In this Stand-up Maths' video, the great Matt Parker compares the volumes of a prism and an antiprism. Considering they have a square base, what is the antiprism/prism ratio of their volumes?

Following Matt Parker's example, the prism can be one of the Twin Towers from the World Trade Center, and the antiprism can be the approximate shape of the One World Trade Center. We can draw the former in red and the latter in blue+green, with Asymptote. We consider a square base of side \(L\) and height \(h\), and in the figure we use \(h/L=4\). Click on the image to access an interactive WebGL image:

In that video, the exact answer is not provided, so I tried to calculate it. The volume of the prism is simply \(V=L^2h\), but that of the antiprism is not so easy. Here is a GeoGebra applet of the horizontal section of the antiprism as a function of \(z/h\), where \(z\) is the elevation. Click on the image to access the applet. Then use the slider in the middle to change its value from 0 to 1.

In the applet we can see that the section is an irregular octagon, shown in red. We want to calculate its area as a function of \(z/L\). This polygon can be divided into 2 different triangles, each one appearing 4 times. All these triangles have the sides of the red polygon as bases. The non-tilted ones, i.e. the horizontal and vertical sides, have length \(l_1=L(1-\frac{z}{h})\). On the other hand, the tilted bases have length \(l_2=L\frac{z}{h}\).

The triangles with base \(l_1\) have height \(d_1=\frac{L}{2}(1+(\sqrt{2}-1)\frac{z}{h})\). The other triangles have height \(h_2=\frac{L}{2}(\frac{z}{h}+\sqrt{2}(1-\frac{z}{h}))\). Thus, the area of the polygon is

\[A(z)=\left(1+2(\sqrt{2}-1)\left(\frac{z}{h}-(\frac{z}{h})^2\right)\right)L^2.\]To simplify the expression, we define \(\gamma=2(\sqrt{2}-1)\), so that the area can be written as

\[A(z)=\left(1+2\gamma\left(\frac{z}{h}-(\frac{z}{h})^2\right)\right)L^2.\]The volume of the object can be calculated as the sum (integral) of all these sections, each with a thickness \(dz\). We have

\[V'=\int_{z=0}^{z=h}A(z)dz=\left(\frac{2+\sqrt{2}}{3}\right)hL^2.\]Thus, the ratio of the antiprism/prism volumes is

\[\frac{V'}{V}=\frac{2+\sqrt{2}}{3}\approx 1.138.\]In conclusion, the antiprism has a slight bigger volume than its prism counterpart.