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.
 A: 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. $\qquad\blacksquare$
A: 
Corners of unit square $U(0,0),P(1,1)\;$
Intersection between unit circle and straight line
$$ (x-\frac12)^2+(y-\frac12)^2= \frac14\;;y= \tan v\cdot x; \;$$
gives coordinates $(xI,yI), (xE, yE) $ for points $(I,E); \;$
Area (doubled)  is given by determinant
$$\big[(1,1,1),(xI,xE,1),(yI,yE,1)\big]$$
$$
t= \tan v \;;
\left(\text{xE}(\text{t$\_$})=\frac{t+\sqrt{2} \sqrt{t}+1}{2 \left(t^2+1\right)};\text{yE}(\text{t$\_$})=\frac{1}{2} \left(\frac{t^2}{t^2+1}+\frac{t}{t^2+1}+\frac{\sqrt{2} t^{3/2}}{t^2+1}\right);\right) \left(\text{xI}(\text{t$\_$})=\frac{t-\sqrt{2} \sqrt{t}+1}{2 \left(t^2+1\right)};\text{yI}(\text{t$\_$})=\frac{1}{2} \left(\frac{t^2}{t^2+1}+\frac{t}{t^2+1}-\frac{\sqrt{2} t^{3/2}}{t^2+1}\right);\right)
$$
Expressions of Area ( heavy, not pasted here ) and derivatives were handled by Mathematica . Graph of the derivative crosses x-axis at
$$\tan v \approx  0.267949 = \tan 15^{\circ}$$
Transversal that maximizes area of triangle $PIE$ is inclined at 15 degrees to x-axis. Area by substitution of critical  value is
$$ \Delta PIE=\dfrac14$$
which also agrees with result by Bongocat.

A: 
Note $EI = 2\sqrt{OI^2-OT^2} = 2\sqrt{\frac14-OT^2}$ and $\sin\alpha = \frac{OT}{OU} = \sqrt2 OT$. Then,
\begin{align}
[PIE] & = [PUE]-[PUI] \\
& = \frac12 PU (EU - IU)\sin\alpha = \frac{\sqrt2}2EI\sin\alpha \\
&=2OT \sqrt{\frac14-OT^2}
\le \frac14
\end{align}
where the inequality $2\sqrt{xy}\le x+y$ is applied in the last step. Thus, the maximum area is $[PIE]_{max} = \frac14$.
A: I would provide a solution without trigonometry, as asked in the OP. Consider the circle as centered in point $(1/2,1/2)$ of a Cartesian plane, so that the  square corners are $L(0,0)$, $U(0,1)$, $M(1,1)$, $P(1,0)$. The circle equation is $(x-1/2)^2+(y-1/2)^2=1/4$. Let us call $k$ the slope of a line passing through $U$ and intersecting the circle. Its equation is then $y=kx+1$. By construction, the line intersects the circle only if $k\leq 0$.

The intersection points $I$ and $E$ between the line and the circle are given by the solutions of the system composed by the equations of the line and the circle. These solutions are
$$I\left(\frac{1-k- \sqrt{-2k}}{2(k^2 + 1)}, \frac{k^2+k+2-k\sqrt{-2k} }{2(k^2 + 1)}\right) $$
$$E\left(\frac{1-k+\sqrt{-2k}}{2(k^2 + 1)}, \frac{k^2+k+2+k\sqrt{-2k} }{2(k^2 + 1)}\right) $$
By the standard formula for the distance between points, after some calculations and simplifications we obtain
$$IE=\frac{\sqrt{-2k}}{\sqrt{k^2 + 1}}$$
Now let us consider the perpendicular to $IE$ drawn from $P(1,0)$. This line must have angular coefficient $-1/k$ and has to satisfy $0=-1/k+r$, which implies $r=1/k$. The line has then equation $y=-x/k+1/k$. The  coordinates of $X$ are the solutions of the system composed by the two lines. Solving the system, it follows
$$X\left( \frac{1-k}{k^2+1}, \frac{1+k}{k^2+1}\right)$$
and using again the  formula for the distance between two points we get
$$PX= \frac{|k+1|}{\sqrt{k^2+1}}$$
Therefore, the area of $\triangle{PIE}$ is
$$ A(\triangle{PIE})\\=    \frac{IE \cdot PX}{2}= \frac{1}{2} \cdot \frac{\sqrt{-2k}}{\sqrt{k^2 + 1}}  \cdot \frac{|k+1|}{\sqrt{k^2+1}}\\
= \frac{|k+1| \sqrt{-2k}}{2(k^2 + 1)} $$
Taking the derivative we have
$$\frac{(k^3 + 3 k^2 - 3 k - 1)}{(2 \sqrt{-2k} (k^2 + 1)^2)} \,\text{for} \,\,k>-1$$
$$-\frac{(k^3 + 3 k^2 - 3 k - 1)}{(2 \sqrt{-2k} (k^2 + 1)^2)} \,\text{for} \,\,k<-1$$
As expected by the symmetry of the problem with respect to the case $k=-1$, setting these equations equal to zero we obtain that the area function has two maxima in $-2+\sqrt{3}$ and $-2-\sqrt{3}$. Substituting these values in the formula of $\triangle{PIE}$ area we conclude that the maximal area is
$$A_{max}(\triangle{PIE})=  \frac{(\sqrt{3} -1)  (\sqrt{ 4-2\sqrt{3}} )}{4 (4-2\sqrt{3} )}\\=\frac{1}{4}\,\,\, \left( \text{for}\,\, k=
-2+\sqrt{3}\right) $$
and
$$A_{max}(\triangle{PIE})=  \frac{(\sqrt{3} +1)  (\sqrt{ 4+2\sqrt{3}} )}{4 (4+2\sqrt{3} )}\\=\frac{1}{4}\,\,\, \left( \text{for}\,\, k=
-2-\sqrt{3}\right) $$
Here is a graph of the area as a function of the slope $k$, as provided by the formula above:

Note that for $k=0$, i.e. the points $I$ and $E$  coincide in $(1/2,1)$ with the midpoint of the upper side of the square, as expected the area formula gives zero. As $k$ decreases, the area increases, achieves its first maximal value of $1/4$ in $k=-2+\sqrt{3}$, and again decreases to zero in $k=-1$ (this is the case where the points $I$, $E$, and $P$ are aligned on the diagonal $UP$). As $k$ further decreases, the area increases again, achieves its second maximal value of $1/4$ in $k=-2-\sqrt{3}$, and progressively decreases tending to zero for $k\rightarrow -\infty$ (this is the case in which the points $I$ and $E$  coincide in $(0,1/2)$ with the midpoint of the left side of the square).
Lastly, note that, if we call $\alpha$ the angle $\angle{MUI}$, the slopes of $-2+\sqrt{3}$ and $-2+\sqrt{3}$ correspond to the the values $\alpha=\pi/12=15°$ and $\alpha=5\pi/12=75°$ easily obtained by the trigonometric approach.
