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title: Platonic Solid (Sacred Geometry)
tags: Interesting Readings
description:
type: slide
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theme: white
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### Platonic Solid
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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">
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#### Tetrahedron (Fire)
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<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 ?
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#### Cube (Earth)
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<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
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#### Octahedron (Air)
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<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?
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#### Icosahedron (water)
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<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.
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#### Dodecahedron (Aether)
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<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?
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#### Euler's Fomular (Euler characteristic)
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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.
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#### They are the only FIVE!
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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.
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#### 1. Angle's clue
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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?
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#### 2. Euler's clue
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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 $\{(3,3), (3,4), (4,3), (3,5), (5,3) \}$
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