add images to platonic-volume

posts/misc/platonic-volume/index.qmd

@@ -5,12 +5,20 @@ description: |
 format:
   html:
     html-math-method: katex
-date: "TODO"
-date-modified: "2025-06-03"
+date: "2021-07-25"
+date-modified: "2025-06-13"
 categories:
   - geometry
 ---
 
+<style>
+.figure-img {
+  max-width: 512px;
+  object-fit: contain;
+  height: 100%;
+}
+</style>
+
 
 The Platonic solids have been known for millennia.
 They bear the name of Plato, who spoke of them in his dialogue *Timaeus*.
@@ -27,19 +35,18 @@ While surface area may be troublesome in the case of the dodecahedron
   and for trigonometry students to calculate for the solids composed of equilateral triangles.
 On the other hand, the volume is somewhat mystical.
 
-The volume itself is a meaningless quantity for comparison unless put in ratio with another volume.
 Fortunately, edge length is the only free variable in a Platonic solid,
   meaning their volumes can be parametrized by this value alone.
-Further, since the cube has such simple expression for its volume (the edge length cubed),
-  it is a natural choice as a base for the comparison[^1].
-Therefore, I will derive this ratio for the solids in question.
+The volume itself is a meaningless quantity for comparison unless put in ratio with another volume,
+  but since the cube has such simple expression for its volume (the edge length cubed),
+  it is a natural choice for this comparison[^1].
 
 [^1]: Though this is a pure mathematical concept, empirical units use the same convention.
 For instance, a cubic centimeter is defined as the volume occupied by a cube which is a
   centimeter long in each dimension, despite being applicable to volumes of any shape.
 
-This post will calculate the volume without using any trigonometric functions (sine, cosine, tangent),
-  and instead opts for a more compass-and-straightedge approach.
+This post will calculate this ratio without using any trigonometric functions (sine, cosine, tangent),
+  instead opting for a more compass-and-straightedge approach.
 Consequently, it becomes more natural to calculate the *square* of the volume to better cooperate
   with the Pythagorean theorem.
 
@@ -62,21 +69,21 @@ There are [many centers of a triangle](https://faculty.evansville.edu/ck6/encycl
     - The distance from a vertex to the circumcenter is called the *circumradius* (*c*).
 - The *incenter* is equidistant from every edge.
   It is the center of a circle which lies tangent to every edge
-  (i.e., radii can be drawn which are are perpendicular to the edge).
-    - It is constructed by finding the intersection of the lines which bisect each angle.
+    - It can be 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*).
 
-::: {}
-![](index_files/triangle centers.png)
+![
+  Constructing the circumcenter and incenter.
+  Angle bisectors in blue, perpendicular bisectors in red, in- and circumradii in green.
+](./triangle_centers.png)
 
-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.
+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.
 
-![](index_files/incenter area.png)
+![
+  &nbsp;
+](./incenter_area.png)
 
 $$
 \begin{align*}
@@ -95,7 +102,9 @@ In fact, the bisection of an angle involves constructing a rhombus, which is mad
 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*).
 
-![](index_files/equilateral center.png)
+![
+  &nbsp;
+](./equilateral_center.png)
 
 This idea of incenters and circumcenters can be extended to other 2D figures
   such as the square and regular pentagon.
@@ -104,7 +113,7 @@ 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.
 
-![](index_files/square pentagon circles.png)
+![](./square_pentagon_circles.png)
 
 
 ### Cubes, Prisms, and Pyramids
@@ -129,12 +138,12 @@ Connecting all other bases to this point produces five pyramids, whose bases all
 Designate these pyramids as "bottom", "left", "right", "front", and "back",
   where left and right correspond to *x* and front and back correspond to *y*.
 
-![](index_files/cube pyramids.png)
+![](./cube_pyramids.png)
 
 $$
 \begin{align*}
   V_\text{cube}
-    &= Bh = V_\text{bottom} + V_\text{left} + V_\text{right} V_\text{front} + V_\text{back}
+    &= 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)
   \\
@@ -149,82 +158,88 @@ This can be generalized to a pyramid based on any prism, where the top point lie
 However, this is beyond the scope of this post.
 
