Five perfect solids — and a corner-angle argument that says there can never be a sixth.
| Shape | V | E | F |
|---|
| Regular polygon | square |
| Its interior angle | 90° |
| How many meet | 3 |
| Angle at the corner | 270° |
| Shape | V | E | F |
|---|---|---|---|
| — | — | — | — |
| now | — | — | — |
| — | — | — | — |
Where's the maths? This is geometry — the geometry of regular polygons and solids that you meet in Sec 1 and Sec 2. A Platonic solid is a solid whose faces are all copies of the same regular polygon, with the same number of faces meeting at every single corner. You might expect there to be a whole family of them, one for each polygon, but there are exactly five and there can never be a sixth. The reason is an angle argument you already have the tools for: you know the interior angle of a regular polygon, and you know that the angles round a point on a flat page add to 360°. At the corner of a solid the angles that meet must add to less than 360°, because the corner has to have somewhere to fold into — and only five combinations of polygon and corner survive that test. The second idea on screen is counting: label the corners V, the edges E and the flat faces F, and for every one of these solids (and for every cut-cornered version of them) the answer to V − E + F is always the same number, 2.
Beyond the syllabus: Euler's formula V − E + F = 2 is enrichment.
The interior angle of a regular polygon. The interior angles of any p-sided polygon add to (p − 2) × 180°, so for a regular one every angle is interior angle = (p − 2) × 180° ÷ p. That gives 60° for a triangle, 90° for a square, 108° for a pentagon and 120° for a hexagon. Nothing else is needed.
The corner test. Put q of those polygons round one corner. If the total is exactly 360° the polygons lie flat and you have a tiling of the page, not a corner: six triangles, four squares or three hexagons. If the total is more than 360° the polygons overlap and no corner exists at all. Only when the total is under 360° is there a gap to pull shut, and the shape lifts off the page into three dimensions. Run through every case and just five survive: 3 triangles (tetrahedron), 4 triangles (octahedron), 5 triangles (icosahedron), 3 squares (cube), 3 pentagons (dodecahedron). Seven triangles or four pentagons never get started, and a heptagon's 128.57° is already too big to use three of.
Euler's formula. Count the corners, edges and faces of any of them and you get V − E + F = 2: the cube gives 8 − 12 + 6, the dodecahedron 20 − 30 + 12, the icosahedron 12 − 30 + 20. It is not a coincidence of regular shapes — slice the corners off with the truncation slider and V, E and F all change wildly, yet the answer stays 2. The soccer ball (12 pentagons and 20 hexagons) gives 60 − 90 + 32 = 2.
Duality. Mark the centre of every face of a cube and join the neighbours: you get an octahedron, and doing it again gives the cube back. The two solids are duals — V and F swap over while E stays the same, which is exactly why Euler's formula treats them alike. Tetrahedron ↔ tetrahedron, cube ↔ octahedron, dodecahedron ↔ icosahedron: five solids, three partnerships. The morph passes through the halfway shape where each edge has shrunk to the point where it crossed its partner's edge, on a sphere that touches every edge of both.
Everything on screen is computed, not typed in: the corners come from exact coordinates (the icosahedron's are (0, ±1, ±φ) and its cyclic shuffles, with the golden ratio φ = (1 + √5)⁄2), the edges are found as the closest pairs of corners, the faces are traced round the corner-to-corner network, and V, E and F are then simply the lengths of those three lists. The surface area and volume readouts are added up from the actual triangles.
Controls: drag to orbit, scroll or pinch to zoom. Sliders set the truncation, the fold and the morph; the buttons pick the solid, the corner and the dual pair. Space bar plays or pauses the morph.