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No, the distance can be as big as you want. Take two isosceles triangles with small bases, which "look" in the opposite directions.

For the updated question, the answer is still "NO". Again take long thin isosceles triangle with small base $\ll\varepsilon$. The distance between the barycentre of triangle and its $\varepsilon$-neigborhood is about $$\tfrac16{\cdot}\mathop{\rm diam}.$$

This is the upper bound for $\mathbb R^2$; for $\mathbb R^n$ you should get $$(\tfrac12-\tfrac1n){\cdot}\mathop{\rm diam}.$$

No, the distance can be as big as you want. Take two isosceles triangles with small bases, which "look" in the opposite directions.

For the updated question, the answer is still "NO". Again take long thin isosceles triangle with small base $\ll\varepsilon$. The distance between the barycentre of triangle and its $\varepsilon$-neigborhood is about $$\tfrac16{\cdot}\mathop{\rm diam}.$$

This is the upper bound for $\mathbb R^2$; for $\mathbb R^n$ you should get $$(\tfrac12-\tfrac1{n+1}){\cdot}\mathop{\rm diam}.$$

No, the distance can be as big as you want. Take two isosceles triangles with small bases, which "look" in the opposite directions.

For the updated question, the answer is still "NO". Again take long thin isosceles triangle with small base $\ll\varepsilon$. The distance between the barycentre of triangle and its $\varepsilon$-neigborhood is about $$\tfrac16{\cdot}\mathop{\rm diam}.$$

This is the upper bound for $\mathbb R^2$; for $\mathbb R^n$ you should get $$(\tfrac16-\tfrac1n){\cdot}\mathop{\rm diam}.$$

No, the distance can be as big as you want. Take two isosceles triangles with small bases, which "look" in the opposite directions.

For the updated question, the answer is still "NO". Again take long thin isosceles triangle with small base $\ll\varepsilon$. The distance between the barycentre of triangle and its $\varepsilon$-neigborhood is about $$\tfrac16{\cdot}\mathop{\rm diam}.$$

This is the upper bound for $\mathbb R^2$; for $\mathbb R^n$ you should get $$(\tfrac12-\tfrac1n){\cdot}\mathop{\rm diam}.$$

No, the distance can be as big as you want. Take two isosceles triangles with small bases, which "look" in the opposite directions.

No, the distance can be as big as you want. Take two isosceles triangles with small bases, which "look" in the opposite directions.

For the updated question, the answer is still "NO". Again take long thin isosceles triangle with small base $\ll\varepsilon$. The distance between the barycentre of triangle and its $\varepsilon$-neigborhood is about $$\tfrac16{\cdot}\mathop{\rm diam}.$$

This is the upper bound for $\mathbb R^2$; for $\mathbb R^n$ you should get $$(\tfrac16-\tfrac1n){\cdot}\mathop{\rm diam}.$$