 
+### Simplifying Units
+
+Though it may make sense to use unit lengths for the edges of 3D figures, bisection of edges
+  is easier if the edges are composed of two units.
+This happens to coincide with Plato's description -- the equilateral triangle is described as
+  being formed from two 30-60-90 triangles.
+That is, the edge length of the equilateral triangle was twice the "unit" length:
+  the shortest side of the 30-60-90 triangle.
+
+Thus, ratios to the volume of the *unit* cube must be adjusted by a factor of 8:
+
+$$
+\begin{align*}
+  { V_\text{solid}[2] \over V_\text{cube}[2] }
+    &= { V_\text{solid}[1] \over V_\text{cube}[1] }
+  \\
+  \implies V_\text{solid}[2]
+    &= { V_\text{cube}[2] \over V_\text{cube}[1] } V_\text{solid}[1]
+  \\
+  &= { 2^3 \over 1 } V_\text{solid}[1]
+\end{align*}
+$$
+
+
 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.
+While "simple" is a bit of a misnomer, their volumes do not require regular pentagons,
+  and are therefore easiest to appreciate.
 
 
 ### 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.
+Since the base of the pyramid is a square, its center is equidistant from the vertices of the base.
+Due to the symmetry of the octahedron, this center is the same no matter which square we pick.
+Thus, the altitude *h* of the square pyramid is simply half of the diagonal of the square.
 
-::: {}
-![](index_files/octahedron squares.png)
+$$
+(2h)^2 = e^2 + e^2 = 2 \cdot 2^2 = 2^3
+$$
 
-Primary square in blue, secondary square in red. Diagonals of both squares shown.
-:::
+![
+  Primary square in blue, secondary square in red. Diagonals of both squares shown.
+](./no-pentagons/octahedron_squares.png)
 
-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:
+Therefore, the volume of an octahedron is:
 
 $$
 \begin{align*}
   B^2
-    &= (1^2)^2
+    &= (e^2)^2
+    = (2^2)^2 = 2^4
   \\
-  (2h)^2
-    &= 4h^2 = 1^2 + 1^2 = 2
+  (2 V_\text{sq.pyr.})^2
+    &= (2^2) \left( {B^2 h^2 \over 3^2} \right)
+    = {B^2 (2h^2) \over 3^2}
+    = {{2^4 \cdot 2^3} \over 3^2}
+    = {2^7 \over 3^2}
   \\
-  V_\text{sq.pyr.}^2
-    &= {B^2 h^2 \over 3^2} = {1 \cdot {2 / 4} \over 3^2}
+  (2^3 \cdot V_\text{oct})^2
+    &= {2^7 \over 3^2}
+    \implies V_\text{oct}^2
+      = {2 \over 3^2}
   \\
-  (1^3 \cdot V_\text{oct})^2
-    &= (2V_\text{sq.pyr})^2 = 4V_\text{sq.pyr}^2
-    = 4 \cdot {2 / 4 \over 3^2} = {2 \over 3^2}
-  \\ \\
-  V_\text{oct} &= {\sqrt{2} \over 3}
+  V_\text{oct}
+    &= {\sqrt{2} \over 3}
 \end{align*}
 $$
 
 
 ### Tetrahedron
 
-From here on out, it becomes convenient to specify the edge length of every solid to be 2,
-  since this simplifies the bisection of edges.
-This happens to coincide with Plato's description, where the equilateral triangle is described as
-  being formed from two 30-60-90 triangles.
-That is, the edge length of the equilateral triangle was twice the "unit" length: the shortest side of the 30-60-90 triangle.
-
-However, we must remember that volumes need to be put in ratio with a cube volume of 8:
-
-$$
-\begin{align*}
-  { V_\text{solid[2]} \over V_\text{cube[2]} }
-    &= { V_\text{solid[1]} \over V_\text{cube[1]} }
-  \\
-  \implies V_\text{solid[2]}
-    &= { V_\text{cube[2]} \over V_\text{cube[1]} } V_\text{solid[1]}
-  \\
-  &= { 2^3 \over 1 } V_\text{solid[1]}
-$$
-
 Since the tetrahedron is itself a pyramid, its volume follows from the earlier formula.
 First, we must calculate the (square of the) area of the base of an equilateral triangle.
 
