<s>[INST] 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.
 [/INST] 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.
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