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Mario Roberts
Mario Roberts
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The setting is a convex cone $C$ in $\mathbb{R}^d$ with the property that if you cut it with $S^d$ the volume of the cut is greater than or equal to $\operatorname{Vol}(S^d)/d+1$. That is, the volume of the cone (i.e. the volume of its cut with $S^d$) is greater than or equal to the volume of the cone from the standard partition from $\mathbb{R}^d$ into $d+1$ cones.

The cut of the boundary of $C$ and $S^d$ determines a $d-1$ hyperplane. It is stated now, that the distance from this hyperplane to the origin is less than or equal to $\frac{1}{2}$.

In dimension $d=2$, this fact is easy to confirm by simple triangle properties. I would be very grateful though for an aproach in arbitrary dimensions.

Edit: I realised that the distance should decrease with increasing dimension, so probably its a good approach to just show that, since for d=2 its confirmed

The setting is a convex cone $C$ in $\mathbb{R}^d$ with the property that if you cut it with $S^d$ the volume of the cut is greater than or equal to $\operatorname{Vol}(S^d)/d+1$. That is, the volume of the cone (i.e. the volume of its cut with $S^d$) is greater than or equal to the volume of the cone from the standard partition from $\mathbb{R}^d$ into $d+1$ cones.

The cut of the boundary of $C$ and $S^d$ determines a $d-1$ hyperplane. It is stated now, that the distance from this hyperplane to the origin is less than or equal to $\frac{1}{2}$.

In dimension $d=2$, this fact is easy to confirm by simple triangle properties. I would be very grateful though for an aproach in arbitrary dimensions.

The setting is a convex cone $C$ in $\mathbb{R}^d$ with the property that if you cut it with $S^d$ the volume of the cut is greater than or equal to $\operatorname{Vol}(S^d)/d+1$. That is, the volume of the cone (i.e. the volume of its cut with $S^d$) is greater than or equal to the volume of the cone from the standard partition from $\mathbb{R}^d$ into $d+1$ cones.

The cut of the boundary of $C$ and $S^d$ determines a $d-1$ hyperplane. It is stated now, that the distance from this hyperplane to the origin is less than or equal to $\frac{1}{2}$.

In dimension $d=2$, this fact is easy to confirm by simple triangle properties. I would be very grateful though for an aproach in arbitrary dimensions.

Edit: I realised that the distance should decrease with increasing dimension, so probably its a good approach to just show that, since for d=2 its confirmed

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Michael Albanese
Michael Albanese
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distance Distance from a hyperplane (determined by the cut of a convex cone with the unit sphere) to the origin

The setting is a convex cone C$C$ in R^d$\mathbb{R}^d$ with the property that if uyou cut it with S^d$S^d$ the Volumevolume of the cut is equal or greater then the Volume(S^d)/d+1 (that meansthan or equal to $\operatorname{Vol}(S^d)/d+1$. That is, the Volumevolume of the Conecone (i.e.volume of the volume of its cut with S^d$S^d$) is greater than or equal thento the Volumevolume of the Conecone from the standard partition from R^d$\mathbb{R}^d$ into d+1 Cones)$d+1$ cones.

The cut of the boundary of C$C$ and S^d$S^d$ determines a d-1 Hyperplane$d-1$ hyperplane. It It is statetstated now, that the distance from this Hyperplanehyperplane to the origin is less/eq 1/2 than or equal to $\frac{1}{2}$.

In dimension d=2$d=2$, this fact is easy to confirm by simple triangle properties. I would be very greatfulgrateful though for aan aproach in abitrary dimensionarbitrary dimensions.

distance from a hyperplane (determined by the cut of a convex cone with the unit sphere) to the origin

The setting is a convex cone C in R^d with the property that if u cut it with S^d the Volume of the cut is equal or greater then the Volume(S^d)/d+1 (that means the Volume of the Cone (i.e.volume of its cut with S^d) is greater or equal then the Volume of the Cone from the standard partition from R^d into d+1 Cones).

The cut of the boundary of C and S^d determines a d-1 Hyperplane. It is statet now, that the distance from this Hyperplane to the origin is less/eq 1/2.

In dimension d=2 this fact is easy to confirm by simple triangle properties. I would be very greatful though for a aproach in abitrary dimension.

Distance from a hyperplane (determined by the cut of a convex cone with the unit sphere) to the origin

The setting is a convex cone $C$ in $\mathbb{R}^d$ with the property that if you cut it with $S^d$ the volume of the cut is greater than or equal to $\operatorname{Vol}(S^d)/d+1$. That is, the volume of the cone (i.e. the volume of its cut with $S^d$) is greater than or equal to the volume of the cone from the standard partition from $\mathbb{R}^d$ into $d+1$ cones.

The cut of the boundary of $C$ and $S^d$ determines a $d-1$ hyperplane. It is stated now, that the distance from this hyperplane to the origin is less than or equal to $\frac{1}{2}$.

In dimension $d=2$, this fact is easy to confirm by simple triangle properties. I would be very grateful though for an aproach in arbitrary dimensions.

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Mario Roberts
Mario Roberts
  • 11
  • 2

distance from a hyperplane (determined by the cut of a convex cone with the unit sphere) to the origin

The setting is a convex cone C in R^d with the property that if u cut it with S^d the Volume of the cut is equal or greater then the Volume(S^d)/d+1 (that means the Volume of the Cone (i.e.volume of its cut with S^d) is greater or equal then the Volume of the Cone from the standard partition from R^d into d+1 Cones).

The cut of the boundary of C and S^d determines a d-1 Hyperplane. It is statet now, that the distance from this Hyperplane to the origin is less/eq 1/2.

In dimension d=2 this fact is easy to confirm by simple triangle properties. I would be very greatful though for a aproach in abitrary dimension.