629 lines
21 KiB
Plaintext
629 lines
21 KiB
Plaintext
---
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title: "On the Volume of the Platonic Solids"
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description: |
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How big are the Platonic solids in relation to one another?
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format:
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html:
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html-math-method: katex
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date: "TODO"
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date-modified: "2025-06-03"
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categories:
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- geometry
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---
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The Platonic solids have been known for millennia.
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They bear the name of Plato, who spoke of them in his dialogue *Timaeus*.
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He describes their "construction" (sans the dodecahedron) from the most basic
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"isosceles and scalene" triangles, or in modern parlance, "45-45-90 and 30-60-90" triangles, respectively.
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However, the construction was not mathematical, and to my knowledge, each solid was first
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rigorously described from first principles in Book XIII of Euclid's Elements.
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In my teenage years, I recall viewing articles on the solids with their volumes
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proudly displayed next to their surface area.
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While surface area may be troublesome in the case of the dodecahedron
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(as the geometry of regular pentagons is not widely taught),
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it is easy enough for grade schoolers to calculate for cubes,
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and for trigonometry students to calculate for the solids composed of equilateral triangles.
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On the other hand, the volume is somewhat mystical.
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The volume itself is a meaningless quantity for comparison unless put in ratio with another volume.
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Fortunately, edge length is the only free variable in a Platonic solid,
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meaning their volumes can be parametrized by this value alone.
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Further, since the cube has such simple expression for its volume (the edge length cubed),
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it is a natural choice as a base for the comparison[^1].
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Therefore, I will derive this ratio for the solids in question.
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[^1]: Though this is a pure mathematical concept, empirical units use the same convention.
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For instance, a cubic centimeter is defined as the volume occupied by a cube which is a
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centimeter long in each dimension, despite being applicable to volumes of any shape.
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This post will calculate the volume without using any trigonometric functions (sine, cosine, tangent),
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and instead opts for a more compass-and-straightedge approach.
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Consequently, it becomes more natural to calculate the *square* of the volume to better cooperate
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with the Pythagorean theorem.
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A Recap of Geometry
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-------------------
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For those with only a vague recollection (or perhaps none at all) of geometry,
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this section is intended as a refresher.
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### Planar Geometry
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There are [many centers of a triangle](https://faculty.evansville.edu/ck6/encyclopedia/etc.html),
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but for us, two are of primary interest:
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- The *circumcenter* is equidistant from every vertex.
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In other words, it is the center of a circle containing all three vertices.
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- It can be constructed by finding the intersection of the edges' perpendicular bisectors.
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- The distance from a vertex to the circumcenter is called the *circumradius* (*c*).
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- The *incenter* is equidistant from every edge.
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It is the center of a circle which lies tangent to every edge
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(i.e., radii can be drawn which are are perpendicular to the edge).
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- It is constructed by finding the intersection of the lines which bisect each angle.
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- The perpendicular distance from an edge to the incenter is called the *inradius* (*a*).
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::: {}
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Constructing the circumcenter and incenter.
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Angle bisectors in blue, perpendicular bisectors in red, in- and circumradii in green.
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:::
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The inradius is special because it is also an altitude for a triangle formed by the inradius and an edge of the larger triangle.
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This means that the area of the larger triangle is the sum of these smaller triangles.
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$$
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\begin{align*}
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A &= \left ({e_1 a \over 2} + {e_2 a \over 2} + {e_3 a \over 2} \right)
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= \left ({a \over 2} \right ) (e_1 + e_2 + e_3)
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\\
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&= {Pa \over 2}
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\end{align*}
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$$
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This gives an expression for the area.
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For an equilateral triangle, these two centers coincide.
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This is because the perpendicular bisectors of the edges *are* the angle bisectors.
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In fact, the bisection of an angle involves constructing a rhombus, which is made up of
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two isosceles triangles (of which the equilateral triangle is a special case).
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In this case, the inradius is also called the *apothem*, and the difference between it and the circumradius
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is immediately apparent and called the *sagitta* (*s*).
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This idea of incenters and circumcenters can be extended to other 2D figures
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such as the square and regular pentagon.
