Geometry Proof to Find Maximum area of $\triangle PIE$ 
Circle $\omega$ is inscribed in unit square $PLUM,$ and points $I$ and $E$ lie on $\omega$ such that
$U, I,$ and $E$ are collinear. Find, with proof, the greatest possible area for $\triangle PIE.$

I'm not sure if there is a solution possible without trigonometry.
Also, for my diagram in my solution, I'm not sure how to center it. Sorry about that.

Let $O$ be the center of circle $\omega.$ Let $X$ be the foot of the altitude from $P$ to $IE,$ and let $Y$ be the foot from $O$ to $IE.$ Denote line segment $\overline{YO}$ as length $d,$ the radius as $r,$ and $\angle XUP$ as $\theta.$

$\textbf{Claim:}$ The greatest area of $\triangle PIE$ is $\frac{1}{4}.$
$\textbf{Proof:}$ To find the area of $\triangle PIE,$ we can find the lengths of $\overline{IE}$ and $\overline{PX},$ and then use the formula of the area of a triangle to conclude. Let's start by finding lengths $\overline{IO}$ and $\overline{YO},$ and then apply the Pythagorean Theorem to get our $\overline{IY},$ then multiply by two to get the base of $\triangle PIE.$ Clearly, segment $\overline{IO}$ is the radius of the circle, which has length $1/2.$ Then, by taking $\sin \theta,$ we have $$\sin \theta = \frac{\overline{YO}}{\overline{UO}}=\frac{d}{\sqrt{2}/2} \implies d = \frac{\sqrt{2}}{2} \cdot \sin \theta.$$
Similarly with $\overline{PX},$ we take $\sin \theta$ and get $$\sin \theta = \frac{\overline{PX}}{\overline{UP}} = \frac{\overline{PX}}{\sqrt{2}} \implies \overline{PX} = \sqrt{2} \cdot \sin \theta.$$
Thus, after finding these two lengths, we know the largest possible area of $\triangle PIE$ is $$\displaystyle{\max\left(\frac{1}{2} \cdot \overline{PX} \cdot \overline{IE}\right) = \max \left(\frac{1}{2} \cdot \sqrt{2} \cdot \sin \theta \cdot 2\sqrt{r^2-d^2}\right)}.$$
Note that $\overline{IE} = 2\sqrt{r^2-d^2}$ by the Pythagorean Theorem, where clearly in the diagram $\overline{IO}$ is the radius and $\overline{YO}$ is distance $d.$ Simplifying our equation above to lowest terms, we get:
\begin{align*}
\max \left(\frac{1}{2} \cdot \sqrt{2} \cdot \sin \theta \cdot 2\sqrt{r^2-d^2}\right) &= \max \left(\frac{1}{2} \cdot \sqrt{2} \cdot \sin \theta \cdot 2\sqrt{\left(\frac{1}{2} \cdot \frac{1}{2}\right)-\left(\frac{1}{2} \cdot \sin^2 \theta\right)}\right) \\
&= \max \left(\frac{1}{2} \cdot \sqrt{2} \cdot \sin \theta \cdot \sqrt{1 - 2 \sin^2 \theta} \right). \\
\end{align*}
Then, let's subsititute $\alpha = \sin \theta.$ Thus, to maximize the area of $\triangle PIE,$ all we need to do is find the maximum of $\max \left(\frac{1}{2} \cdot \sqrt{2} \cdot \alpha \cdot \sqrt{1-2\alpha^2}\right).$
This implies we need to find the value of $\alpha$ that suffices to maximize the following equation: $$\max \left(\alpha^2 \left(1-2\alpha^2\right)\right) = \max \left(\alpha^2 - 2\alpha^4 \right).$$
Taking the derivative of $\alpha^2 - 2\alpha^4,$ we get $$\frac{\mathrm{d}}{\mathrm{d}\alpha} \left(\alpha^2 - 2\alpha^4\right) = 2\alpha - 8\alpha^3.$$
The equation $2\alpha - 8\alpha^3 = 0$ is maximized when $\alpha = \frac{1}{2},$ which we hence get the largest area of $\triangle PIE$ as $\frac{1}{2} \cdot \frac{1}{2} = \boxed{\frac{1}{4}},$ as desired. $\qquad\blacksquare$
$\textbf{Claim:}$ There is such $\theta$ that achieves the maximum area stated above.
$\textbf{Proof:}$ We have found that the maximum of $\sin \theta = \frac{1}{2},$ thus meaning that the $\theta = 30^{\circ},$ in which that is where the maximum area of $\triangle PIE$ occurs. Hence, proven. $\qquad\blacksquare$
