In the rectangle $ABCD$, the sides are $AB=3, BC=2$. $P$ is a point on the side $AB$ such that the line $PD$ touches, at point $E$, the circle with diameter $BC$. The line passing through the center of the circle and $E$ intersects the side $AB$ at $Q$. What is the area of the triangle $PQE$?
Since the lengths of two tangent segments drawn from a point outside a circle are equal, $D E=D C=3$ and $P E=P B$. If the length of these latter segments is $x$, then in the right triangle $D A P$, by the Pythagorean theorem,

$$
A D^{2}+A P^{2}=D P^{2}, \quad \text { that is } \quad 2^{2}+(3-x)^{2}=(3+x)^{2}
$$

From this equation, we get $x=\frac{1}{3}$.

The triangle $Q E P$ is similar to the triangle $D A P$, because the angle at $P$ is common, and the angles at $A$ and $E$ are right angles (the angle at $E$ is a right angle because $Q E$ is the extension of the radius of the circle at $E$, which is perpendicular to the tangent at $E$). Therefore, the ratio of the corresponding sides of the triangles is the same:

$$
\frac{E Q}{P E}=\frac{A D}{A P}
$$

From this, we get $E Q=\frac{P E \cdot A D}{A P}=\frac{x \cdot 2}{3-x}=\frac{1}{4}$.

The area of the right triangle $Q E P$ is therefore $\frac{1}{2} \cdot P E \cdot E Q=\frac{1}{24}$ square units.

Based on the work of Gergő Takács (Tatabánya, Árpád Gymnasium and Pedagogical College, 11th grade)