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added a counter-example for the general question
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Sergei Ivanov
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The answer to the title question is yes (well, I assume that by a "surface" you mean something reasonable, like a boundary of a convex set).

Let $AB$ be the longest segment with endpoints on the surface. We may assume that its length equals 2 and its midpoint is the origin. Consider projections to the planes that contain $AB$. Since projections do not increase distances, $AB$ is a diameter of each projection. Hence all projections to this family of planes are unit discs centered at the origin. The intersection of the corresponding cylinders is the unit ball, hence the result.

Added. In general, we cannot determine a convex body from the set of shadows (if we don't know the correspondence between shadows and directions of projections).

Take a unit ball and cut off three identical tiny caps whose centers form a regular triangle on the sphere and are not on one great circle. Looking at shadows, you cannot tell whether all three or only two caps are removed, because each projection shows you no more than two of them.

The same construction works for polyhedra if you start with an icosahedron rather than a ball.

The answer to the title question is yes (well, I assume that by a "surface" you mean something reasonable, like a boundary of a convex set).

Let $AB$ be the longest segment with endpoints on the surface. We may assume that its length equals 2 and its midpoint is the origin. Consider projections to the planes that contain $AB$. Since projections do not increase distances, $AB$ is a diameter of each projection. Hence all projections to this family of planes are unit discs centered at the origin. The intersection of the corresponding cylinders is the unit ball, hence the result.

The answer to the title question is yes (well, I assume that by a "surface" you mean something reasonable, like a boundary of a convex set).

Let $AB$ be the longest segment with endpoints on the surface. We may assume that its length equals 2 and its midpoint is the origin. Consider projections to the planes that contain $AB$. Since projections do not increase distances, $AB$ is a diameter of each projection. Hence all projections to this family of planes are unit discs centered at the origin. The intersection of the corresponding cylinders is the unit ball, hence the result.

Added. In general, we cannot determine a convex body from the set of shadows (if we don't know the correspondence between shadows and directions of projections).

Take a unit ball and cut off three identical tiny caps whose centers form a regular triangle on the sphere and are not on one great circle. Looking at shadows, you cannot tell whether all three or only two caps are removed, because each projection shows you no more than two of them.

The same construction works for polyhedra if you start with an icosahedron rather than a ball.

Source Link
Sergei Ivanov
  • 32.4k
  • 2
  • 98
  • 154

The answer to the title question is yes (well, I assume that by a "surface" you mean something reasonable, like a boundary of a convex set).

Let $AB$ be the longest segment with endpoints on the surface. We may assume that its length equals 2 and its midpoint is the origin. Consider projections to the planes that contain $AB$. Since projections do not increase distances, $AB$ is a diameter of each projection. Hence all projections to this family of planes are unit discs centered at the origin. The intersection of the corresponding cylinders is the unit ball, hence the result.