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For a square, the center is simply the intersection of the diagonals (i.e., the diagonals' common midpoint).
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The pentagon is trickier and will be discussed later.
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Regardless, the expression for the area ${Pa \over 2}$ still works,
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since the polygon can be triangulated through the center in a similar way.
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### Cubes, Prisms, and Pyramids
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Now we speak of 3D geometry.
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The volume of a prism is equal to the height times the area of the base,
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where the "height" is orthogonal to the plane of the base.
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Pyramids with the same height and base have one-third this area.
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$$
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V_\text{prism} = Bh,~~ V_\text{pyramid} = {Bh \over 3}
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$$
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This volume formula can be made more intuitive by considering the cube.
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The pyramid formed by one of the faces and an edge perpendicular to it will contain
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one square and two half-squares, or two squares in total.
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Therefore three pyramids are needed to recreate all six faces of the cube.
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For a slightly more detailed explanation, consider a point inside the face on top of the cube.
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Its (perpendicular) distance from one edge is *x* and its distance to an edge adjacent to that is *y*.
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Connecting all other bases to this point produces five pyramids, whose bases all have the same area.
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Designate these pyramids as "bottom", "left", "right", "front", and "back",
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where left and right correspond to *x* and front and back correspond to *y*.
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$$
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\begin{align*}
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V_\text{cube}
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&= Bh = V_\text{bottom} + V_\text{left} + V_\text{right} V_\text{front} + V_\text{back}
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\\
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&= rBh + rBx + rB(h-x) + rBy + rB(h-y)
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\\
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&= rBh + rBh + rBh \implies 1 = 3r
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\\
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r &= {1 \over 3}
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\end{align*}
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$$
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This can be generalized to a pyramid based on any prism, where the top point lies
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in the plane of one of the bases.
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However, this is beyond the scope of this post.
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Simple Solids: the Octahedron and the Tetrahedron
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-------------------------------------------------
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While "simple" is a bit of a misnomer, their volumes are easiest to appreciate,
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since they do not need regular pentagons.
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### Octahedron
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The octahedron can be thought of as two square pyramids joined end-on-end,
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with uniform edge length throughout.
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Since the base is a square, its center is equidistant from the vertices of the base.
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An alternative, congruent square can be noticed by the symmetry of the octahedron,
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meaning the center is also equidistant from the top of the square pyramid,
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and that the segment connecting the two is an altitude of the pyramid.
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::: {}
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Primary square in blue, secondary square in red. Diagonals of both squares shown.
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:::
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The length of this altitude is simply half of the diagonal of the square.
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Therefore, the volume of an octahedron (calculated using edge length 1) is:
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$$
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\begin{align*}
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B^2
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&= (1^2)^2
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\\
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(2h)^2
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&= 4h^2 = 1^2 + 1^2 = 2
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\\
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V_\text{sq.pyr.}^2
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&= {B^2 h^2 \over 3^2} = {1 \cdot {2 / 4} \over 3^2}
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\\
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(1^3 \cdot V_\text{oct})^2
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&= (2V_\text{sq.pyr})^2 = 4V_\text{sq.pyr}^2
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= 4 \cdot {2 / 4 \over 3^2} = {2 \over 3^2}
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\\ \\
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V_\text{oct} &= {\sqrt{2} \over 3}
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\end{align*}
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$$
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### Tetrahedron
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From here on out, it becomes convenient to specify the edge length of every solid to be 2,
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since this simplifies the bisection of edges.
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This happens to coincide with Plato's description, where the equilateral triangle is described as
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being formed from two 30-60-90 triangles.
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That is, the edge length of the equilateral triangle was twice the "unit" length: the shortest side of the 30-60-90 triangle.
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However, we must remember that volumes need to be put in ratio with a cube volume of 8:
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$$
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\begin{align*}
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{ V_\text{solid[2]} \over V_\text{cube[2]} }
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&= { V_\text{solid[1]} \over V_\text{cube[1]} }
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\\
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\implies V_\text{solid[2]}
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&= { V_\text{cube[2]} \over V_\text{cube[1]} } V_\text{solid[1]}
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\\
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&= { 2^3 \over 1 } V_\text{solid[1]}
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$$
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Since the tetrahedron is itself a pyramid, its volume follows from the earlier formula.