-:::: {layout-ncol="2"}
+:::: {layout-ncol="2" layout-valign="center"}
 
-::: {.column width="49%"}
-![](index_files/equilateral triangle area.png)
-:::
+![](./no-pentagons/equilateral_triangle_area.png)
 
-::: {.column width="49%"}
+::: {.column}
 $$
 \begin{align*}
   d_\text{altitude}^2
@@ -243,13 +258,11 @@ Next, bisect this triangle through any edge, then use it to bisect the tetrahedr
 This forms an isosceles triangle containing an edge and the altitudes of two faces.
 Bisecting the angle where the two alittudes meet (perpendicularly) bisects the edge.
 
-:::: {layout-ncol="2"}
+:::: {layout-ncol="2" layout-valign="center"}
 
-::: {.column width="49%"}
-![](index_files/tetrahedron bisect.png)
-:::
+![](./no-pentagons/tetrahedron_bisect.png)
 
-::: {.column width="49%"}
+::: {.column}
 $$
 \begin{align*}
   \textcolor{blue}{d_\text{length}}^2
@@ -270,13 +283,14 @@ $$
 Since *h* is known, we can calculate the volume.
 
 $$
-  ({ 2^3 \cdot V_\text{tet[1]} })^2
+\begin{align*}
+  ({ 2^3 \cdot V_\text{tet} })^2
     &= {B^2 h^2 \over 3^2} = {3 \cdot (8/3) \over 3^2}
   \\
-  V_\text{tet[1]}^2
+  V_\text{tet}^2
     &= {8 \over 2^6 \cdot 3^2} = {1 \over 2^3 \cdot 3^2} = {1 \over 2 \cdot 6^2}
   \\
-  V_\text{tet[1]}
+  V_\text{tet}
     &= \sqrt{1 \over 6^2 \cdot 2} = {1 \over 6\sqrt 2}
 \end{align*}
 $$
@@ -297,19 +311,19 @@ Since the diagonal is parallel to one of the sides, a parallelogram can be forme
 More specifically, this parallelogram is a rhombus, since the segments must have equal lengths
   to the sides.
 
-::: {}
-![](index_files/pentagram rhombus.png)
-
-Left: Pentagram in regular pentagon;
-Middle: Isosceles trapezoid, with parallel lines marked in blue;
-Right: Rhombus in regular pentagon
-:::
+![
+  Left: Pentagram in regular pentagon; \\
+  Middle: Isosceles trapezoid, with parallel lines marked in blue; \\
+  Right: Rhombus in regular pentagon
+](./pentagram_rhombus.png)
 
 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*.
 
-![](index_files/pentagon measurements.png)
+![
+  &nbsp;
+](./pentagon_measurements.png)
 
 $$
 \begin{align*}
@@ -326,7 +340,7 @@ $$
 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*:
+This gives us enough information to deduce *d*:
 
 $$
 \begin{align*}
@@ -342,74 +356,78 @@ $$
 \end{align*}
 $$
 
-This is the minimal polynomial of the golden ratio *φ*.
+The (positive) quantity satisfying this polynomial is the golden ratio *φ*.
 It is half the length of the diagonal, so the ratio of a diagonal to a side is also *φ*.
 
 To make calculations easier, some conversions will be made to base *φ*, or phinary.
 If you are not familiar already with phinary, I have already written at length about it [here](
   /posts/polycount/1
 ).
-Finally, the apothem *a* and height *l* can be calculated by similar triangles.
 
