Volume of a great icosahedron

Question

This is the image of a Great Icosahedron that I obtained starting from the coordinates of the vertices, as $A$ ,$B$ $C$, etc.. Now I want to calculate the volume of the solid. In internet (as in this page of Wolfram) I can find a formula for this volume, $$ V=\frac{l^3}{4}(25+9\sqrt{5}) $$

but I can't understand how it is obtained. Some one can explain how it come from.

Answer 1

I'll find the point $B$ from the following illustration using the regular icosahedron from this answer.

(sorry to every reader for the godawful "graphic design"...)

Namely, it can be taken as the intersection of planes $P_1$, $P_2$, $P_3$, where $$(0,-p,1),(-1,0,-p),(p,1,0) \in P_1,$$ $$(0,-p,1),(1,0,-p),(-p,1,0) \in P_2,$$ $$(0,p,1),(-p,-1,0),(p,-1,0) \in P_3$$ (three equilateral triangles with edge length $2 p = 1 + \sqrt{5}$).

We have a system of linear equations $$\lambda_1 (-1, p, -p-1) + \mu_1 (p, p+1,-1) =$$ $$=\lambda_2 (1,p,-p-1) + \mu_2 (-p,p+1,-1) =$$ $$=(0,2p, 0) + \lambda_3 (-p,-p-1,-1) + \mu_3 (p, -p-1, -1).$$

CAS computation says $\lambda_3 = \mu_3 = \dfrac{9+4\sqrt{5}}{20+9\sqrt{5}}$ and therefore $B = (0,p,1) - 2\frac{8p+5}{18p+11}(0,p+1,1) = \frac{1}{18p+11}(0,-13p-8,2p+1)$.

Now we just find the tetrahedron volume by the |determinant|/6 formula. The red-blue intersection vertex is at $(p-2, -1, 0)$ (from another $3 \times 2$ system of intersecting lines), the green-blue one is, symmetrically, at $(2-p, -1, 0)$. Again by CAS the volume is $\tilde V_2 = 14/(6 \sqrt{5}) - 1$.

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It is not hard to see that the dodecahedron in our problem has edge length $2/p^2$, so we'll rescale and use $V_2 = \tilde V_2 \frac{8}{p^6}$ in the final calculation.

The regular dodecahedron with edge $1$ has volume $V_0 = (15+7\sqrt{5})/4$.

The upper part of the "pentagrammatic pyramid" has height $\sqrt{1 + 2/\sqrt{5}}$, so the volume $V_1$ of its convex hull is that times $\frac{1}{3} \sqrt{25 + 10 \sqrt{5}}/4$, or in other words, $V_1 = \frac{5 + 2 \sqrt{5}}{12}$.

Now we need only calculate $V_0 + 12 (V_1 - 5 V_2)$. This is a pleasant exercise for the reader...

Answer 2

Let us rescale the great icosahedron so that its true vertices have as coordinates cyclic permutations of $(\pm1,0,\pm\varphi)$ where $\varphi$ is the golden ratio $\frac{\sqrt5+1}2$, so that its true edge length is $2\varphi$. Then we have the following decomposition:

The orange, red and blue vertices in the positive (larger) tetrahedron have coordinates $A=(1,0,\varphi)$, $B=(0,2-\varphi,1)$ and $C=(0,\varphi-2,1)$ respectively; the fourth vertex is the origin. The negative (smaller) tetrahedron shares the first three vertices and its fourth vertex $D$ – black in the above images – lies at the intersection of the two lines between red balls and blue balls in the second image: $$(1-t)(0,2-\varphi,1)+t(2-\varphi,-1,0)=(1-u)(0,\varphi-2,1)+u(2-\varphi,1,0)\implies t=u=\frac{\varphi-2}{\varphi-3}$$ $$D=\left(\frac{7-4\varphi}5,0,\frac{2+\varphi}5\right)$$ Then the two tetrahedron volumes may be computed as sixths of determinants; they are $\frac{4-2\varphi}6$ for the positive tetrahedron and $\frac{28\varphi-44}{30}$ for the negative one. Thus the volume of a great icosahedron with true edge length $2\varphi$ is $$60\cdot\frac{4-2\varphi}6-60\cdot\frac{28\varphi-44}{30}=128-76\varphi$$ Finally, using MathWorld's scaling where the tetrahedron red–blue distance is $1$, the volume is $$\frac{128-76\varphi}{(4-2\varphi)^3}=\frac{9\varphi+8}2=\frac{25+9\sqrt5}4$$

Answer 3

First of all, @ChrisLewis is totally right. The area of a great icosahedron can be described as the area of a small stellated dodecahedron with $5\cdot 12=60$ congruent irregular tetrahedra carved out from it. For further reference call the side length of the regular dodecahedron $e$.

