A doughnut is not a ball
Tonight's idea
Take a cube. Count its corners (vertices, \(V\)), edges (\(E\)) and faces (\(F\)): \(8 - 12 + 6 = 2\). You met this in grade 5. Euler's formula says every solid shaped like a ball gives 2, however it is cut into faces.
Two moves show why the cutting does not matter. Put a new corner on an edge: \(V\) and \(E\) both go up by 1. Draw a new edge across a face: \(E\) and \(F\) both go up by 1. Either way \(V - E + F\) stays put.
Stretching changes no count either. So the number belongs to the surface, not to the drawing on it. One rule: a face must be a patch with no hole in it.
\(V - E + F = 2\) on anything shaped like a ball. On a doughnut it is 0.
Suppose a doughnut (a torus) could be squashed into a ball without tearing. A map drawn on it would ride along, giving a ball with number 0. Every map on a ball gives 2. So it cannot be done. A count detects the hole.
Watch one: a doughnut made from a sheet
Rule a stretchy sheet into a 3 × 3 grid of squares. Glue the top side to the bottom side to make a tube. Bend the tube round and glue its two ends. That is a torus, with a map already drawn on it.
Corners: the letters A to I, so \(V = 9\). Edges: each corner has one edge going right and one going down, so \(E = 2 \times 9 = 18\). Faces: \(F = 9\) squares. So \(V - E + F = 9 - 18 + 9 = 0\).
Your turn
- Find \(V - E + F\) for (a) a tetrahedron (a pyramid on a triangle); (b) a pyramid on a square; (c) a triangular prism.
- Glue a 4 × 5 grid of squares in the same way. Find \(V\), \(E\), \(F\) and \(V - E + F\). What happens for an \(m \times n\) grid?
- The picture frame has four faces on top, four underneath, four outer walls and four inner walls. (a) Count \(V\), \(E\) and \(F\), and find \(V - E + F\). (b) A friend rubs out the eight slanting edges and counts the top as one ring-shaped face, and the underside too. What number does she get? Which rule did she break?
- A solid shaped like a ball has 20 faces, all triangles. How many edges and how many corners does it have? Start from the fact that every edge is shared by exactly two faces.
Stretch: two holes
Take two doughnuts with grids on them. Cut one square face out of each. Glue the doughnuts together along the rims of the two gaps, corner to corner and edge to edge. Each doughnut started at 0. What is \(V - E + F\) for the two-holed surface? What would a three-holed pretzel give?
Answers
- (a) \(4 - 6 + 4 = 2\). (b) \(5 - 8 + 5 = 2\). (c) \(6 - 9 + 5 = 2\). All are shaped like a ball.
- \(V = 20\), \(E = 40\), \(F = 20\), so \(20 - 40 + 20 = 0\). For \(m \times n\): \(V = mn\), \(E = 2mn\), \(F = mn\), and \(mn - 2mn + mn = 0\) every time. The number does not depend on the grid.
- (a) \(V = 16\): eight corners on top, eight underneath. \(E = 32\): twelve on top (four outer, four inner, four slanting), twelve underneath, eight upright. \(F = 16\). So \(16 - 32 + 16 = 0\): a frame is a doughnut. (b) \(E = 32 - 8 = 24\) and \(F = 16 - 6 = 10\), so \(16 - 24 + 10 = 2\). That is wrong, because a ring-shaped face has a hole in it. Faces must be patches without holes.
- Twenty triangles have \(20 \times 3 = 60\) sides, and each edge is two of those sides, so \(E = 30\). Then \(V - 30 + 20 = 2\) gives \(V = 12\). (This is the icosahedron.)
Stretch. Before gluing, the two totals add to \(0 + 0 = 0\). Removing two faces takes away 2. Gluing the rims merges four corners with four corners and four edges with four edges, so \(V\) drops by 4 and \(E\) drops by 4, which cancel in \(V - E + F\). The answer is \(-2\). Each extra hole takes off another 2, so three holes give \(-4\).