2

Ref: Shadows and planar sections of polyhedra

By shadow we mean the orthogonal projection of a convex 3D body C onto a 2D plane, for example, the shadow on the xy-plane, with C above (z>0) that plane and the light at L=(0,0,+∞). C an be freely rotated as it hovers above the xy-plane. Are the following claims easy to prove/counter?

  1. If a convex 3D body C has all its shadows (orthogonal projected onto any plane) to be of equal area, then, C is a sphere.

  2. If a convex 3D body C has all its shadows to be of equal perimeter, then, C is a sphere.

Note added on 6th August 2023: This question has been answered in comments below.

Nandakumar R
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    your question #1 seems to be a duplicate of https://mathoverflow.net/questions/138525/a-question-about3-3-dimensional-convex-bodies/138534 – Yoav Kallus Jul 06 '23 at 13:52
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    and @alvarezpaiva's comment on my answer to that question should give you a good hint for question #2, if you replace the square of the radial function by the support function – Yoav Kallus Jul 06 '23 at 13:56
  • Thank you for those pointers. I guess, despite the repetition, this post adds a little bit of value in the second claim (if you think otherwise, do comment). A further claim has also been posted: https://mathoverflow.net/questions/450310/a-claim-on-planar-sections-of-3d-convex-bodies – Nandakumar R Jul 06 '23 at 19:38
  • you're welcome. Feel free to write up an answer to #2 if you think it adds value – Yoav Kallus Jul 07 '23 at 01:51
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    As given in https://en.wikipedia.org/wiki/Girth_(geometry), the girth of a geometric object, in a certain direction, is the perimeter of its parallel projection in that direction (perimeter of the shadow).

    From Hilbert and Cohn-Vossen's 'Geometry and the Imagination', page 216-7, Minkowski has proved that all convex surfaces of constant girth (ie. all its shadows have constant perimeter), then the surface is one of constant width - not necessarily a sphere. This answers question 2.

     – Nandakumar R Aug 06 '23 at 12:00
  • In above, 'girth' should be viewed not merely as parallel projection but orthogonal projection. The argument remains valid. – Nandakumar R Aug 07 '23 at 11:11

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