372 lines
20 KiB
Plaintext
372 lines
20 KiB
Plaintext
---
|
|
format:
|
|
html:
|
|
html-math-method: katex
|
|
---
|
|
|
|
|
|
On the Volume of the Platonic Solids
|
|
====================================
|
|
|
|
|
|
The Platonic solids have been known for millennia. They bear the name of Plato, who spoke of them in his dialogue *Timaeus*. He describes their "construction" (sans the dodecahedron) from the most basic "isosceles and scalene" triangles, or in modern parlance, the "45-45-90 and 30-60-90" triangles. However, the construction was not mathematical, and to my knowledge, each solid was first rigorously described from principles in Book XIII of Euclid's Elements.
|
|
|
|
In my teenage years, I recall viewing articles on the solids with their volumes so proudly displayed along with their surface area. While the latter quantity may be troublesome in the case of the dodecahedron (as the geometry of regular pentagons is not widely taught), it is easy for any student of trigonometry to calculate the surface area of the solids made of equilateral triangles, and easy for any child who knows of squares to do so for the cube. On the other hand, the volume is somewhat mystical.
|
|
|
|
There is only one free variable in a Platonic solid, its edge length, which means that their volumes are parametrized by this value alone. To be a true unitless value, a volume must be in a ratio with another volume. The cube has the simplest expression for its volume; it is simply the side length cubed. Therefore in the following post, I will derive the ratio of each solid's volume to the volume of the cube formed by any side.
|
|
|
|
This post will calculate the volume without using any trigonometric functions (sine, cosine tangent), opting instead for a more compass-and-straightedge approach. For this reason, *square* of the volume will be calculated initially to better cooperate with the Pythagorean theorem. Additionally, except for the octahedron, the edge length of every solid will be 2 to simplify the bisection of edges. This happens to coincide with Plato's description; two 30-60-90 triangles were used to make an equilateral triangle, meaning that its edge length was twice the "unit" length: the shortest side of the 30-60-90 triangle.
|
|
|
|
|
|
A Recap of Geometry
|
|
-------------------
|
|
|
|
For those with only a vague recollection (or perhaps none at all) of geometry, this section is intended as a refresher.
|
|
|
|
|
|
### Planar Geometry
|
|
|
|
There are [many centers of a triangle](https://faculty.evansville.edu/ck6/encyclopedia/etc.html), but there are two of primary interest:
|
|
|
|
- The *circumcenter* is the center of the circle containing every vertex, meaning that it is equidistant from every vertex.
|
|
- It is constructed by finding the intersection of the edges' perpendicular bisectors.
|
|
- The distance from a vertex to the circumcenter is called the *circumradius* (*c*).
|
|
- The *incenter* is equidistant from every edge, meaning that the length of the perpendicular segment connecting an edge and the incenter is the same for all edges.
|
|
- It is constructed by finding the intersection of the lines which bisect each angle.
|
|
- The perpendicular distance from an edge to the incenter is called the *inradius* (*a*).
|
|
|
|
::: {}
|
|
|
|
|
|
Constructing the circumcenter and incenter. Angle bisectors in blue, perpendicular bisectors in red, in- and circumradii in green.
|
|
:::
|
|
|
|
The inradius is special because it is also an altitude for a triangle formed by the inradius and an edge of the larger triangle. This means that the area of the larger triangle is the sum of these smaller triangles.
|
|
|
|
|
|
|
|
$$
|
|
\begin{align*}
|
|
A &= \left ({e_1 a \over 2} + {e_2 a \over 2} + {e_3 a \over 2} \right) =
|
|
\left ({a \over 2} \right ) (e_1 + e_2 + e_3) \\
|
|
&= {Pa \over 2}
|
|
\end{align*}
|
|
$$
|
|
|
|
This gives an expression for the area. For an equilateral triangle, these two centers coincide. This is because the perpendicular bisectors of the edges *are* the angle bisectors. In fact, the bisection of an angle involves constructing a rhombus, which is made up of two isosceles triangles (of which the equilateral triangle is a special case). In this case, the inradius is also called the *apothem*, and the difference between it and the circumradius is immediately apparent and called the *sagitta* (*s*).
