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I will try to answer your question 2. The answer is negative. There must be at least two antipodal points with parallel tangent planes. Here is a sketch of a proof. The idea of that proof is that points you are looking for have variational nature. Proof works in any dimension.

Denote the boundary of your body by $\Gamma$. Now we fix a Euclidean structure on $\mathbb R^n$. Let $f\colon S^{n-1}\to \mathbb R$ be a support function of $\Gamma$, here $S^{n-1}$ is a unit sphere and $f(x)=\max_{v\in \Gamma} (x,v)$, $(,)$ is the scalar product. Denote by $v(x) \in \Gamma$ such a point that $f(x)=(x,v(x))$. The tangent plane to $\Gamma$ at the point $v(x)$ is orthogonal to $x$. Then two points $v(x)$ and $v(-x)$ are antipodal if and only if there exists $\lambda>0$ such that $x$ is a critical point of the function $f(y)-\lambda f(-y)$ with zero critical value. The functions $f(y)$ and $f(-y)$ are positive (here we use the fact that the origin belongs to the interior of $\Gamma$), hence for some $\lambda>0$ zero is a critical value of $f(y)-\lambda f(-y)$. That is it.

I will try to answer your question 2. The answer is negative. There must be at least two antipodal points with parallel tangent planes. Here is a sketch of a proof. The idea of that proof is that points you are looking for have variational nature. Proof works in any dimension.

Denote the boundary of your body by $\Gamma$. Now we fix a Euclidean structure on $\mathbb R^n$. Let $f\colon S^{n-1}\to \mathbb R$ be a support function of $\Gamma$, here $S^{n-1}$ is a unit sphere and $f(x)=\max_{v\in \Gamma} (x,v)$, $(,)$ is the scalar product. Denote by $v(x) \in \Gamma$ such a point that $f(x)=(x,v(x))$. The tangent plane to $\Gamma$ at the point $v(x)$ is orthogonal to $x$. Then two points $v(x)$ and $v(-x)$ are antipodal if and only if there exists $\lambda>0$ such that $x$ is a critical point of the function $f(y)-\lambda f(-y)$ with zero critical value. The functions $f(y)$ and $f(-y)$ are positive (here we use the fact that the origin belongs to the interior of $\Gamma$), hence for some $\lambda>0$ zero is a critical value of $f(y)-\lambda f(-y)$. That is it.

One can easily generalize that statement - Let two compact convex bodies both contains origin in the interiors. Than there exist at least two rays passing through the origin, such that support planes of their intersections with boundaries are parallel.

I will try to answer your question 2. The answer is negative. There must be at least two antipodal points with parallel tangent planes. Here is a sketch of a proof.

Denote the boundary of your body by $\Gamma$. Now we fix a Euclidean structure on $\mathbb R^n$. Let $f\colon S^{n-1}\to \mathbb R$ be a support function of $\Gamma$, here $S^{n-1}$ is a unit sphere and $f(x)=\max_{v\in \Gamma} (x,v)$, $(,)$ is the scalar product. Denote by $v(x) \in \Gamma$ such a point that $f(x)=(x,v(x))$. The tangent plane to $\Gamma$ at the point $v(x)$ is orthogonal to $x$. Then two points $v(x)$ and $v(-x)$ are antipodal if and only if there exists $\lambda>0$ such that $x$ is a critical point of the function $f(y)-\lambda f(-y)$ with zero critical value. The functions $f(y)$ and $f(-y)$ are positive (here we use the fact that the origin belongs to the interior of $\Gamma$), hence for some $\lambda>0$ zero is a critical value of $f(y)-\lambda f(-y)$. That is it.

I will try to answer your question 2. The answer is negative. There must be at least two antipodal points with parallel tangent planes. Here is a sketch of a proof. The idea of that proof is that points you are looking for have variational nature. Proof works in any dimension.

Denote the boundary of your body by $\Gamma$. Now we fix a Euclidean structure on $\mathbb R^n$. Let $f\colon S^{n-1}\to \mathbb R$ be a support function of $\Gamma$, here $S^{n-1}$ is a unit sphere and $f(x)=\max_{v\in \Gamma} (x,v)$, $(,)$ is the scalar product. Denote by $v(x) \in \Gamma$ such a point that $f(x)=(x,v(x))$. The tangent plane to $\Gamma$ at the point $v(x)$ is orthogonal to $x$. Then two points $v(x)$ and $v(-x)$ are antipodal if and only if there exists $\lambda>0$ such that $x$ is a critical point of the function $f(y)-\lambda f(-y)$ with zero critical value. The functions $f(y)$ and $f(-y)$ are positive (here we use the fact that the origin belongs to the interior of $\Gamma$), hence for some $\lambda>0$ zero is a critical value of $f(y)-\lambda f(-y)$. That is it.

I will try to answer your question 2. The answer is negative. There must be at least two points with parallel tangent planes. Here is a sketch of a proof.

Denote your convex hypersurface by $\Gamma$. Now we fix a Euclidean structure on $\mathbb R^n$. Let $f\colon S^{n-1}\to \mathbb R$ be a support function of $\Gamma$, here $S^{n-1}$ is a unit sphere and $f(x)=\max_{v\in \Gamma} (x,v)$, $(,)$ is the scalar product. Denote by $v(x) \in \Gamma$ such a point that $f(x)=(x,v(x))$. The tangent plane to $\Gamma$ at the point $v(x)$ is orthogonal to $x$. Then two points $v(x)$ and $v(-x)$ are antipodal if and only if there exists $\lambda>0$ such that $x$ is a critical point of the function $f(y)-\lambda f(-y)$ with zero critical value. The functions $f(y)$ and $f(-y)$ are positive (here we use the fact that the origin belongs to the interior of $\Gamma$), hence for some $\lambda>0$ zero is a critical value of $f(y)-\lambda f(-y)$. That is it.

I will try to answer your question 2. The answer is negative. There must be at least two antipodal points with parallel tangent planes. Here is a sketch of a proof.

Denote the boundary of your body by $\Gamma$. Now we fix a Euclidean structure on $\mathbb R^n$. Let $f\colon S^{n-1}\to \mathbb R$ be a support function of $\Gamma$, here $S^{n-1}$ is a unit sphere and $f(x)=\max_{v\in \Gamma} (x,v)$, $(,)$ is the scalar product. Denote by $v(x) \in \Gamma$ such a point that $f(x)=(x,v(x))$. The tangent plane to $\Gamma$ at the point $v(x)$ is orthogonal to $x$. Then two points $v(x)$ and $v(-x)$ are antipodal if and only if there exists $\lambda>0$ such that $x$ is a critical point of the function $f(y)-\lambda f(-y)$ with zero critical value. The functions $f(y)$ and $f(-y)$ are positive (here we use the fact that the origin belongs to the interior of $\Gamma$), hence for some $\lambda>0$ zero is a critical value of $f(y)-\lambda f(-y)$. That is it.