
2. On the unit square $A B C D$ is given point $E$ on $C D$ in such a way, that $|\angle B A E|=$ $60^{\circ}$. Further let $X$ be an arbitrary inner point of the segment $A E$. Finally let $Y$ be the intersection of a line, perpendicular to $B X$ and containing $X$, with the line $B C$. What is the least possible length of $B Y$ ?

(Michal Rolínek)


Solution. Let us consider the Thales circle over $B Y$, which is circumscribed to $B Y X$. This circle contains $X$ and touches $A B$ in $B$. Of all such circles is the one which touches $A E$ (and it has to be in $X$ ) obviously the one with the least diameter (let us call the circle $k$ ). This circle is thus inscribed to the equilateral triangle $A A^{\prime} F$ where $A^{\prime}$ is the image of $\mathrm{A}$ in the point symmetry with respect to $B$ and $F$ lies on the half line $B C$ (see Fig. 1; there you can see one of the circles with smaller diameter than $k$ as well). The center of $k$ is the center of mass of the triangle $A A^{\prime} F$, equilateral triangle with sides of length 2 , thus the diameter of $k$ is $B Y=\frac{2}{3} \sqrt{3}$ and the corresponding $X$ is a center of $A F$, that is it belongs to the segment $A E$, since $|A X|=1<|A E|$.

Answer. The least possible length of $B Y$ is $\frac{2}{3} \sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_d252d571aa2423c54010g-13.jpg?height=501&width=563&top_left_y=115&top_left_x=421)

Fig. 1