|
|
|
|
|
|
|
|
This idea of incenters and circumcenters can be extended to other 2D figures such as the square and regular pentagon. For a square, the center is simply the intersection of the diagonals (i.e., the diagonals' common midpoint). The pentagon is trickier, and will be discussed later. Regardless, the expression for the area ${Pa \over 2}$ still works, since the polygon can be triangulated through the center in a similar way.
|
|
|
|
|
|
|
|
|
|
### Cubes, Prisms, and Pyramids
|
|
|
|
Now we speak of 3D geometry. The volume of a prism is equal to the height times the area of the base, where the "height" is orthogonal to the plane of the base. Pyramids with the same height and base have one-third this area.
|
|
|
|
$$
|
|
V_\text{prism} = Bh,~~ V_\text{pyramid} = {Bh \over 3}
|
|
$$
|
|
|
|
This volume formula can be made more intuitive by considering the cube. The pyramid formed by one of the faces and an edge perpendicular to it will contain one square and two half-squares, or two squares in total. Therefore three pyramids are needed to recreate all six faces of the cube.
|
|
|
|
For a slightly more detailed explanation, consider a point inside the face on top of the cube. Its (perpendicular) distance from one edge is *x* and its distance to an edge adjacent to that is *y*. Connecting all other bases to this point produces five pyramids, whose bases all have the same area. Designate these pyramids as "bottom", "left", "right", "front", and "back", where left and right correspond to *x* and front and back correspond to *y*.
|
|
|
|
|
|
|
|
$$
|
|
\begin{align*}
|
|
V_\text{cube} &= Bh = V_\text{bottom} + V_\text{left} + V_\text{right}
|
|
+ V_\text{front} + V_\text{back} \\
|
|
&= rBh + rBx + rB(h-x) + rBy + rB(h-y) \\
|
|
&= rBh + rBh + rBh \implies 1 = 3r \\
|
|
r &= {1 \over 3}
|
|
\end{align*}
|
|
$$
|
|
|
|
This can be generalized to a pyramid based on any prism, where the top point lies in the plane of one of the bases. However, this is beyond the scope of this post.
|
|
|
|
|
|
Simple Solids: the Octahedron and the Tetrahedron
|
|
-------------------------------------------------
|
|
|
|
While "simple" is a bit of a misnomer, their volumes are easiest to appreciate, since they do not need regular pentagons.
|
|
|
|
|
|
### Octahedron
|
|
|
|
The octahedron can be thought of as two square pyramids joined end-on-end, with uniform edge length throughout. Since the base is a square, its center is equidistant from the vertices of the base. An alternative, congruent square can be noticed by the symmetry of the octahedron, meaning the center is also equidistant from the top of the square pyramid, and that the segment connecting the two is an altitude of the pyramid.
|
|
|
|
::: {}
|
|
|
|
|
|
Primary square in blue, secondary square in red. Diagonals of both squares shown.
|
|
:::
|
|
|
|
The length of this altitude is simply half of the diagonal of the square. Therefore, the volume of an octahedron (calculated using edge length 1) is:
|
|
|
|
$$
|
|
\begin{align*}
|
|
B^2 &= (1^2)^2 \\
|
|
(2h)^2 &= 4h^2 = 1^2 + 1^2 = 2 \\
|
|
V_\text{sq.pyr.}^2 &= {B^2 h^2 \over 3^2} =
|
|
{1 \cdot {2 / 4} \over 3^2} \\
|
|
(1^3 \cdot V_\text{oct})^2 &= (2V_\text{sq.pyr})^2 = 4V_\text{sq.pyr}^2 =
|
|
4 \cdot {2 / 4 \over 3^2} = {2 \over 3^2} \\ \\
|
|
V_\text{oct} &= {\sqrt{2} \over 3}
|
|
\end{align*}
|
|
$$
|
|
|
|
|
|
### Tetrahedron
|
|
|
|
The tetrahedron is itself a pyramid. First, the (square of the) area of the base of an equilateral triangle must be known. As a reminder, this and all future solids will have edge length 2.
