Monoxdifly said:
I want to prove the formula of regular dodecahedron's volume. For this, I need to split the dodecahedron to 12 pentagonal pyramid with the same size. However, I don't know how to determine the height of those pentagonal pyramids. Any references I had encountered involving either golden ratio or incircle formula. Can we just determine their height using pure Pythagorean theorem? Trigonometry is also welcomed, but I still have some doubts about determining the angles where 3 faces of the dodecahedron meet because they provided no proof. Any help will be appreciated. Thanks.
This is another tough one.
Step 1 is to check that $\tan54^\circ = \dfrac{1+\sqrt5}{\sqrt{10-2\sqrt5}}.$ That is essentially what you asked in http://mathhelpboards.com/trigonometry-12/ask-tan-36-a-18156.html#post83543, or you can find a detailed proof of it
here. You can simplify that expression in two ways. First, multiply top and bottom of the fraction by $\sqrt{10+2\sqrt5}.$ Second, use the fact that $\bigl(1+\sqrt5\bigr)^2 = 6+2\sqrt5$ to write it as $$\tan54^\circ = \frac{1+\sqrt5}{\sqrt{10-2\sqrt5}} = \frac{\sqrt{(6+2\sqrt5)(10+2\sqrt5)}}{\sqrt{80}} = \frac{\sqrt{80+32\sqrt5}}{4\sqrt5} = \frac{\sqrt{5+2\sqrt5}}{\sqrt5}.$$
Step 2 is to find the area of a regular pentagon with side $a$. This consists of five "pizza slice" triangles with base $a$ and height $a\tan54^\circ$. Using Step 1, their total area comes out as $A = \sqrt{5\bigl(5+2\sqrt5\bigr)}a^2.$
Step 3 is to use coordinates to represent a regular dodecahedron. The
standard choice is to take the $20$ vertices to be the points $$(\pm1,\pm1,\pm1),\quad (0,\pm\tau^{-1},\pm\tau),\quad (\pm\tau,0,\pm\tau^{-1}), \quad (\pm\tau^{-1},\pm\tau,0),$$ where $\tau$ is the golden ratio $\frac12(1+\sqrt5).$ Select one face of this dodecahedron, for example by choosing the five vertices $$(1,1,1),\quad (0,\tau^{-1},\tau),\quad (-1,1,1), \quad (-\tau^{-1},\tau,0), \quad (\tau^{-1},\tau,0).$$ The centre point of this face (obtained by taking the means of the coordinates of the vertices) is the point $$\bigl(0, \tfrac{5+3\sqrt5}{10}, \tfrac{5+\sqrt5}{10} \bigr).$$ The distance $h_0$ from that point to the centre of the dodecahedron (which is the origin) is given by $$h_0^2 = \tfrac1{100}\bigl((5+3\sqrt5)^2 + (5+\sqrt5)^2\bigr),$$ which simplifies to $h_0^2 = \frac15\bigl(5+2\sqrt5\bigr).$ The edges of this canonical dodecahedron have length $\sqrt5-1$. So in a dodecahedron with side $a$ the distance corresponding to $h_0$ would be $$h = \frac{\sqrt{\frac 15\bigl(5+2\sqrt5 \bigr)}}{\sqrt5-1}a.$$
Step 4. We can now get the volume of the dodecahedron as the union of $12$ pentagonal pyramids with base area $A$ and height $h$, namely $$V = 12\times\frac13 \times \sqrt{5\bigl(5+2\sqrt5\bigr)}a^2 \times \frac{\sqrt{\frac15\bigl(5+2\sqrt5\bigr)}}{\sqrt5-1}a.$$ Using the same trick as in Step 1 to rationalise the fraction (here, multiply top and bottom by $\sqrt5 + 1$), this simplifies to $$V = \tfrac14\bigl(15 + 7\sqrt5\bigr)a^3.$$
