This week’s Riddler Classic is a short problem 3D geometry. Here we go! (I paraphrased the question)
A polyhedron has six edges. Five of the edges have length $1$. What is the largest possible volume?
Here is my solution
[Show Solution]
The only polyhedron with six edges is a tetrahedron, which is a pyramid with a triangular base. Two of the faces will be equilateral triangles that share a common edge. This accounts for the five edges of length 1. The length of the sixth edge is determined by the angle between the faces, which we will call $\theta$. Here is an animation showing the different tetrahedra you get as you vary $\theta$:
In this diagram, $AB=BC=AC=AD=BD=1$ and $OD=OC=\frac{\sqrt{3}}{2}$.
The volume is equal to
\begin{align}
V&=\frac{1}{3}(\text{Area of base})\cdot(\text{altitude}) \\
&= \frac{1}{3}(\text{Area ABC})\cdot(DG) \\
&= \frac{1}{3}\left( \frac{1}{2} (AB)(OC) \right) \cdot \left( (OD) \sin\theta \right) \\
&= \frac{1}{3} \cdot \frac{1}{2}\cdot 1 \cdot \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} \cdot \sin\theta \\
&= \frac{1}{8}\cdot\sin\theta
\end{align}Therefore, the maximum volume is $\frac{1}{8}$ and it occurs when $\theta=90^\circ$. This is intuitive because the area of the base is fixed, so the largest volume occurs when the altitude is as large as possible.