3. Given a triangle $ABC$ with centroid $T$. On the segments $BT$ and $CT$, right isosceles triangles $BTK$ and $CTL$ are constructed as shown in the figure. Let $D$ be the midpoint of side $BC$ and $E$ the midpoint of segment $KL$. Determine all possible values of the ratio $|AT|:|DE|$.

(Michal Rolínek)

![](https://cdn.mathpix.com/cropped/2024_04_17_968d90939233ae120dcfg-05.jpg?height=294&width=457&top_left_y=230&top_left_x=1368)
SOLUTION. We prove that the ratio $|A T|:|D E|$ has only one possible value $2 \sqrt{2}$.

First, we derive that for the lengths of segments $T D$ and $T E$, it holds that $|T D| /|D E|=\sqrt{2}$.* We focus on triangles $B T C$ and $K T L$, which are shaded in the figure. Their angles at the common vertex $T$ are equal, because

$$
|\angle B T C|=|\angle B T K|+|\angle K T C|=45^{\circ}+|\angle K T C|=|\angle K T C|+|\angle C T L|=|\angle K T L| .
$$

Moreover, $|B T| /|T C|=|K T| /|T L|$ due to the similarity of triangles $B K T$ and $C L T$. Therefore, according to the SAS (Side-Angle-Side) similarity criterion, our triangles $B T C$ and $K T L$ are similar, and the ratio is $|B T|:|K T|$ with the value $\sqrt{2}$ (thanks to the right isosceles triangle $B K T$). For the lengths of the corresponding medians $T D$ and $T E$ of the marked triangles, it indeed holds that $|T D| /|T E|=$ $=\sqrt{2}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_968d90939233ae120dcfg-05.jpg?height=546&width=808&top_left_y=1189&top_left_x=630)

The derived result remains to be connected with the fact that for the centroid $T$ of triangle $A B C$, it holds that $|A T|=2|T D|$. We thus obtain

$$
\frac{|A T|}{|D E|}=\frac{2|T D|}{|D E|}=2 \sqrt{2}
$$

This completes the proof of the statement in the first sentence of the solution.

Remark. We solved the problem in the situation as shown in the given figure, according to which the triangle $B T K$ is drawn in the half-plane $B T C$ and the triangle $C T L$ in the half-plane $C T A$. The necessary equality of angles $B T C$ and $K T L$ was derived in the solution under the assumption that point $K$ lies in the angle $B T C$ as in the figure from the problem statement. This assumption is only satisfied for those initial triangles $A B C$ in which the mentioned angle $B T C$ has a measure of at least $45^{\circ}$. However, this is not the case, for example, in the following figure. Why the angles $B T C$ and $K T L$ are equal *always*, can be justified as follows: Due to the problem statement, the oriented angles $B T K$ and $C T L$ are equal (both have a measure of $45^{\circ}$ and the same orientation as the angle $B T C$), so in the rotation determined by them with center $T$, the image of the angle $B T C$ is precisely the angle $K T L$.

![](https://cdn.mathpix.com/cropped/2024_04_17_968d90939233ae120dcfg-06.jpg?height=648&width=1128&top_left_y=230&top_left_x=470)