First Round of the 73rd Czech and Slovak Mathematical Olympiad (take-home part)

We shall prove that the ratio has to be equal to $2\sqrt{2}$.

First, we shall observe that the triangles $BTC$ and $KTL$ (coloured turquoise and yellow in the diagram) are similar, since
$$\angle BTC = \angle BTK + \angle KTC = 45^\circ + \angle KTC = \angle KTC + \angle CTL = \angle KTL,$$
and by similarity of the triangles $BKT$ and $CLT$, we have $\frac{BT}{CT} = \frac{KT}{LT}$. The ratio of similarity has to be $\sqrt{2}$, since $\frac{BT}{KT} = \sqrt{2}$.
Since $D$ is a midpoint of $BC$ and $E$ is the midpoint of $KL$, the triangles $BTD$ and $KTE$ are also similar. This tells us that $\angle BTD = \angle KTE$ and $\frac{BT}{TD} = \frac{KT}{TE}$. Subtracting $\angle KTD$ from the equality gives us $\angle BTK = \angle DTE$ and the equality of ratios can be rearranged to $\frac{BT}{KT} = \frac{TD}{TE}$, so the

triangles BTK and DTE are similar. Therefore, the triangle DTE is also a right-angled isosceles triangle.
Finally, recall that since T is the centroid, we have AT = 2*TD*, so the desired ratio can be computed simply as
$$\frac{AT}{DE} = 2\frac{DT}{DE} = 2\sqrt{2},$$
as desired.