-::: {}
-![](index_files/pentagon apothem.png)
-:::
+To calculate the apothem, we can calculate the sagitta *s* and height *l* by similar triangles.
+
+![
+  &nbsp;
+](./pentagon_apothem.png)
 
 $$
 \begin{align*}
-  \textcolor{blue}{c \over a}
-    &= \textcolor{brown}{2 \over \phi},~ a^2 + 1^2 = c^2
-    \implies 1 = c^2 -\ a^2 = (c + a)(c -\ a)
+  &\textcolor{blue}{c \over a}
+    = \textcolor{brown}{2 \over \phi}
   \\
-  l
-    &= c + a = {2a \over \phi} + a = a{2 + \phi \over \phi}
+  l &= c + a = {2a \over \phi} + a = a{2 + \phi \over \phi}
     = a{12_\phi \over 10_\phi}
     = a{2\bar{1}0_\phi \over 10_\phi} = a(2\bar{1}_\phi)
-  \\ \\
-  s
-    &= c -\ a = {2a \over \phi} -\ a
+  \\
+  s &= c -\ a = {2a \over \phi} -\ a
     = a{2 -\ \phi \over \phi} = a{\bar{1}2_\phi \over 10_\phi}
     = a{2\bar{3}0_\phi \over 10_\phi} = a(2\bar{3}_\phi)
-  \\ \\ \\
-  1
-    &= ls = a^2(2\bar{1}_\phi)(2\bar{3}_\phi)
+\end{align*}
+$$
+
+Finally, the Pythagorean theorem tells us that the product of these quantities is 1,
+  allowing us to calculate *a*.
+
+$$
+\begin{align*}
+  1 &= c^2 -\ a^2 = (c + a)(c -\ a) = ls
+  \\
+  &= a^2(2\bar{1}_\phi)(2\bar{3}_\phi)
     = a^2(4\bar{8}3_\phi)
     = a^2(\bar{4}7_\phi) = a^2(3.\bar{4}_\phi)
   \\
-  a^2
-    &= {1 \over 3.\bar{4}_\phi} \cdot {43_\phi \over 43_\phi}
+  a^2 &= {1 \over 3.\bar{4}_\phi} \cdot {43_\phi \over 43_\phi}
     = {43_\phi \over [12]\bar{7}.[\bar{12}]_\phi}
     = {3 + 4\phi \over 5}
-  \\ \\
-  \implies l^2
-    &= a^2(2\bar{1}_\phi)^2
+  \\
+  \implies l^2 &= a^2(2\bar{1}_\phi)^2
     = {3 + 4\phi \over 5} \cdot (4\bar{4}1_\phi)
     = {3 + 4\phi \over 5} \cdot 5 = 3 + 4\phi
 \end{align*}
 $$
 
 The last few steps in solving for $a^2$ are somewhat tricky.
-The conjugate of *φ* is $-{1 \over \phi}$.
-Since the digit in the *φ*^^-1^^ place value is negative, its conjugate has a positive value in the *φ* place value,
-  i.e., $3.\bar{4}_{\phi^*} = 43_\phi$.
+The conjugate of *φ* is $\varphi^* = -{1 \over \varphi}$.
+Since the digit in the *φ*^-1^ place value is negative, its conjugate has a positive value in the *φ* place value,
+  i.e., $3.\bar{4}_{\varphi^*} = 43_\varphi$.
 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.
+  which means that the scary quantity $[12]\bar{7}.[\bar{12}]_\varphi$ resolves cleanly to 5.
 
 The division can also be done explicitly in phinary:
 
