle $\Delta\ ABC$ of side length $1$. When $A'\in [AB]$ with $|A-A'|=\varepsilon$, then $|A-C|+|C-B| -|A-B|=1$. $$|A-C|+|C-A'|-|A-A'|$$ is close to $2$. Hence we do have the condition, but there is $1$-Lipschitz map.

ii) I conjectured that the following is equivalent condition : $ \Delta A'B'C' $ is contained in $ \Delta ABC$

Proof : Consider the following case : $A=A',\ B=B'$ and $\angle \ ABC < \angle\ ABC'$.

When $A$ has a foot $A_f$ in $[BC]$ and $A$ has a foot $A_f'$ in $[BC']$, then note that $|A-A_f|<|A-A_f'|$ which shows that there is no $1$-Lipschitz map.

Answer 2

The "strong triangle inequality" is not necessary. I use a modification of HK Lee example.

Let $T=ABC$ be an equilateral triangle with unit side length. Consider the point $B'$ on the side $AB$ at distance $\varepsilon$ of $B$, and consider the triangle $T'=AB'C$.

First, $T$ majorizes $T'$, because the projection of $T$ onto the (convex) $T'$ is a 1-Lipschitz map which sends vertices to vertices and sides to sides (for the obvious labeling).

Moreover, the strong triangle inequality fails. Remark that $AB'=1-\varepsilon$ and $CB' = 1-\varepsilon + \varepsilon^2$. We have $AC+CB-AB=1$ and $AC+CB'-AB' = 1 + \varepsilon^2$.