B2. In the square $A B C D$, a circle $\mathcal{K}$, a semicircle $\mathcal{P}$, and a quarter circle $\mathcal{Q}$ are inscribed. The quarter circle $\mathcal{Q}$ has its center at vertex $A$ and contains points $B$ and $D$. The semicircle $\mathcal{P}$ has its center at the midpoint of side $A D$ and contains points $A$ and $D$. The circle $\mathcal{K}$ touches the semicircle $\mathcal{P}$ at point $E$, the quarter circle $\mathcal{Q}$ at point $F$, and side $A B$ of the square $A B C D$ at point $G$.

(a) Express the radius of the circle $\mathcal{K}$ in terms of the side length of the square $A B C D$.

(b) Prove that the line $E F$ is parallel to side $A B$.
B2.

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(a) Let the side length of the square $ABCD$ be $a$, and the radius of the circle $\mathcal{K}$ be $r$. Let $S$ be the center of the semicircle $\mathcal{P}$, $T$ the center of the circle $\mathcal{K}$, and $H$ the orthogonal projection of point $T$ onto side $AD$. Let the length of segment $AG$ be $x$. Since $|AT| = |AF| - |FT| = a - r$, by the Pythagorean theorem for triangle $AGT$, we have $x^2 + r^2 = (a - r)^2$, which simplifies to $x^2 = a^2 - 2ar$. Since $AGTH$ is a rectangle, $|HT| = x$ and $|SH| = |SA| - |AH| = \frac{a}{2} - r$. Also, $|ST| = |SE| + |ET| = \frac{a}{2} + r$, so by the Pythagorean theorem for triangle $THS$, we have $x^2 + \left(\frac{a}{2} - r\right)^2 = \left(\frac{a}{2} + r\right)^2$, which simplifies to $x^2 = 2ar$. From both equations, we get $a^2 - 2ar = 2ar$ or $r = \frac{a}{4}$.

(b) From the result in part (a), we have $|AH| = \frac{a}{4}$ and thus $|HS| = |SA| - |AH| = \frac{a}{2} - \frac{a}{4} = \frac{a}{4}$, so triangle $ATS$ is isosceles with the vertex at $T$. Similarly, triangle $FTE$ is isosceles with the vertex at $T$. From this, we derive

$$
\angle EFT = \frac{\pi - \angle FTE}{2} = \frac{\angle ETA}{2} = \angle HTA = \angle GAT
$$

thus line $EF$ is parallel to side $AB$.

Identifying one of the collinearities $A, T, F$ and $S, E, T$ 1 point.

Obtaining one of the equations $x^2 + r^2 = (a - r)^2$ and $x^2 + \left(\frac{a}{2} - r\right)^2 = \left(\frac{a}{2} + r\right)^2 \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ 1 point.

Writing the remaining equation and the remaining collinearity ........................................................... 1 point.

![](https://cdn.mathpix.com/cropped/2024_06_07_e347356273e0c1397da7g-19.jpg?height=52&width=1727&top_left_y=2047&top_left_x=176)

![](https://cdn.mathpix.com/cropped/2024_06_07_e347356273e0c1397da7g-19.jpg?height=49&width=1722&top_left_y=2100&top_left_x=178)

Proof of the parallelism of $AB$ and $EF$ ............................................................ 2 points.