### Platonic Solid </br> <font size = 5> More than 2000 years ago, we (human-being) started to develop concrete examples to represent some abstract ideas... One great example was from Plato (around 360 B.C.), He connected five different solids with five "elements": <img src="https://i.ibb.co/f29ZH4K/platonic-solids-5elements.jpg" width="800" height="180"> </font> --- #### Tetrahedron (Fire) <font size = 5> <img src="https://i.ibb.co/TBmd8X0/Tetrahedron-fire.jpg" width="200" height="300"> - the heat of fire feels sharp and stabbing; - **(4,4,6)**: 4 faces, 4 vertices, 6 edges; - all faces are regular polygons (equilateral triangle) - **(3, 3)**: 3 edges per face, 3 faces meet at each vertex - How many rotational symmetry ? How many symmatry plane (for reflective symmetry) ? How many total symmetries ? </font> --- #### Cube (Earth) <font size = 5> <img src="https://i.ibb.co/c81hqF6/Cube-Earth.png" width="300" height="300"> - cube is the only normal solid that tessellates Euclidean space; - **(6,8,12)**: 6 faces, 8 vertices, 12 edges; - all faces are regular polygons (Square) - **(4, 3)**: 4 edges per face, 3 faces meet at each vertex - How many body diagonals? How many rotational symmetry? - 3 classes of edges, each class with 4 parallal edges </font> --- #### Octahedron (Air) <font size = 5> <img src="https://i.ibb.co/j48mXfG/Octahedron-Air.jpg" width="230" height="280"> - its minuscule elements are so smooth that one can scarcely feel them; - **(8,6,12)**: 8 faces, 6 vertices, 12 edges; - all faces are regular polygons (equilateral triangle) - **(3, 4)**: 3 edges per face, 4 faces meet at each vertex - How many body diagonals? How many rotational symmetry? - Can you see "relationship" between the Cube and Octahedron? </font> --- #### Icosahedron (water) <font size = 5> <img src="https://i.ibb.co/tC2HMzT/Isocahedron-Water.jpg" width="230" height="300"> - flows out of one's hand as though it were made of tiny little balls; - **(20,12,30)**: 20 faces, 12 vertices, 30 edges; - all faces are regular polygons (equilateral triangle) - **(3, 5)**: 3 edges per face, 5 faces meet at each vertex - It has the greatest volume for its surface area of any platonic solid. - It has the greatest number of faces of any platonic solid. </font> --- #### Dodecahedron (Aether) <font size = 5> <img src="https://i.ibb.co/Fsn6TZ8/Aether-Dodecahedron.jpg" width="230" height="300"> - flows out of one's hand as though it were made of tiny little balls; - **(12,20,30)**: 12 faces, 20 vertices, 30 edges; - all faces are regular polygons (pentagonal) - **(5, 3)**: 5 edges per face, 3 faces meet at each vertex - Can you see "relationship" between Icosahedron and Dodecahedron? </font> --- #### Euler's Fomular (Euler characteristic) <font size = 5> Let's sum-up those numbers about face, edge, and vertex: <img src="https://i.ibb.co/Gczpz33/Euler-Characristics.png" width="530" height="300"> Euler's cheracteristic applies for all convex polyhedron. </font> --- #### They are the only FIVE! <font size = 6> An obvious question is, **is there any more regular polyhedron?** The answer is negative. We have only these five regular polyhedron, no more! It is time to show some **Mathematics**: there are at least two ways of showing the conclusion above, let's call them "Angle's clue" and "Euler's clue" respectively. </font> --- #### 1. Angle's clue <font size = 5> 1. For each angle in regular m-polygon, $size\ of\ angle = (1-\frac{2}{m})\pi$, (**?why?**) 2. so if we have n polygons meet at one vertex, at each vertex, the sum of angles is, $$n(1-\frac{2}{m})\pi$$ 3. the sum of angle in 2 is less than $2\pi$, (**?why?**), so we have $$n(1-\frac{2}{m}) \pi \lt 2\pi$$ 4. Rearrange above inequality, we have $(n-2)(m-2)\lt 4$ (**Try it yourself**) The only pairs of integers (n, m) satisfy the inequality above are $\{(3,3), (3,4), (4,3), (3,5), (5,3) \}$ - Note: Why not some number less than 3? </font> --- #### 2. Euler's clue <font size = 5> 1. We still use (n,m) as n m-polygons meet at one vertex to represent the regular polyhedron; 2. So we have $mF = nV = 2E$ (**?why?**) 3. From Euler's Charecteristic, $V-E+F = 2$, we have $$\frac{2E}{n} - E + \frac{2E}{m} = 2$$ 4. Divide $2E$ on both side and rearrange, we have $\frac{1}{n} + \frac{1}{m} = \frac{1}{2} + \frac{1}{E}$ 5. Then $\frac{1}{n} + \frac{1}{m} \gt \frac{1}{2}$ (**?why?**) - The only posible integer pairs are $\{(3,3), (3,4), (4,3), (3,5), (5,3) \}$ </font>
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