This is from the exercise problems(not the live contest) for regional math competition.
Let triangle $ABC$ satisfy $\angle C > 90^\circ$ and let its incenter be $I$. Let $\ell$ be the line parallel to $BC$ passing through $I$, and call $\ell \cap AB = D$, $\ell \cap BC = E$. We know that $AI = 3$ and the radius of inscribed circle of the triangle $ABC$ is $1$. Finally, $(\text{area of }ABC) = 5\sqrt{2}$ is given. The problem is to find $DE$.
Below is my attempt to do something but I am stuck. How can I proceed to get $DE$?
Thank you in advance for any form of help, hint, or solution.
You've made a good start. Using your diagram, we have that
$$2\alpha + 2\beta + 2\gamma = 180^{\circ} \;\to\; \alpha + \beta + \gamma = 90^{\circ} \tag{1}\label{eq1A}$$
Thus, using your $x + y = 3\sqrt{2}$ result, plus the Pythagorean theorem to get $\lvert BI\rvert = \sqrt{1 + y^2}$ and $\lvert CI\rvert = \sqrt{1 + x^2}$, results in
$$\begin{equation}\begin{aligned} \sin(\alpha + \beta) & = \sin(90^{\circ} - \gamma) \\ \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta) & = \cos(\gamma) \\ \frac{1}{\sqrt{1 + y^2}}\left(\frac{x}{\sqrt{1 + x^2}}\right) + \frac{y}{\sqrt{1 + y^2}}\left(\frac{1}{\sqrt{1 + x^2}}\right) & = \frac{2\sqrt{2}}{3} \\ x + y & = \frac{2\sqrt{2}}{3}(\sqrt{1 + x^2})(\sqrt{1 + y^2}) \\ 3\sqrt{2} & = \frac{2\sqrt{2}}{3}(\sqrt{1 + x^2})(\sqrt{1 + y^2}) \\ \frac{9}{2} & = (\sqrt{1 + x^2})(\sqrt{1 + y^2}) \\ \frac{81}{4} & = (1 + x^2)(1 + y^2) \\ \frac{81}{4} & = 1 + x^2 + y^2 + x^2y^2 \\ x^2 + y^2 & = \frac{77}{4} - x^2y^2 \end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Squaring both sides of your $x + y = 3\sqrt{2}$ result gives that
$$\begin{equation}\begin{aligned} x^2 + 2xy + y^2 & = 18 \\ \left(\frac{77}{4} - x^2y^2\right) + 2xy & = 18 \\ -x^2y^2 + 2xy + \frac{77}{4} - \frac{72}{4} & = 0 \\ (xy)^2 - 2xy - \frac{5}{4} & = 0 \end{aligned}\end{equation}\tag{3}\label{eq3A}$$
This is a quadratic equation in $xy$ so, using the quadratic formula and that $xy \gt 0$, we get
$$\begin{equation}\begin{aligned} xy & = \frac{2 \pm \sqrt{4 - 4\left(-\frac{5}{4}\right)}}{2} \\ & = \frac{2 \pm \sqrt{4 + 5}}{2} \\ & = \frac{2 \pm 3}{2} \\ & = \frac{5}{2} \end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Since $DI \parallel BC$, letting $F$ be the point where the vertical line from $I$ touches $AB$, we have $\measuredangle IDF = \measuredangle CBF = 2\alpha$. Thus, from $\triangle IDF$, we have
$$\begin{equation}\begin{aligned} \lvert DI\rvert & = \frac{1}{\sin(2\alpha)} \\ & = \frac{1}{2\sin(\alpha)\cos(\alpha)} \\ & = \frac{1}{2\left(\frac{1}{\sqrt{1+y^2}}\right)\left(\frac{y}{\sqrt{1+y^2}}\right)} \\ & = \frac{1 + y^2}{2y} \\ & = \frac{1}{2}\left(\frac{1}{y} + y\right) \end{aligned}\end{equation}\tag{5}\label{eq5A}$$
Similarly, we get $\lvert IE\rvert = \frac{1}{2}\left(\frac{1}{x} + x\right)$. Adding these $2$ results, and using \eqref{eq4A}, gives
$$\begin{equation}\begin{aligned} \lvert DE\rvert & = \lvert DI\rvert + \lvert IE\rvert \\ & = \frac{1}{2}\left(\frac{1}{x} + \frac{1}{y} + x + y\right) \\ & = \frac{1}{2}\left(\frac{y}{xy} + \frac{x}{xy} + x + y\right) \\ & = \frac{1}{2}\left(\frac{1}{xy}(x + y) + (x + y)\right) \\ & = \frac{x + y}{2}\left(\frac{1}{xy} + 1\right) \\ & = \frac{3\sqrt{2}}{2}\left(\frac{2}{5} + 1\right) \\ & = \frac{3}{\sqrt{2}}\left(\frac{7}{5}\right) \\ & = \frac{21}{5\sqrt{2}} \end{aligned}\end{equation}\tag{6}\label{eq6A}$$