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The Lumenary
Wonder TrailOptional · never gradedStarts from Shapes

Corners, edges and faces

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The question

Count the corners, edges and faces of a cube, then work out corners − edges + faces. Try a pyramid and a tetrahedron too. Why do you keep getting the same number — and can you find a shape that gives a different one?

Try the experiment before reading on. There’s no wrong way to explore.

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Experiment

Vertices, edges and faces

Count V, E and F on each solid and work out V − E + F. Then see what holes do.

Drag to turn it. Dashed lines are edges at the back.

Tetrahedron

Count them yourself — turning the solid helps. Click in a box to highlight what you are counting.

Your results
SolidVEFV − E + F
Tetrahedron????
Cube????
Octahedron????
Dodecahedron????
Icosahedron????
Square pyramid????
Picture frame (one hole)????

Counted 0 of 7. Every solid without holes gives 2; holes change the answer.

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Hints, if you want them

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    What’s going on

    What’s going on

    For a cube, 8−12+6=28 - 12 + 6 = 2. For a tetrahedron, 4−6+4=24 - 6 + 4 = 2. For a square pyramid, 5−8+5=25 - 8 + 5 = 2. Leonhard Euler noticed in the 1750s that

    V−E+F=2V - E + F = 2

    for every convex solid with flat faces, however many corners it has.

    Here is the idea of a proof. Take away one face and stretch what is left flat on a table, like a net. Now remove edges one at a time. Removing an edge that separates two faces lowers EE and FF by one each; removing an edge that dangles at a corner lowers EE and VV by one each. Neither move changes V−E+FV - E + F. At the end only a single point is left, with 1−0+0=11 - 0 + 0 = 1; adding back the face we took away gives 22. Making each step watertight takes real care — that care is the story of Imre Lakatos's book Proofs and Refutations.

    The picture-frame solid has a hole through it, and for it V−E+F=0V - E + F = 0. A solid with two holes gives −2-2. The number does not depend on how the surface is cut into faces — only on its shape up to stretching and bending, without tearing or gluing. Mathematicians call it the Euler characteristic χ\chi; for a closed surface with gg holes, χ=2−2g\chi = 2 - 2g. It is one of the first topological invariants, and it reaches surprisingly far: the Gauss–Bonnet theorem says that the total curvature of a closed surface is 2πχ2\pi\chi.

    Note

    Counting examples is evidence, not proof. The sketch above is the idea of a proof for convex solids; showing that χ\chi is the same for every way of cutting a surface into faces needs the tools of algebraic topology.

    Computationally testedAI-drafted · awaiting human review
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    Where this leads