The area can thus be divided into three subquestions:

1. The volume of a regular dodecahedron with edge length $e$ which is $$\frac{e^3}{4}(15+7\sqrt{5}).$$
2. The volume of a pentagonal pyramid with height $e\sqrt{\frac{1}{5} (5+2\sqrt{5})}$, this height is due to your Wolfram Link. This volume is due to this page and equals $$\frac{1}{12} e^3\sqrt{25 + 10\sqrt{5}} \sqrt{\frac{1}{5} (5+2\sqrt{5})}$$
3. Now, for the more difficult part we require the volume of 1 of the 60 irregular tetrahedra. A plot of this is at the bottom of this post. We define the regular pentagon $ABCDE$ with coordinates $A=\left(0, \sqrt{\frac{5+\sqrt{5}}{10}}e, 0\right)$, $B=\left(\frac{1+\sqrt{5}}{4}e, \sqrt{\frac{5-\sqrt{5}}{40}}e, 0\right)$, $C=\left(\frac{e}{2}, -\sqrt{\frac{5+2\sqrt{5}}{20}}e, 0\right)$, $D=\left(-\frac{e}{2}, -\sqrt{\frac{5+2\sqrt{5}}{20}}e, 0\right)$, $E=\left(-\frac{1+\sqrt{5}}{4}e, \sqrt{\frac{5-\sqrt{5}}{40}}e, 0\right)$. Also define the point $H$ to be the dimple/bottom of the pentagrammic pyramids and define point $P$ to be the top of the pentagrammic pyramids, we have $H=\left(0, 0, -\sqrt{\frac{5-\sqrt{5}}{10}}e\right)$ and $P=\left(0, 0, \sqrt{\frac{5+2\sqrt{5}}{5}}e\right)$. We draw a line through the points $B$ and $D$ and call the intersection with the y-axis point $F$, we can calculate the coordinates of the point $F$ to be $F = \left(0, -\sqrt{\frac{1}{5}(5-2\sqrt{5})}e, 0\right)$, the line through $P$ and $F$ will intersect the equilateral triangle $CHD$ in another point which we call $G$. The plane spanned by $CHD$ has equation: $$\underbrace{\sqrt{\frac{5-\sqrt{5}}{10}}}_{\alpha}y + \underbrace{\sqrt{\frac{5+2\sqrt{5}}{20}}}_{\beta}z + \alpha \beta e = 0$$ We can parameterize the line through $P$ and $F$ by: $$\bigg( 0, 0, \underbrace{\sqrt{\frac{5+2\sqrt{5}}{5}}}_{\eta} e \bigg) + \lambda \bigg( 0, \underbrace{\sqrt{\frac{5-2\sqrt{5}}{5}}}_{\delta}e, \eta e\bigg).$$ Equating the line and the plane gives the equality $z=-\alpha e - \frac{\alpha}{\beta} \delta e \lambda = \eta (1+\lambda) e$, which implies that $\lambda = -\frac{1}{10}(5+3\sqrt{5})$, substituting this gives $G=\left(0, -\frac{1}{5}\sqrt{\frac{5+\sqrt{5}}{2}}e,-\frac{1}{5}\sqrt{\frac{5-\sqrt{5}}{2}}e \right)$. The irregular tetrahedron is formed by points $GCDP$, the volume of such a solid will be one sixth times $e$ times the distance of $G$ to $CD$ times the distance of $P$ to $GCD$. We get a volume of \begin{align} \frac{1}{6} e^2 &\sqrt{\left(-\frac{1}{5}\sqrt{\frac{5+\sqrt{5}}{2}}+\sqrt{\frac{5+2\sqrt{5}}{20}} \right)^2 + \left(\frac{1}{5}\sqrt{\frac{5+\sqrt{5}}{2}}\right)^2} \cdot \textrm{dist}(P, HCD) \\ &=\frac{\sqrt{15}}{60}e^2 \cdot \textrm{dist}(P, HCD) \\ &=\frac{\sqrt{15}}{60}e^2 \frac{\beta \sqrt{\frac{5+2\sqrt{5}}{5}}e + \alpha \beta e}{\sqrt{\alpha^2 + \beta^2}}\\ &=\frac{\sqrt{15}}{60}e^3 \frac{\frac{1}{10}(5+2\sqrt{5}) + \frac{1}{20}(5+\sqrt{5})}{\sqrt{\frac{5-\sqrt{5}}{10} + \frac{5+2\sqrt{5}}{20}}} \\ &=\frac{\sqrt{15}}{60}e^3 \frac{\frac{1}{4}(3+\sqrt{5})}{\sqrt{3}/2} =\frac{\sqrt{5}}{120}e^3(3+\sqrt{5}) = \frac{5+3\sqrt{5}}{120} e^3 \end{align}

The total volume of a great icosahedron is thus \begin{align} \frac{e^3}{4}(15+7\sqrt{5}) + &12 \cdot \frac{1}{12} e^3\sqrt{25 + 10\sqrt{5}} \sqrt{\frac{1}{5} (5+2\sqrt{5})} - 60 \cdot \frac{5+3\sqrt{5}}{120} e^3\\ =\frac{e^3}{4}(25 + 9 \sqrt{5}). \end{align}