# Area/Volume of a Triangle, Tetrahedron, 4-Simplex, 5-Simplex, ....

Posted 3 years ago
5747 Views
|
2 Replies
|
5 Total Likes
|
 The area of a unit-edge triangle is $\sqrt3/4$. The volume of a unit-edge tetrahedron is $1/(6\sqrt2)$. What happens beyond that? At the entry simplex we can find the Cayley-Menger Determinant. Since all those $d$ values will be 1, we end up taking the determinant of the inverse of the identity matrix. So we end up with reciprocals of the following for dimensions 2 to 6. 1/Table[Sqrt[(-1)^(n + 1)/(2^n (n!)^2) MatrixForm[(1 - IdentityMatrix[n + 2])]], {n, 2, 6}] The bottom (actually top) part winds up being $\sqrt{3},\sqrt{4},\sqrt{5},\sqrt{6},\sqrt{7}$ for dimensions 2 to 6, or dimension+1. The top (actually bottom) part is $\sqrt{2^n} n!$. The volume of the $n$-simplex turns out to be ${\large \frac{\sqrt{2^n} n!}{\sqrt{n+1}}}$. Table[Sqrt[n + 1]/(Sqrt^n n!), {n, 2, 12}] It seems that the reciprocals of dimensions 7 and 8 seem to be integers. The code below finds the sequence A155946: 7, 8, 17, 24, 31, 48, 49, 71, 80, 97, ... . First /@ Select[Table[{n, (Sqrt^n n!)/Sqrt[n + 1]}, {n, 2, 1000}], IntegerQ[Last[#]] &] For another check, (16) at Triangle Area and (1) at Tetrahedron suggests a sequence of matrices that appends 1 to each coordinate and takes the determinant. If all but one of the coordinates is some permutation of $\left(0,0,0, ..., \frac{1}{\sqrt{2}} \right)$, those $n-1$ edges will all be distance 1 from each other. For the last coordinate point, we can find a point $\left( a,a,a, ..., a \right)$ at distance 1 from the other points. FullSimplify[(a /. Solve[Sqrt[(n - 1) a^2 + (a - 1/Sqrt)^2] == 1, a]) ] That gives us $a$ values of $\frac{1-\sqrt{n+1}}{n\sqrt{2}}$ or $\frac{1+ \sqrt{n+1}}{n\sqrt{2}}$. For dimension 4, the former value is $\frac{1-\sqrt{5}}{4 \sqrt{2}} = \frac{\sqrt{3-\sqrt{5}}}{-4} = \frac{1-\phi }{\sqrt8}$, where $\phi$ is the golden ratio. That leads to the following matrix which happens to give the predicted volume. $${\small V = \frac{\sqrt5 }{96} = \frac{1}{4!} \left| \begin{matrix} 0 & 0 & 0 & \frac{1}{\sqrt{2}} & 1 \\\\ 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 1 \\\\ 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 1 \\\\ \frac{1}{\sqrt{2}} & 0 & 0 & 0 & 1 \\\\ \frac{1-\phi }{\sqrt8} & \frac{1-\phi }{\sqrt8} & \frac{1-\phi }{\sqrt8} & \frac{1-\phi }{\sqrt8} & 1 \\\\ \end{matrix} \right| }$$ The square, cube, hypercube and so on are easier to calculate. Answer
2 Replies
Sort By:
Posted 3 years ago - Congratulations! This post is now a Staff Pick as distinguished on your profile! Thank you, keep it coming! Answer
Posted 1 year ago
 Thank you for this post! I would like to compute the area of the probability simplex. The difference between the probability simplex and the simplex studied in this post is that the former has different side length. Is it possible to change these formula for such simplicies? Any help is highly appreciated! Answer