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Consider the convex hull $L_\varepsilon$ of the points $(\pm 1,0)$ and the $\varepsilon$-disc centered at the origin for small $\varepsilon>0$.

If there is a $(\varepsilon,1)$-bi-Lipschitz map from the unit disc onto $L_\varepsilon$ then all the angles of curve bounding $L_\varepsilon$ has to be at least $\pi{\cdot}\varepsilon$ (see the sketch below).

On the other hand, $L_\varepsilon$ has two corners with angles about $2{\cdot} \varepsilon$.

The sketch. Assume $f$ maps the half-plane $H$ to the angle $A$ and has Lipschitz constants $\varepsilon$ and $1$, yet assume that $f(0)=0$ and $0$ is the tip of $A$.

Essentially we need to show that angle measure of $A$ is at least $\varepsilon{\cdot}\pi$. Note that if $B_r(x)$ lies in $H$ then $B_{\varepsilon\cdot r}(f(x))$ lies in $A$. It follows that the image of an $\alpha$-angle in $H$ with tip at $0$ intersects unit circle along an arc of length $\alpha+o(\alpha)$. Hence the result follows.

Consider the convex hull $L_\varepsilon$ of the points $(\pm 1,0)$ and the $\varepsilon$-disc centered at the origin for small $\varepsilon>0$.

If there is a $(\varepsilon,1)$-bi-Lipschitz map from the unit disc onto $L_\varepsilon$ then all the angles of curve bounding $L_\varepsilon$ has to be at least $\pi{\cdot}\varepsilon$ (see the sketch below).

On the other hand, $L_\varepsilon$ has two corners with angles about $2{\cdot} \varepsilon$.

The sketch. Assume $f$ maps the half-plane $H$ to the angle $A$ and has Lipschitz constants $\varepsilon$ and $1$, yet assume that $f(0)=0$ and $0$ is the tip of $A$.

Essentially we need to show that angle measure of $A$ is at least $\varepsilon{\cdot}\pi$. Note that if $B_r(x)$ lies in $H$ then $B_{\varepsilon\cdot r}(f(x))$ lies in $A$. It follows that the image of an $\alpha$-angle in $H$ with tip at $0$ intersects unit circle along an arc of length $\varepsilon{\cdot}\alpha+o(\alpha)$. Hence the result follows.

Consider the convex hull $L_r$ of the points $(\pm 1,0)$ and the $r$-disc centered at the origin for small $r>0$.

If there is a $(r,1)$-bi-Lipschitz map from the unit disc onto $L_r$ then all the angles of curve bounding $L_r$ has to be at least $\pi{\cdot}r$.

On the other hand, $L_r$ has two corners with angles about $2{\cdot} r$.

Consider the convex hull $L_\varepsilon$ of the points $(\pm 1,0)$ and the $\varepsilon$-disc centered at the origin for small $\varepsilon>0$.

If there is a $(\varepsilon,1)$-bi-Lipschitz map from the unit disc onto $L_\varepsilon$ then all the angles of curve bounding $L_\varepsilon$ has to be at least $\pi{\cdot}\varepsilon$ (see the sketch below).

On the other hand, $L_\varepsilon$ has two corners with angles about $2{\cdot} \varepsilon$.

The sketch. Assume $f$ maps the half-plane $H$ to the angle $A$ and has Lipschitz constants $\varepsilon$ and $1$, yet assume that $f(0)=0$ and $0$ is the tip of $A$.

Essentially we need to show that angle measure of $A$ is at least $\varepsilon{\cdot}\pi$. Note that if $B_r(x)$ lies in $H$ then $B_{\varepsilon\cdot r}(f(x))$ lies in $A$. It follows that the image of an $\alpha$-angle in $H$ with tip at $0$ intersects unit circle along an arc of length $\alpha+o(\alpha)$. Hence the result follows.

Consider convex hull $L_r$ of the points $(\pm 1,0)$ and the $r$-disc centered at the origin for small.

If there is a $(r,1)$-bi-Lipschitz map from the unit disc onto $L_r$ then all the angles of curve bounding $L_r$ has to be at least $\pi\cdot r$.

On the other hand, $L_r$ has two corners with angles about $2\cdot r$.

Consider the convex hull $L_r$ of the points $(\pm 1,0)$ and the $r$-disc centered at the origin for small $r>0$.

If there is a $(r,1)$-bi-Lipschitz map from the unit disc onto $L_r$ then all the angles of curve bounding $L_r$ has to be at least $\pi{\cdot}r$.

On the other hand, $L_r$ has two corners with angles about $2{\cdot} r$.