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First, we must calculate the (square of the) area of the base of an equilateral triangle.
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:::: {layout-ncol="2"}
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::: {.column width="49%"}
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:::
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::: {.column width="49%"}
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$$
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\begin{align*}
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d_\text{altitude}^2
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&= \textcolor{orange}{2}^2 -\ \textcolor{green}{1}^2 = 3
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\\
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B^2
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&= \left ( {2 \cdot d_\text{altitude} \over 2} \right )^2 = 3
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\end{align*}
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$$
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:::
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::::
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Next, bisect this triangle through any edge, then use it to bisect the tetrahedron through
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the plane containing this line and the remaining edge.
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This forms an isosceles triangle containing an edge and the altitudes of two faces.
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Bisecting the angle where the two alittudes meet (perpendicularly) bisects the edge.
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:::: {layout-ncol="2"}
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::: {.column width="49%"}
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:::
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::: {.column width="49%"}
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$$
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\begin{align*}
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\textcolor{blue}{d_\text{length}}^2
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&= d_\text{altitude}^2 -\ \textcolor{green}{1}^2 = 2
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\\
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(2A_\text{center})^2
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&= (2d_\text{length})^2 = (\textcolor{red}{h} d_\text{altitude})^2
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\\
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&= 4 \cdot 2 = 3h^2
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\\
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h^2 &= 8 / 3
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\end{align*}
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$$
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:::
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::::
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Since *h* is known, we can calculate the volume.
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$$
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({ 2^3 \cdot V_\text{tet[1]} })^2
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&= {B^2 h^2 \over 3^2} = {3 \cdot (8/3) \over 3^2}
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\\
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V_\text{tet[1]}^2
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&= {8 \over 2^6 \cdot 3^2} = {1 \over 2^3 \cdot 3^2} = {1 \over 2 \cdot 6^2}
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\\
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V_\text{tet[1]}
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&= \sqrt{1 \over 6^2 \cdot 2} = {1 \over 6\sqrt 2}
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\end{align*}
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$$
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### Returning to 2D: Regular Pentagons
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Both of the icosahedron and dodecahedron contain regular pentagons.
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Thus, it is necessary to examine them in detail.
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The regular pentagon has five diagonals, which form a pentagram.
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Since all angles in a regular pentagon are equal, the trapezoid formed by three consecutive edges
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and one diagonal is isosceles.
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This means the diagonal is parallel to one of the edges, which applies to all diagonals by symmetry.
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Since the diagonal is parallel to one of the sides, a parallelogram can be formed from
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two sides and segements from two diagonals.
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More specifically, this parallelogram is a rhombus, since the segments must have equal lengths
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to the sides.
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::: {}
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Left: Pentagram in regular pentagon;
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Middle: Isosceles trapezoid, with parallel lines marked in blue;
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Right: Rhombus in regular pentagon
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:::
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Bisect the pentagon vertically and let the length of half of the diagonal of a pentagon be *d*,
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half the length of the other diagonal of a rhombus be *h*,
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and the remaining height of the pentagon be *g*.
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$$
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\begin{align*}
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\textcolor{orange}{d}^2 + \textcolor{red}{h}^2
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&= \textcolor{blue}{2}^2
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\\
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2\textcolor{darkblue}{A_\text{blue}}
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&= 2\textcolor{green}{g} = h(\textcolor{magenta}{d -\ (2 -\ d)}) = h (2d -\ 2)
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\\
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\implies g &= {h(2d -\ 2) \over 2} = h(d -\ 1)
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\end{align*}
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$$
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Notice that the center of a pentagram contains a regular pentagon.
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This means that the ratio of its height to the side is equal to the ratio of
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the larger pentagon's height to its side.
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This is enough information to deduce *d*:
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$$
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\begin{align*}
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{\textcolor{red}{h} \over 2(\textcolor{brown}{2 -\ d})}
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&= {2\textcolor{red}{h} + \textcolor{green}{g} \over \textcolor{blue}{2}}
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= {2h + h(d-1) \over 2} = {h(1 + d) \over 2 }
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\\
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2h &= 2h(1 + d)(2 -\ d)
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\\
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1 &= (1 + d)(2 -\ d) = 2 -\ d + 2d -\ d^2
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\\
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0 &= d^2 -\ d -\ 1
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\end{align*}
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$$
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This is the minimal polynomial of the golden ratio *φ*.
