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The 2D Circle eliminates option B.

The next two 2D graphs seem to work for both option A and C...

How can I be 100% sure about which one it is?

StopReadingThisUsername
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    Hint: If you cut A and C at $z=1$, for which of those do you get a point in the center, like the picture for $z=1$? – rogerl Dec 22 '17 at 18:02
  • How does C work with the third crossection? And what do you understand by "A", "B" and "C"? – Caran-d'Ache Dec 22 '17 at 18:02

1 Answers1

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Call the surfaces, in order left-to-right, "nappes", "saddle", and "hourglass".

The $(1,y,z)$ plane cuts the nappes at a circle, cuts the saddle in a curve that can be the graph of a function (possibly a parabola opening downward), and cuts the hourglass in a circle. Therefore, the first graph does not go with the saddle.

The $(x,-1.5,z)$ plane cuts the nappes at a hyperbola (the conic section produced by a plane parallel to the axis of the nappes), cuts the saddle in a curve that can be the graph of a function (possibly a parabola opening upward), and cuts the hourglass in a mildly distorted hyperbola. Therefore, the second graph does not go with the saddle.

The $(x,y,1)$ plane cuts the nappes at a hyperbola, cuts the saddle in a distorted hyperbola, and cuts the hourglass in a (possibly distorted) degenerate hyperbola. Thus, the third graph goes with the hourglass.

Therefore, the three planar cuts are from the hourglass.

The shortest way to explain this to someone else is starting with the third planar cut, explaining that it is compatible only with the hourglass, then showing the first two planar cuts are also compatible with the hourglass.

Eric Towers
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