 $$
 \begin{align*}
-  {1 \over 3.\bar{4}_\phi}
-    &= {1 \over 0.\bar{1}3_\phi}
-    = {500_\phi \over 5 (\bar{1}3_\phi)}
-    = {233_\phi + (0 = \textcolor{red}{4\bar{4}}\bar{4}0_\phi
-    = \textcolor{red}{26\bar{2}}\bar{4}0_\phi) \over 5 (\bar{1}3_\phi)}
+  {1 \over 3.\bar{4}_\varphi}
+    &= {1 \over 0.\bar{1}3_\varphi}
+    = {500_\varphi \over 5 (\bar{1}3_\varphi)}
+    = {233_\varphi + (0 = \textcolor{red}{4\bar{4}}\bar{4}0_\varphi
+    = \textcolor{red}{26\bar{2}}\bar{4}0_\varphi) \over 5 (\bar{1}3_\varphi)}
   \\
-  &= {\bar{2}60\bar{1}3_\phi \over 5 (\bar{1}3_\phi)}
-    = {2001_\phi \over 5}
-    = {221_\phi \over 5}
-    = {43_\phi \over 5}
-    = {3 + 4\phi \over 5}
+  &= {\bar{2}60\bar{1}3_\varphi \over 5 (\bar{1}3_\varphi)}
+    = {2001_\varphi \over 5}
+    = {221_\varphi \over 5}
+    = {43_\varphi \over 5}
+    = {3 + 4\varphi \over 5}
 \end{align*}
 $$
 
@@ -419,7 +437,6 @@ The Remaining Solids
 
 With the diagonal length and apothem of a regular pentagon in tow, the geometry of
   the final two solids may be explored.
-As a reminder, these solids will have edge length 2, so they will be put in ratio with a cube volume of 8.
 
 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 regular polygon its center.
@@ -433,47 +450,44 @@ These will become important shortly.
 
 ### The Icosahedron: an Antiprism in Profile
 
-The icosahedron may also be thought of as two pentagonal pyramids connected to
+The icosahedron can be decomposed into 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%"}
-![](index_files/icosahedron with antiprism.png)
-
-Icosahedron with pentagonal antiprism in blue
-:::
-
-::: {.column width="49%"}
-![](index_files/equilateral triangles.png)
-
-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
-  ${3 \over 9} = {1 \over 3}$ .
+![
+  Icosahedron with pentagonal antiprism in blue
+](./pentagons/icosahedron_with_antiprism.png)
 
 The pentagonal antiprism may be bisected bisected along the plane containing the altitudes
-  of two triangles opposite one another.
+  of two antipodal triangles.
 This forms a parallelogram with side lengths of the altitude of an equilateral triangle
-  and height of a pentagon.
+  and height of a pentagon (*l*).
 
-![](index_files/antiprism measures.png)
+A segment connecting the centers of two antipodal faces is a diameter of the insphere (2*h*).
+This diameter can be translated along the parallelogram by *a*, the apothem of an equilateral trangle.
+This forms a right triangle with *a* and 2*h* as legs and *l* as the hypotenuse.
+
+::: {layout-ncol="2" layout-valign="center"}
+![
+  &nbsp;
+](./pentagons/antiprism_measures.png)
+
+![
+  Construction showing $2a = c$
+](./pentagons/equilateral_triangles.png)
+:::
+
+For an edge length of 2, we know from the tetrahedron that $d_{altitude}^2 = (a + c)^2 = 3$.
+From the above construction, $(3a)^2 = 3 \implies a^2 = 1 / 3$.
+So,
 
 $$
 \begin{align*}
   (\textcolor{green}{2h})^2
-    &= \textcolor{red}{l}^2 -\ {1 \over 3}
-    = 3 + 4\phi -\ {1 \over 3}
+    &= \textcolor{red}{l}^2 - a^2
+    = \textcolor{red}{3 + 4\phi}
+      -\ {1 \over 3}
     = {3(3 + 4\phi) -\ 1 \over 3}
   \\
   h^2
@@ -484,17 +498,23 @@ $$
   32_\phi
     &= 210_\phi = 1100_\phi = 10000_\phi
     = \phi^4
-  \\ \\
-  (2^3 \cdot V_\text{ico[1]})^2
+\end{align*}
+$$
+
+With the inradius in tow, we can calculate the volume of the icosahedron.
+
+$$
+\begin{align*}
+  (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
+  V_\text{ico}^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}
+  V_\text{ico} &= {5 \phi^2 \over 6}
 \end{align*}
 $$
 
@@ -503,31 +523,40 @@ $$
 
 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.
 