|
|
|
|
:::: {layout-ncol="2"}
|
|
::: {.column width="49%"}
|
|
|
|
:::
|
|
|
|
::: {.column width="49%"}
|
|
$$
|
|
\begin{align*}
|
|
d_\text{altitude}^2 &=
|
|
\textcolor{orange}{2}^2 -\ \textcolor{green}{1}^2 = 3 \\
|
|
B^2 &= \left ( {2 \cdot d_\text{altitude} \over 2} \right )^2 = 3
|
|
\end{align*}
|
|
$$
|
|
:::
|
|
::::
|
|
|
|
Next, bisect the tetrahedron through one edge and the altitudes of two faces. The altitudes form the legs of an isosceles triangle, so bisecting the angle where they meet (perpendicularly) bisects the remaining edge.
|
|
|
|
:::: {layout-ncol="2"}
|
|
::: {.column width="49%"}
|
|
|
|
:::
|
|
|
|
::: {.column width="49%"}
|
|
$$
|
|
\begin{align*}
|
|
\textcolor{blue}{d_\text{length}}^2 &=
|
|
d_\text{altitude}^2 -\ \textcolor{green}{1}^2 = 2 \\
|
|
(2A_\text{center})^2 &= (2d_\text{length})^2 = (\textcolor{red}{h} d_\text{altitude})^2 \\
|
|
&= 4 \cdot 2 = 3h^2 \\
|
|
h^2 &= 8 / 3
|
|
\end{align*}
|
|
$$
|
|
:::
|
|
::::
|
|
|
|
Since *h* is known, we can calculate the volume. Note that the volume is multiplied by the cube of the side length to produce the correct ratio to a unit cube.
|
|
|
|
$$
|
|
\begin{align*}
|
|
({2^3 \cdot V_\text{tet}})^2 &= {B^2 h^2 \over 3^2} = {3 \cdot (8/3) \over 3^2} \\
|
|
V_\text{tet}^2 &= {8 \over 2^6 \cdot 3^2} = {1 \over 2^3 \cdot 3^2} = {1 \over 2 \cdot 6^2} \\
|
|
V_\text{tet} &= \sqrt{1 \over 6^2 \cdot 2} = {1 \over 6\sqrt 2}
|
|
\end{align*}
|
|
$$
|
|
|
|
|
|
### Returning to 2D: Regular Pentagons
|
|
|
|
Both of the icosahedron and dodecahedron contain regular pentagons. Thus, it is necessary to examine them in detail.
|
|
|
|
The regular pentagon has five diagonals, which form a pentagram. Since all angles in a regular pentagon are equal, the trapezoid formed by three consecutive edges and one diagonal is isosceles. This means the diagonal is parallel to one of the edges, which applies to all diagonals by symmetry. Since the diagonal and side are parallel, this means that any two adjacent edges form a parallelogram with part of the diagonal. More specifically, it is a rhombus and those "parts of diagonals" have length equal to the side.
|
|
|
|
::: {}
|
|
|
|
|
|
Left: Pentagram in regular pentagon; Middle: Isosceles trapezoid, with parallel lines marked in blue; Right: Rhombus in regular pentagon
|
|
:::
|
|
|
|
Bisect the pentagon vertically and let the length of half of the diagonal of a pentagon be *d*, half the length of the other diagonal of a rhombus be *h*, and the remaining height of the pentagon be *g*.
|
|
|
|
|
|
|
|
$$
|
|
\begin{align*}
|
|
\textcolor{orange}{d}^2 + \textcolor{red}{h}^2 &= \textcolor{blue}{2}^2 \\
|
|
2\textcolor{darkblue}{A_\text{blue}} &= 2\textcolor{green}{g} = h(\textcolor{magenta}{d -\ (2 -\ d)}) = h (2d -\ 2) \\
|
|
\implies g &= {h(2d -\ 2) \over 2} = h(d -\ 1)
|
|
\end{align*}
|
|
$$
|
|
|
|
Notice that the center of a pentagram contains a regular pentagon. This means that the ratio of its height to the side is equal to the ratio of the larger pentagon's height to its side. This is enough information to deduce *d*:
|
|
|
|
\begin{align*}
|
|
{\textcolor{red}{h} \over 2(\textcolor{brown}{2 -\ d})} &=
|
|
{2\textcolor{red}{h} + \textcolor{green}{g} \over \textcolor{blue}{2}} =
|
|
|
|
{2h + h(d-1) \over 2} = {h(1 + d) \over 2 } \\
|
|
2h &= 2h(1 + d)(2 -\ d) \\
|
|
1 &= (1 + d)(2 -\ d) = 2 -\ d + 2d -\ d^2 \\
|
|
0 &= d^2 -\ d -\ 1
|
|
\end{align*}
|
|
|
|
This is the minimal polynomial of the golden ratio $\phi$; it is half the length of the diagonal, so the ratio of a diagonal to a side is also $\phi$. To make calculations easier, some conversions will be made to base $\phi$, or phinary. If you are not familiar already with phinary, I have already written at length about it [here](). Finally, the apothem *a* and height *l* can be calculated by similar triangles.
