5 added 7 characters in body

I am looking for an elementary way to prove the following theorem.

Theorem. Let $\alpha$ and $\beta$ be two simple convex closed curves in $\mathbb R^2$. Assume $$\mathop{\rm length} \alpha=\mathop{\rm length}\beta$$ and there is a 1-Lipschitz bijecction $f\colon\alpha\to\beta$. Then $f$ is an isometry.

It would be better if the same proof would work for Lobachevsky plane and unit sphere (for the sphere one has to assume that the length of the curves is $<2{\cdot}\pi$).

The proof I know is simple, but it use Alexandrov geometry quite a bit: If we cut from the plane the region bounded by $\alpha$ and glue instead the region bounded by $\beta$ then the obtained space will have curvature $\ge0$ in the sence of Alexandrov and it is easy to show that it has to be isometric to the Euclidean plane. Hence the result.

P.S. This morning I realized that this also follows from the following continuos version of Cauchy's Arm Lemma:

Let $\alpha,\beta\colon[0,\ell]\to\mathbb R^2$ be closed convex curves with unit-speed parameter. Assume that for any $t$ in a subinterval $[a,b]\subset [0,\ell]$, the curvature of $\alpha$ at $\alpha(t)$ is at most the curvature $\beta$ at $\beta(t)$. Then $|\alpha(a)-\alpha(b)|\ge|\beta(a)-\beta(b)|$ and equality holds only if the resriction $\alpha|[a,b]$ is isometric to the resriction $\beta|[a,b]$.

4 added 36 characters in body

I am looking for an elementary way to prove the following theorem.

Theorem. Let $\alpha$ and $\beta$ be two simple convex curves in $\mathbb R^2$. Assume $$\mathop{\rm length} \alpha=\mathop{\rm length}\beta$$ and there is a 1-Lipschitz bijecction $f\colon\alpha\to\beta$. Then $f$ is an isometry.

It would be better if the same proof would work for Lobachevsky plane and unit sphere (for the sphere one has to assume that the length of the curves is $<2{\cdot}\pi$).

The proof I know is simple, but it use Alexandrov geometry quite a bit: If we cut from the plane the region bounded by $\alpha$ and glue instead the region bounded by $\beta$ then the obtained space will have curvature $\ge0$ in the sence of Alexandrov and it is easy to show that it has to be isometric to the Euclidean plane. Hence the result.

P.S. This morning I realized that this also follows from a the following continuos version of Cauchy's Arm Lemma; so really simple solution is very unlikely.

Namely, assume :

Let $\alpha,\beta\colon[0,\ell]\to\mathbb R^2$ are a be closed convex curves with unit-speed parameter. Assume that for any $t$ in a subinterval $[a,b]\subset [0,\ell]$, the curvature of $\alpha$ at $\alpha(t)$ is at most the curvature $\beta$ at $\beta(t)$ (In the non-smooth case the curvature is measure and this inequality still has sense). \beta(t)$. Then$\alpha$|\alpha(a)-\alpha(b)|\ge|\beta(a)-\beta(b)|$ and equality holds only if the resriction $\beta$ are \alpha|[a,b]$is isometric .to the resriction$\beta|[a,b]$. 3 added 48 characters in body I am looking for an elementary way to prove the following theorem. Theorem. Let$\alpha$and$\beta$be two simple convex curves in$\mathbb R^2$. Assume $$\mathop{\rm length} \alpha=\mathop{\rm length}\beta$$ and there is a 1-Lipschitz bijecction$f\colon\alpha\to\beta$. Then$f$is an isometry. It would be better if the same proof would work for Lobachevsky plane and unit sphere (for the sphere one has to assume that the length of the curves is$<2{\cdot}\pi$). The proof I know is simple, but it use Alexandrov geometry quite a bit: If we cut from the plane the region bounded by$\alpha$and glue instead the region bounded by$\beta$then the obtained space will have curvature$\ge0$in the sence of Alexandrov and it is easy to show that it has to be isometric to the Euclidean plane. Hence the result. P.S. This morning I realized that this also follows from a version of Cauchy's Arm Lemma; so really simple solution is very unlikely. Namely, assume$\alpha,\beta\colon[0,\ell]\to\mathbb R^2$are a closed convex curves with unit-speed parameter. Assume that for any$t$curvature of$\alpha$at$\alpha(t)$is at most the curvature$\beta$at$\beta(t)$(In the non-smooth case the curvature is measure and this inequality still has sense). Then$\alpha$and$\beta\$ are isometric.