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It is half the length of the diagonal, so the ratio of a diagonal to a side is also *φ*.
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To make calculations easier, some conversions will be made to base *φ*, or phinary.
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If you are not familiar already with phinary, I have already written at length about it [here](
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/posts/polycount/1
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).
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Finally, the apothem *a* and height *l* can be calculated by similar triangles.
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::: {}
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:::
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$$
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\begin{align*}
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\textcolor{blue}{c \over a}
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&= \textcolor{brown}{2 \over \phi},~ a^2 + 1^2 = c^2
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\implies 1 = c^2 -\ a^2 = (c + a)(c -\ a)
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\\
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l
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&= c + a = {2a \over \phi} + a = a{2 + \phi \over \phi}
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= a{12_\phi \over 10_\phi}
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= a{2\bar{1}0_\phi \over 10_\phi} = a(2\bar{1}_\phi)
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\\ \\
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s
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&= c -\ a = {2a \over \phi} -\ a
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= a{2 -\ \phi \over \phi} = a{\bar{1}2_\phi \over 10_\phi}
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= a{2\bar{3}0_\phi \over 10_\phi} = a(2\bar{3}_\phi)
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\\ \\ \\
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1
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&= ls = a^2(2\bar{1}_\phi)(2\bar{3}_\phi)
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= a^2(4\bar{8}3_\phi)
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= a^2(\bar{4}7_\phi) = a^2(3.\bar{4}_\phi)
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\\
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a^2
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&= {1 \over 3.\bar{4}_\phi} \cdot {43_\phi \over 43_\phi}
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= {43_\phi \over [12]\bar{7}.[\bar{12}]_\phi}
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= {3 + 4\phi \over 5}
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\\ \\
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\implies l^2
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&= a^2(2\bar{1}_\phi)^2
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= {3 + 4\phi \over 5} \cdot (4\bar{4}1_\phi)
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= {3 + 4\phi \over 5} \cdot 5 = 3 + 4\phi
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\end{align*}
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$$
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The last few steps in solving for $a^2$ are somewhat tricky.
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The conjugate of *φ* is $-{1 \over \phi}$.
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Since the digit in the *φ*^^-1^^ place value is negative, its conjugate has a positive value in the *φ* place value,
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i.e., $3.\bar{4}_{\phi^*} = 43_\phi$.
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Multiplying a quadratic root by its conjugate produces an integer value,
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which means that the scary quantity $[12]\bar{7}.[\bar{12}]_\phi$ resolves cleanly to 5.
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The division can also be done explicitly in phinary:
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$$
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\begin{align*}
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{1 \over 3.\bar{4}_\phi}
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&= {1 \over 0.\bar{1}3_\phi}
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= {500_\phi \over 5 (\bar{1}3_\phi)}
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= {233_\phi + (0 = \textcolor{red}{4\bar{4}}\bar{4}0_\phi
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= \textcolor{red}{26\bar{2}}\bar{4}0_\phi) \over 5 (\bar{1}3_\phi)}
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\\
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&= {\bar{2}60\bar{1}3_\phi \over 5 (\bar{1}3_\phi)}
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= {2001_\phi \over 5}
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= {221_\phi \over 5}
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= {43_\phi \over 5}
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= {3 + 4\phi \over 5}
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\end{align*}
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$$
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The Remaining Solids
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--------------------
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With the diagonal length and apothem of a regular pentagon in tow, the geometry of
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the final two solids may be explored.
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As a reminder, these solids will have edge length 2, so they will be put in ratio with a cube volume of 8.
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The icosahedron and dodecahedron are easiest to dissect as many pyramids joined to a single center.
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This is reminiscent of the area formula which uses the triangulation of a regular polygon its center.
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The altitude (*h*) of any one pyramid is the radius of the *insphere* of the solid,
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which is tangent to the plane of every face.
