---
# System prepended metadata

title: Platonic Solid
tags: [Geometry_Junior, ' Interesting Readings']

---

---
title: Platonic Solid (Sacred Geometry)
tags: Interesting Readings
description: 
type: slide
slideOptions:
 theme: white
 
---

### 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>