|
|
|
|
::: {}
|
|
|
|
:::
|
|
|
|
$$
|
|
\begin{align*}
|
|
\textcolor{blue}{c \over a} &= \textcolor{brown}{2 \over \phi},~
|
|
a^2 + 1^2 = c^2 \implies 1 = c^2 -\ a^2 = (c + a)(c -\ a) \\
|
|
l &= c + a = {2a \over \phi} + a = a{2 + \phi \over \phi} = a{12_\phi \over 10_\phi} =
|
|
a{2\bar{1}0_\phi \over 10_\phi} = a(2\bar{1}_\phi) \\ \\
|
|
s &= c -\ a = {2a \over \phi} -\ a = a{2 -\ \phi \over \phi} = a{\bar{1}2_\phi \over 10_\phi} =
|
|
a{2\bar{3}0_\phi \over 10_\phi} = a(2\bar{3}_\phi) \\ \\ \\
|
|
1 &= ls = a^2(2\bar{1}_\phi)(2\bar{3}_\phi) = a^2(4\bar{8}3_\phi) =
|
|
a^2(\bar{4}7_\phi) = a^2(3.\bar{4}_\phi) \\
|
|
a^2 &= {1 \over 3.\bar{4}_\phi} \cdot {43_\phi \over 43_\phi} =
|
|
{43_\phi \over [12]\bar{7}.[\bar{12}]_\phi} = {3 + 4\phi \over 5} \\ \\
|
|
\implies l^2 &= a^2(2\bar{1}_\phi)^2 = {3 + 4\phi \over 5} \cdot (4\bar{4}1_\phi)
|
|
= {3 + 4\phi \over 5} \cdot 5 = 3 + 4\phi
|
|
\end{align*}
|
|
$$
|
|
|
|
The last few steps in solving for $a^2$ are somewhat tricky. The conjugate of $\phi$ is $-{1 \over \phi}$. Since the digit in the $\phi^{-1}$ place value is negative, its conjugate has a positive value in the $\phi$ place value; i.e., $3.\bar{4}_\phi^* = 43_\phi$. Multiplying a quadratic root by its conjugate produces an integer value, which means that the scary quantity $[12]\bar{7}.[\bar{12}]_\phi$ resolves cleanly to 5.
|
|
|
|
The division can also be done explicitly in phinary:
|
|
|
|
$$
|
|
\begin{align*}
|
|
{1 \over 3.\bar{4}_\phi} &= {1 \over 0.\bar{1}3_\phi} =
|
|
{500_\phi \over 5 (\bar{1}3_\phi)} =
|
|
{233_\phi + (0 = \textcolor{red}{4\bar{4}}\bar{4}0_\phi =
|
|
\textcolor{red}{26\bar{2}}\bar{4}0_\phi) \over 5 (\bar{1}3_\phi)} \\
|
|
&= {\bar{2}60\bar{1}3_\phi \over 5 (\bar{1}3_\phi)} =
|
|
{2001_\phi \over 5} =
|
|
{221_\phi \over 5} =
|
|
{43_\phi \over 5} =
|
|
{3 + 4\phi \over 5}
|
|
\end{align*}
|
|
$$
|
|
|
|
Finally, with the length and apothem of a regular pentagon in tow, the geometry of the final two solids may be explored.
|
|
|
|
|
|
The Remaining Solids
|
|
--------------------
|
|
|
|
The icosahedron and dodecahedron are easiest to dissect as many pyramids joined to a single center. This is reminiscent of the area formula which uses the triangulation of a polygon through the incenter.
|
|
|
|
The altitude (*h*) of any one pyramid is the radius of the *insphere* of the solid, which is tangent to the plane of every face. Similarly, the *circumsphere* (circumradius, *r*) contains all vertices, and the *midsphere* (midradius, $\rho$) is tangent to every edge. These will become important shortly.
