Posssible pentagons in 3D A non-planar pentagon in $\mathbb{R}^3$ has equal sides and four right angles. What are the possible values for the fifth angle?
My attempt
It was quite easy to find an example for $60^\circ$: $A=(1, 0, 1), B=(1, 1, 1), C=(0, 1, 1), D=(0, 0, 1)$ and then $\triangle DEA$ is equilateral. I couldn't find any other.
We can assume WLOG that the first three vertices are the same as A,B,C in the above example, by rotating the pentagon (which is an isometry and therefore preserves angles), but there are two vertices left to determine, which are $6^\circ$ of freedom - so analytic solution won't work.
 A: Let the pentagon be formed by placing five unit vectors $v_1, \ldots, v_5 \in \mathbb{R}^3$ head to tail, with right angles between two successive vectors. Then $$v_1+v_2+v_3+v_4+v_5 = 0.$$
Let $v_{km} = \langle v_k, v_m \rangle$ denote the inner product between two vectors. In particular $$v_{11}=v_{22}=v_{33}=v_{44}=v_{55}=1 \\
v_{12} = v_{23} = v_{34} = v_{45} = 0.
$$
From $0 = \langle v_k, 0 \rangle = \langle v_k, v_1+\ldots+v_5 \rangle$ we get the linear system
$$
\begin{pmatrix}
1&1&1&1&0&0&0\\
1&0&0&0&1&1&0\\
1&1&0&0&0&0&1\\
1&0&1&0&1&0&0\\
1&0&0&1&0&1&1
\end{pmatrix} \begin{pmatrix}
1\\
v_{13}\\
v_{14}\\
v_{15}\\
v_{24}\\
v_{25}\\
v_{35}
\end{pmatrix} = 0.
$$
The kernel of this matrix is spanned by the columns of $$\begin{pmatrix}-2&0\\1&0\\0&1\\1&-1\\2&-1\\0&1\\1&0\end{pmatrix}.$$
Therefore the Gram matrix $(v_{km})$ must be of the form
$$\begin{pmatrix}
1&0&-\tfrac{1}{2}&x&-\tfrac{1}{2}-x\\
0&1&0&-1-x&x\\
-\tfrac{1}{2}&0&1&0&-\tfrac{1}{2}\\
x&-1-x&0&1&0\\
-\tfrac{1}{2}-x&x&-\tfrac{1}{2}&0&1
\end{pmatrix}
$$
for some $x\in\mathbb{R}$.  This matrix must be positive semi-definite and of rank three.  This is the case if and only if $x=0$ or $x=-\tfrac{6}{7}$.  The fifth angle $\alpha$ satisfies $\cos(\alpha) = -v_{15}$.  So $\cos(\alpha)=\tfrac{1}{2}$ or $\cos({\alpha}) = -\tfrac{5}{14}$.
A: For maximal symmetry of the configuration, wlog. the "middle" line $BC$ is given by $B=(-\frac12,0,0)$, $C=(\frac12,0,0)$ and $A$, $D$ lie symmetrically to the $y=0$ plane, i.e. $A=(-\frac12,-y,z)$, $D=(\frac12,y,z)$ with $y^2+z^2=1$.
We may also assume wlog. that $y,z\ge 0$.
Then $E$ must lie on the plane orthogonal to $AB$ throgh $A$ and the plane orthogonal to $CD$ through $D$. By symmetry, the unique line intersection of these planes (exception: if $y=0$ then these planes coincide, see below) is in the $y=0$ plane, so $E=(u,0,w)$. 
Since $AE\perp AB$ we find 
$$-y^2+(w-z)z =0$$
so that $wz=y^2+z^2=1$.
The condition $|AE|=|ED|=1$ gives us $$1=\left(u+\frac12\right)^2+y^2+(w-z)^2=\left(u-\frac12\right)^2+y^2+(w-z)^2 $$
hence $u=0$ and we can continue with
$$ 1=\frac14+y^2+(w-z)^2=\frac14+\underbrace{y^2+z^2}_{=1}+w^2-2\underbrace{wz}_{=1}$$
so that $w^2=\frac74$ and hence $z^2=\frac47$ and $y^2=\frac37$.
The angle $\alpha$ at $E$ can be obtained with the scalar product:
$$\begin{align}\cos \alpha &= (A-E)\cdot(D-E)\\&=\left(-\frac12,-y,z-w\right)\cdot\left(\frac12,y,z-w\right)\\&=-\frac14-y^2+z^2-2wz+w^2\\
&=\frac14-\frac37+\frac47-2+\frac74\\
&=\frac17.\end{align}$$
Thus the only possibility with $y\ne0$ is $\alpha=\arccos\frac17$.
In the exceptional case $y=0$, we can argue in the plane through $A,D$ orthogonal to both $AB$ and $CD$ and see that $DEA$ is an equilateral triangle, i.e. $\alpha=\frac\pi3$.
