Existence of $1$-Lipschitz map between triangles

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 ?

• Here triangle is 1-dimensional ? I mean that $T$ is union of three sides with an induced metric $d$. That is $d$ is not intrinsic metric. – HK Lee May 20 '19 at 6:14
• $T$ is the union of the sides with the metric induced by the Euclidean plane. – Florentin MB May 20 '19 at 6:38
• Posted also on MathOverflow: Existence of $1$-Lipschitz map between triangles. (I though that adding a link - at least in a comment - might be useful for others who see this post.) – Martin Sleziak Jun 12 '19 at 18:10

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.

• Your counter-example doesn't work, there is no 1-Lipschitz map between $ABC$ and $AA'C$. Consider $P$, the middle point of $AB$. Then the length $CP$ is lower than all the length $CP'$ for $P \in [AA']$, therefore one cannot map the point $P$ to the side $AA'$. In particular, your condition ($A'B'C'$ is cointained in $ABC$) is not sufficient. – Florentin MB May 17 '19 at 6:56

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 $$AC+CB'-AB' = 1 + \varepsilon^2$$.