2. Given a square $A B C D$ with side length 1. On its side $C D$, choose a point $E$ such that $|\angle B A E|=60^{\circ}$. Further, choose any internal point on the segment $A E$ and denote it as $X$. Draw a perpendicular from point $X$ to the line $B X$ and denote its intersection with the line $B C$ as $Y$. What is the minimum possible length of the segment $B Y$?
2. Let $X$ and $Y$ be any two points described in the problem. Consider the Thales circle over the diameter $B Y$ circumscribed around the right triangle $B Y X$. This circle contains the point $X$ on the segment $A E$ and simultaneously touches the line $A B$ at point $B$. Among all such circles, the one with the smallest diameter is the circle $k$, which has $\mathrm{s}$ as a segment of $A E$ and is tangent to it, and is therefore inscribed in the equilateral triangle $A A^{\prime} F$, where $A^{\prime}$ is the image of point $A$ under central symmetry with respect to vertex $B$ and $F$ lies on the ray $B C$ (see Fig. 1, where one of the circles with a tangent $A B$ at point $B$, which has a smaller radius than circle $k$ and thus does not "reach" the segment $A E$, is also drawn). Since the center of circle $k$ is also the centroid of triangle $A A^{\prime} F$ with side lengths of 2, its diameter has length $|B Y|=\frac{2}{3} \sqrt{3}$ and the corresponding point $X$ is the midpoint of side $A F$, so it indeed belongs to the segment $A E$, since $|A X|=1<|A E|$.

![](https://cdn.mathpix.com/cropped/2024_04_17_b6718dc1fa139fe24a99g-3.jpg?height=694&width=782&top_left_y=861&top_left_x=637)

Fig. 1

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

ALTERNATIVE SOLUTION. Take any admissible points $X$ and $Y$, let $X_{0}$ be the foot of the perpendicular from point $X$ to $A B$, and set $|A X|=2 x$. Triangle $A X X_{0}$ has internal angles of $30^{\circ}, 60^{\circ}$, and $90^{\circ}$, so $\left|A X_{0}\right|=x,\left|X_{0} B\right|=1-x$ and $\left|X X_{0}\right|=\sqrt{3} x$. According to the Pythagorean theorem for triangle $X X_{0} B$, we have $|B X|^{2}=4 x^{2}-2 x+1$. Furthermore, the right triangles $X X_{0} B$ and $B X Y$ are similar (angles $X_{0} X B$ and $Y B X$ are equal alternate interior angles of the transversal $B X$ of the parallel lines $B Y$ and $X X_{0}$), and we can express the length of segment $B Y$ as

$$
|B Y|=|B X| \cdot \frac{|B X|}{\left|X X_{0}\right|}=\frac{4 x^{2}-2 x+1}{\sqrt{3} x}=\frac{\sqrt{3}}{3}\left(4 x+\frac{1}{x}-2\right) .
$$

By the AM-GM inequality, we also have $4 x+1 / x \geqq 4$, so $|B Y| \geqq \frac{2}{3} \sqrt{3}$. Since equality holds when $4 x=1 / x$ or $x=\frac{1}{2}$, and $|A X|=2 x=1<|A E|$, so point $X$ is indeed a point on the segment $A E$, the value $|B Y|=\frac{2}{3} \sqrt{3}$ is achievable and thus the desired minimum.

For a complete solution, award 6 points. For determining points $X$ and $Y$ such that $|B Y|=\frac{2}{3} \sqrt{3}$, award 2 points, but if the solver completely omits mentioning why the found point $X$ lies on the segment $A E$, award only 1 point. Award the remaining 4 points for proving that lower values cannot be achieved. If the solver proceeds numerically, award 2 points for expressing the length of segment $B Y$ in terms of any parameter determining the position of point $X$ (length $A X$, angle $A B X$, etc.).