## Task A-2.2.

Square $A B C D$ has a side length of 1. Let point $X$ be on side $\overline{A B}$, and point $Y$ be on side $\overline{A D}$ such that $\varangle C X Y=90^{\circ}$. Determine the position of point $X$ for which the area of triangle $C D Y$ is the smallest possible.
## First Solution.

Let $x=|A X|$, and $y=|A Y|$. Triangles $A X Y$ and $B C X$ are similar: both are right-angled, and $\varangle A X Y = \varangle B C X$ because they are angles with perpendicular legs.

![](https://cdn.mathpix.com/cropped/2024_05_30_9a2ee33c697c48e6349cg-10.jpg?height=653&width=711&top_left_y=1678&top_left_x=615)

We conclude that

$$
\frac{|B C|}{|B X|}=\frac{|A X|}{|A Y|}, \quad \text { i.e. } \quad \frac{1}{1-x}=\frac{x}{y}
$$

From this, it follows that $y=x(1-x)=x-x^{2}$.

Now we can express the area of triangle $C D Y$ as a function of $x$:

$$
P(C D Y)=\frac{1}{2}\left(x^{2}-x+1\right)
$$

Therefore, we need to determine $x \in [0,1]$ for which the function $P(x)=\frac{1}{2}\left(x^{2}-x+1\right)$ achieves its minimum possible value.

We see that $P(x)$ is a quadratic function with a positive leading coefficient and its vertex at $x=\frac{1}{2}$. From this, we conclude that the minimum possible area is achieved for $x=\frac{1}{2}$, i.e., when $X$ is the midpoint of segment $\overline{A B}$.