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We refer to a subset S of the surface of a convex solid C as geodesically convex if the shortest path along the surface of C joining any two points in S lies entirely within S (and if there are more than one such shortest paths, at least one such lies fully within S).

Question: Given a convex polyhedron P and an integer n. Without dismantling P, we need to mark out its surface into n equal area portions that are geodesically convex. Given P and n, an algorithm is needed to decide if P allows such a partition and if so, to find it.

The surface of a sphere can obviously be marked out into n geodesically convex equal area pieces for any n - by n great arcs that connect the poles. The surface of any axis-symmetric solid appears to allow division into n equal area geodesically convex pieces. However, I don't see how the surface of a regular tetrahedron can be given such a partition for n=5.

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  • $\begingroup$ Interesting question! Note that the natural partition the surface of a unit cube into three $2 \times 1$ rectangles (Front+Top, Right+Back, Left+Bottom) fails to establish $n=3$ because those rectangles are not geodesically convex. $\endgroup$
    – Joseph O'Rourke
    Commented Sep 27 at 19:42
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    $\begingroup$ Thank you very much. So dividing the surface of a cube into 3 seems nontrivial - and might even be impossible. Will every convex polyhedral surface admit such a partition for n=2? I am not sure. $\endgroup$
    – Nandakumar R
    Commented Oct 3 at 1:05
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    $\begingroup$ It appears that even without the equal area requirement, the question of deciding whether a polyhedral surface can be marked out into n geodesically convex regions remains nontrivial $\endgroup$
    – Nandakumar R
    Commented Oct 6 at 19:31
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    $\begingroup$ Note: reg prev comment, the ‘without area requirement’ variant can be interesting only when number of divisions is LESS than number of faces of the polyhedron. $\endgroup$
    – Nandakumar R
    Commented Oct 7 at 7:30

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