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 = 2TD, so the desired ratio can be computed simply as $$\frac{AT}{DE}=2\frac{DT}{DE}=2\sqrt{2},$$ as desired.