Take an isosceles triangle with legs length $1$ and base angle $\theta$. Take the apex and move it to a leg vertex. Describe the ratio of the new folded shape's area to its previous as a function of $\theta$, and determine the domain of $\theta$ for which this is operation is possible.
In my efforts, I've solved the ratio to be $1+\frac1 4\sec2\theta$, but I can't find and prove any strict bound. I know $\pi/3\le\theta$ since it will fold over exactly at that point, and obviously $\theta<\pi/2$, but I think there may be a stricter bound.
$$\begin{gathered}
A = \frac{1}{2}{1^2}\sin \left( {180^\circ - 2\vartheta } \right) = \frac{1}{2}\sin 2\vartheta \hfill \\
A' = \frac{1}{2} \cdot \frac{1}{2}1h = \frac{1}{8}\tan 2\vartheta \hfill \\
h = \frac{1}{2}\left|\tan \left( {180^\circ - 2\vartheta } \right)\right| > 0 \to 90^\circ > \vartheta > 45^\circ \hfill \\
\frac{{A'}}{A} = \frac{{\frac{1}{8}\left|\tan 2\vartheta \right|}}{{\frac{1}{2}\sin 2\vartheta }} = \frac{1}{4}\left|\sec 2\vartheta\right| \hfill \\
\end{gathered} $$
The base angle must be larger than $45°$ otherwise you have an obtuse angle in the vertex and your folding goes out of the original triangle