|
|
|
|
|
|
### The Icosahedron: an Antiprism in Profile
|
|
|
|
The icosahedron may also be thought of as two pentagonal pyramids connected to either base of a pentagonal *antiprism*. An antiprism is a figure similar to a prism, but with the one of the bases twisted relative to the other and with (equilateral) triangles joining them.
|
|
|
|
:::: {layout-ncol="2"}
|
|
::: {.column width="49%"}
|
|
|
|
|
|
Icosahedron with pentagonal antiprism in blue
|
|
:::
|
|
|
|
::: {.column width="49%"}
|
|
|
|
|
|
Construction showing $2a = c$
|
|
:::
|
|
::::
|
|
|
|
A segment connecting the centers of two antipodal faces is a diameter of the insphere. The altitude of one of these faces will be cut into circumradius and inradius. By similar triangles, the circumradius of an equilateral triangle is exactly twice the length of the inradius. This means the inradius is 1/3 of the altitude, or 1/9 of the square of the altitude. With edge length 2, the square of the altitude is 3, so the square of the inradius is ${3 \over 9} = {1 \over 3}$ .
|
|
|
|
The pentagonal antiprism may be bisected bisected along the plane containing the altitudes of two triangles opposite one another. This forms a parallelogram with side lengths of the altitude of an equilateral triangle and height of a pentagon.
|
|
|
|
|
|
|
|
$$
|
|
\begin{align*}
|
|
(\textcolor{green}{2h})^2 &= \textcolor{red}{l}^2 -\ {1 \over 3} = 3 + 4\phi -\ {1 \over 3} =
|
|
{3(3 + 4\phi) -\ 1 \over 3} \\
|
|
h^2 &= {8 + 12\phi \over 3 \cdot 4} = {2 + 3\phi \over 3} = {32_\phi \over 3} \\
|
|
32_\phi &= 210_\phi = 1100_\phi = 10000_\phi = \phi^4 \\ \\
|
|
(2^3 \cdot V_\text{ico})^2 &= \left ( 20 \cdot {Bh \over 3} \right )^2 =
|
|
{20^2 B^2 h^2 \over 3^2} = {5^2 \cdot 4^2 \cdot 3 \cdot {\phi^4 \over 3} \over 3^2} =
|
|
{5^2 \cdot 2^4 \cdot \phi^4 \over 3^2} \\
|
|
V^2 &= {5^2 \cdot 2^4 \cdot \phi^4 \over 2^6 \cdot 3^2} =
|
|
{5^2 \cdot \phi^4 \over 2^2 \cdot 3^2} \\
|
|
V &= {5 \phi^2 \over 6}
|
|
\end{align*}
|
|
$$
|
|
|
|
|
|
### The Dodecahedron
|
|
|
|
The dodecahedron is a bit trickier. It belongs to a class of polyhedra known as *truncated trapezohedra*. However, the bisection trick from the icosahedron still works. Begin by bisecting the solid through antipodal altitudes. This produces an oblong hexagon made up of four pentagon heights and two edges.
|
|
|
|
|
|
|
|
The segment connecting the antipodal midpoints bisects the hexagon into two (isosceles) trapezoids, and is a diameter of the midsphere. Additionally, it is parallel to the two edges. A second midradius is perpendicular to this one, bisecting the trapezoid.
|
|
|
|
|
|
|
|
The inradius is the altitude of a triangle formed by the length of a pentagon (its base), a midradius, and a circumradius. However, the altitude with respect to the midradius is another midradius. This means that the height can be found by equating areas and completing the square.
