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newbie here.

I'd like to ask you, if you know some brief, but somewhat solid proof of a convex polyhedron and a sphere centered at one of its vertices (with small enough radius, so it intersects only with the edges adjacent to the said vertex), creating convex spherical polygon? Or, can you recommend me some places to look for proofs without getting totally lost?

Thank you for your answers!

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    $\begingroup$ To me it is unclear what you are asking, because the definition of a convex vertex of a polyhedron is that the intersection with a sufficiently small sphere (centered on the vertex) is convex. $\endgroup$
    – Joseph O'Rourke
    Commented Apr 1, 2021 at 22:10
  • $\begingroup$ @JosephO'Rourke Oh really? Didn't know that, thanks! (Not sarcasm). $\endgroup$
    – McDuck
    Commented Apr 2, 2021 at 8:22
  • $\begingroup$ @McDuck Can you explain the context, and your definition for a convex polyhedron and a spherical polygon? Is this all in $\mathbb{R}^3$ or is the dimension intended to be general? $\endgroup$
    – Geva Yashfe
    Commented Apr 2, 2021 at 10:03
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    $\begingroup$ I know no reference, here is an idea for a proof: intersect the sphere with the boundary of the polyhedron (I assume everything lives in $\mathbb{R}^3$). The intersection is the union of sets of the form $S^2 \cap (\text{plane through the center of the sphere})$, each of which is a geodesic on the sphere. You obtain a geodesic spherical polygon. To show convexity look at the angles of this polygon and prove they equal the angles between the corresponding planes. Then look at the intersection of the sphere with the interior of the polyhedron, and show it equals the interior of the polygon. $\endgroup$
    – Geva Yashfe
    Commented Apr 2, 2021 at 10:06
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    $\begingroup$ It is much shorter if you are willing to define convex spherical polygons to be finite intersections of closed hemispheres. In this case it follows directly from the fact that a halfspace passing through the center of the sphere intersects the sphere in a closed hemisphere, and a convex polyhedron is a finite intersection of closed halfspaces... I guess this is equivalent to what @JosephO'Rourke wrote above. $\endgroup$
    – Geva Yashfe
    Commented Apr 2, 2021 at 10:13

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