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Similarly, the *circumsphere* (circumradius, *r*) contains all vertices,
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and the *midsphere* (midradius, *ρ*) is tangent to every edge.
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These will become important shortly.
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### The Icosahedron: an Antiprism in Profile
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The icosahedron may also be thought of as two pentagonal pyramids connected to
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either base of a pentagonal *antiprism*.
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An antiprism is a figure similar to a prism, but with the one of the bases twisted relative
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to the other and with (equilateral) triangles joining them.
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:::: {layout-ncol="2"}
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::: {.column width="49%"}
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Icosahedron with pentagonal antiprism in blue
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:::
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::: {.column width="49%"}
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Construction showing $2a = c$
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:::
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::::
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A segment connecting the centers of two antipodal faces is a diameter of the insphere.
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The altitude of one of these faces will be cut into circumradius and inradius.
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By similar triangles, the circumradius of an equilateral triangle is
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exactly twice the length of the inradius.
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This means the inradius is 1/3 of the altitude, or 1/9 of the square of the altitude.
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With edge length 2, the square of the altitude is 3, so the square of the inradius is
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${3 \over 9} = {1 \over 3}$ .
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The pentagonal antiprism may be bisected bisected along the plane containing the altitudes
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of two triangles opposite one another.
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This forms a parallelogram with side lengths of the altitude of an equilateral triangle
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and height of a pentagon.
|
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$$
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\begin{align*}
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(\textcolor{green}{2h})^2
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&= \textcolor{red}{l}^2 -\ {1 \over 3}
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= 3 + 4\phi -\ {1 \over 3}
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= {3(3 + 4\phi) -\ 1 \over 3}
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\\
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h^2
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&= {8 + 12\phi \over 3 \cdot 4}
|
||
= {2 + 3\phi \over 3}
|
||
= {32_\phi \over 3}
|
||
\\
|
||
32_\phi
|
||
&= 210_\phi = 1100_\phi = 10000_\phi
|
||
= \phi^4
|
||
\\ \\
|
||
(2^3 \cdot V_\text{ico[1]})^2
|
||
&= \left ( 20 \cdot {Bh \over 3} \right )^2
|
||
= {20^2 B^2 h^2 \over 3^2} = {5^2 \cdot 4^2 \cdot 3 \cdot {\phi^4 \over 3} \over 3^2}
|
||
= {5^2 \cdot 2^4 \cdot \phi^4 \over 3^2}
|
||
\\
|
||
V^2
|
||
&= {5^2 \cdot 2^4 \cdot \phi^4 \over 2^6 \cdot 3^2}
|
||
= {5^2 \cdot \phi^4 \over 2^2 \cdot 3^2}
|
||
\\
|
||
V &= {5 \phi^2 \over 6}
|
||
\end{align*}
|
||
$$
|
||
|
||
|
||
### 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.
|
||
|
||
$$
|
||
\begin{align*}
|
||
B^2
|
||
&= \left( {Pa \over 2} \right)^2
|
||
= (5a)^2 = 25a^2 = 5(43_\phi)
|
||
\\
|
||
5h^2
|
||
&= [11]7_\phi = 740_\phi = 4300_\phi
|
||
= (43_\phi)(100_\phi)
|
||
\\
|
||
(2^3 \cdot V_\text{dodec[1]})^2
|
||
&= \left (12 \cdot {Bh \over 3} \right )^2 = 4^2 B^2 h^2
|
||
= 2^4 \cdot 5(43_\phi) \cdot {(43_\phi)(100_\phi) \over 5}
|
||
\\
|
||
V^2
|
||
&= {2^4 \cdot (43_\phi)^2 \cdot (100_\phi) \over 2^6}
|
||
= {(43_\phi)^2(10_\phi)^2 \over 2^2}
|
||
\\
|
||
V
|
||
&= {(43_\phi)(10_\phi) \over 2} = {(430_\phi) \over 2} = {(74_\phi) \over 2}
|
||
= {4 + 7\phi \over 2}
|
||
\end{align*}
|
||
$$
|
||
|
||
|
||
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 one of my sets, 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 cheating in coordinates 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.
|