|
|
|
|
$$
|
|
\begin{align*}
|
|
a^2 + \textcolor{green}{h}^2 &= \textcolor{blue}{\rho}^2,~~
|
|
\textcolor{orange}{l}\textcolor{green}{h} =
|
|
\textcolor{blue}{\rho \rho} \implies \textcolor{orange}{l}^2\textcolor{green}{h}^2 =
|
|
(\textcolor{orange}{(2\bar{1}_\phi) a})^2 h^2 = 5a^2h^2 = \textcolor{blue}{\rho}^4 \\
|
|
5a^2h^2 &= (a^2 + h^2)^2 = a^4 + 2a^2h^2 + h^4 \\ \\
|
|
0 &= a^4 -\ 3a^2h^2 + h^4 = (h^2 -\ x)^2 + y = h^4 -\ 2xh^2 + x^2 + y \\
|
|
-2x &= -3a^2 \implies x = {3a^2 \over 2},~~
|
|
x^2 + y = {9a^4 \over 4} + y = a^4 \\
|
|
y &= {4a^4 \over 4} -\ {9a^4 \over 4} = -{5a^4 \over 4} =
|
|
-{(2\bar{1}_\phi)^2 a^4 \over 4} = -\left ( {(2\bar{1}_\phi) a^2 \over 2} \right )^2
|
|
\\ \\
|
|
(h^2 -\ {3a^2 \over 2})^2 &= -y = \left ( {(2\bar{1}_\phi) a^2 \over 2} \right )^2 \\
|
|
h^2 -\ {3a^2 \over 2} &= {(2\bar{1}_\phi) a^2 \over 2} \\
|
|
h^2 &= {(2\bar{1}_\phi) a^2 \over 2} + {3a^2 \over 2} = {(22_\phi) a^2 \over 2} =
|
|
(11_\phi) a^2 \\
|
|
&= {(11_\phi)(43_\phi) \over 5} = {473_\phi \over 5} =
|
|
{[11]7_\phi \over 5} = {7 + 11\phi \over 5}
|
|
\end{align*}
|
|
$$
|
|
|
|
With the square of the height known, all that is left to do is find the volume.
|
|
|
|
$$
|
|
\begin{align*}
|
|
B^2 &= \left( {Pa \over 2} \right)^2 = (5a)^2 = 25a^2 = 5(43_\phi) \\
|
|
5h^2 &= [11]7_\phi = 740_\phi = 4300_\phi = (43_\phi)(100_\phi) \\
|
|
(2^3 \cdot V_\text{dodec})^2 &= \left (12 \cdot {Bh \over 3} \right )^2 =
|
|
4^2 B^2 h^2 = 2^4 \cdot 5(43_\phi) \cdot {(43_\phi)(100_\phi) \over 5} \\
|
|
V^2 &= {2^4 \cdot (43_\phi)^2 \cdot (100_\phi) \over 2^6} =
|
|
{(43_\phi)^2(10_\phi)^2 \over 2^2} \\
|
|
V &= {(43_\phi)(10_\phi) \over 2} = {(430_\phi) \over 2} = {(74_\phi) \over 2} =
|
|
{4 + 7\phi \over 2}
|
|
\end{align*}
|
|
$$
|
|
|
|
|
|
Closing
|
|
-------
|
|
|
|
Since each of these volumes has been calculated algebraically, there have been no approximate decimal forms. Ordered by size, the volumes of each of the solids are:
|
|
|
|
| Solid | Volume | Approximation | Length of Side with Unit Volume |
|
|
|--------------|-----------------------|-----------------|---------------------------------|
|
|
| Tetrahedron | ${1 \over 6\sqrt 2}$ | 0.1178511302... | 2.039648903... |
|
|
| Octahedron | ${\sqrt{2} \over 3}$ | 0.4714045208... | 1.284898293... |
|
|
| Cube | $1^3$ | 1 | 1 |
|
|
| Icosahedron | ${5\phi^2 \over 6}$ | 2.181694991... | 0.7710253465... |
|
|
| Dodecahedron | ${4 + 7\phi \over 2}$ | 7.663118961... |0.5072220724... |
|
|
|
|
The dodecahedron being so much larger than the icosahedron surprised me, to be honest. When one glances at a set of dice (as one does), it seems like the d20 is larger than the d12, albeit with smaller edges. However, at least in my set, the edges of the d20 are in fact about 1.5 times as long as those of the d12, implying their volumes are roughly equal.
|
|
|
|
***
|
|
|
|
I tried to use as much coordinate-free geometry as I could in producing these diagrams. GeoGebra lacks a tool for producing Platonic solids other than cubes and tetrahedra, so I ended up using approximations for the octahedron and icosahedron diagrams. On the other hand, the hexagon I described in the dodecahedron is of such importance to its construction that I ended up constructing it from scratch. I am rather proud of this because I did so without looking up someone else's. After having written this post, I feel much more competent with compass-and-straightedge constructions.
|
|
|
|
All diagrams made with GeoGebra.
|