Editing Antiprism (section)

==== Derivation ====
The circumradius of the horizontal circumcircle of the regular <math>n</math>-gon at the base is
:<math>
R(0) = \frac{l}{2\sin\frac{\pi}{n}}.
</math>
The vertices at the base are at
:<math>\left(\begin{array}{c}R(0)\cos\frac{2\pi m}{n} \\ R(0)\sin\frac{2\pi m}{n} \\ 0\end{array}\right),\quad  m=0..n-1;</math>
the vertices at the top are at
:<math>\left(\begin{array}{c}R(0)\cos\frac{2\pi (m+1/2)}{n}\\R(0)\sin\frac{2\pi (m+1/2)}{n}\\h\end{array}\right), \quad m=0..n-1.</math>
Via linear interpolation, points on the outer triangular edges of the antiprism that connect vertices at the bottom with vertices at the top
are at
:<math>\left(\begin{array}{c}
\frac{R(0)}{h}[(h-z)\cos\frac{2\pi m}{n}+z\cos\frac{\pi(2m+1)}{n}]\\
\frac{R(0)}{h}[(h-z)\sin\frac{2\pi m}{n}+z\sin\frac{\pi(2m+1)}{n}]\\
\\z\end{array}\right), \quad 0\le z\le h, m=0..n-1</math>
and at
:<math>\left(\begin{array}{c}
\frac{R(0)}{h}[(h-z)\cos\frac{2\pi (m+1)}{n}+z\cos\frac{\pi(2m+1)}{n}]\\
\frac{R(0)}{h}[(h-z)\sin\frac{2\pi (m+1)}{n}+z\sin\frac{\pi(2m+1)}{n}]\\
\\z\end{array}\right), \quad 0\le z\le h, m=0..n-1.</math>
By building the sums of the squares of the <math>x</math> and <math>y</math> coordinates in one of the previous two vectors,
the squared circumradius of this section at altitude <math>z</math> is
:<math>
R(z)^2 = \frac{R(0)^2}{h^2}[h^2-2hz+2z^2+2z(h-z)\cos\frac{\pi}{n}].
</math> 
The horizontal section at altitude <math>0\le z\le h</math> above the base is a <math>2n</math>-gon (truncated <math>n</math>-gon)
with <math>n</math> sides of length <math>l_1(z)=l(1-z/h)</math> alternating with <math>n</math> sides of length <math>l_2(z)=lz/h</math>.
(These are derived from the length of the difference of the previous two vectors.)
It can be dissected into <math>n</math> isoceless triangles of edges <math>R(z),R(z)</math> and <math>l_1</math> (semiperimeter <math>R(z)+l_1(z)/2</math>)
plus <math>n</math>
isoceless triangles of edges <math>R(z),R(z)</math> and <math>l_2(z)</math> (semiperimeter <math>R(z)+l_2(z)/2</math>).
According to Heron's formula the areas of these triangles are
:<math>
Q_1(z) = \frac{R(0)^2}{h^2} (h-z)\left[(h-z)\cos\frac{\pi}{n}+z\right] \sin\frac{\pi}{n}
</math>
and
:<math>
Q_2(z) = 
\frac{R(0)^2}{h^2} z\left[z\cos\frac{\pi}{n}+h-z\right] \sin\frac{\pi}{n}
. 
</math>
The area of the section is <math>n[Q_1(z)+Q_2(z)]</math>, and the volume is
:<math>
V = n\int_0^h [Q_1(z)+Q_2(z)] dz 
= \frac{nh}{3}R(0)^2\sin\frac{\pi}{n}(1+2\cos\frac{\pi}{n})
= \frac{nh}{12}l^2\frac{1+2\cos\frac{\pi}{n}}{\sin\frac{\pi}{n}}
. 
</math>

The volume of a right {{mvar|n}}-gonal [[Prism (geometry)|prism]] with the same {{mvar|l}} and {{mvar|h}} is:
<math display=block>V_{\mathrm{prism}}=\frac{nhl^2}{4} \cot\frac{\pi}{n}</math>
which is smaller than that of an antiprism.