+The bisection trick from the icosahedron mostly still works.
 Begin by bisecting the solid through antipodal altitudes.
 This produces an oblong hexagon made up of four pentagon heights and two edges.
 
-![](index_files/dodecahedron with hemisphere.png)
+![
+  &nbsp;
+](./pentagons/bisected_dodecahedron.png)
 
 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.
+It is parallel to the two edges, so dropping an altitude from this edge to the diameter
+  produces a midradius.
 
-![](index_files/dodecahedron measures.png)
+![
+  &nbsp;
+](./pentagons/dodecahedron_measures.png)
 
-The inradius is the altitude of a triangle formed by the length of a pentagon (its base),
+The inradius *h* 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.
+The inradius meets the pentagon at its center, so we can use its apothem to relate
+  the inradius and midradius.
+
+$$
+a^2 + \textcolor{green}{h}^2 = \textcolor{blue}{\rho}^2
+$$
+
+From the trapezoid, we know that the altitude with respect to the midradius is another midradius.
+This means that the height can be found by equating areas.
 
 $$
 \begin{align*}
-  a^2 + \textcolor{green}{h}^2
-    &= \textcolor{blue}{\rho}^2,~~
-    \textcolor{orange}{l}\textcolor{green}{h}
-    = \textcolor{blue}{\rho \rho} \implies \textcolor{orange}{l}^2\textcolor{green}{h}^2
+  \textcolor{orange}{l}\textcolor{green}{h}
+    &= \textcolor{blue}{\rho \rho} \implies \textcolor{orange}{l}^2\textcolor{green}{h}^2
     = (\textcolor{orange}{(2\bar{1}_\phi) a})^2 h^2
     = 5a^2h^2
     = \textcolor{blue}{\rho}^4
@@ -535,28 +564,32 @@ $$
   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
+\end{align*}
+$$
+
+All terms in this polynomial are square, so we can solve for *h* in terms of *a* by completing the square.
+Note that *x* and *y* are auxiliary terms in this process.
+
+$$
+\begin{align*}
+  0 &= h^4 -\ 3a^2h^2 + a^4
+    = (h^2 -\ x)^2 - y^2
   \\
-  -2x
-    &= -3a^2
-    \implies x = {3a^2 \over 2},~~
-    x^2 + y = {9a^4 \over 4} + y = a^4
+  &= h^4 -\ 2xh^2 + x^2 - y^2
   \\
-  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
+  -2x &= -3a^2
+    \implies x = {3a^2 \over 2}
+  \\
+    x^2 - y^2 &= {9a^4 \over 4} - y^2 = a^4
+  \\
+  \implies y^2 &= {9a^4 \over 4} - {4a^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\bar{1}_\phi) a^2 \over 2}
+    &= y
+    = {(2\bar{1}_\phi) a^2 \over 2}
   \\
   h^2
     &= {(2\bar{1}_\phi) a^2 \over 2} + {3a^2 \over 2}
@@ -568,7 +601,7 @@ $$
 \end{align*}
 $$
 
-With the square of the height known, all that is left to do is find the volume.
+With the square of the inradius known, all that is left to do is find the volume.
 
 $$
 \begin{align*}
@@ -580,15 +613,15 @@ $$
     &= [11]7_\phi = 740_\phi = 4300_\phi
     = (43_\phi)(100_\phi)
   \\
-  (2^3 \cdot V_\text{dodec[1]})^2
+  (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
+  V_\text{dodec}^2
     &= {2^4 \cdot (43_\phi)^2 \cdot (100_\phi) \over 2^6}
     = {(43_\phi)^2(10_\phi)^2 \over 2^2}
   \\
-  V
+  V_\text{dodec}
     &= {(43_\phi)(10_\phi) \over 2} = {(430_\phi) \over 2} = {(74_\phi) \over 2}
     = {4 + 7\phi \over 2}
 \end{align*}
@@ -613,13 +646,14 @@ The dodecahedron being so much larger than the icosahedron surprised me, to be h
 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.
+  as those of the d12, implying their volumes are roughly equal (which would be handy for a manufacturer).
 
 ***
 
 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.