## Task A-2.4.

Point $P$ is the midpoint of segment $\overline{A B}$ with length 2. Let $T$ be the point of tangency of the tangent from point $A$ to the circle with diameter $\overline{P B}$. Determine the length $|P T|$.
## Second Solution.

Let $O$ be the midpoint of the segment $\overline{P B}$. From the conditions of the problem, it is clear that $|A P|=|B P|=1$, $|P O|=|O B|=\frac{1}{2}$.

Triangle $A T O$ is a right triangle because the radius is perpendicular to the tangent. Therefore, by the Pythagorean theorem, we have $|A T|^{2}+|O T|^{2}=|A O|^{2}$. Substituting $|O T|=1$ and $|A O|=\frac{3}{2}$, we get

$$
|A T|=\sqrt{2}
$$

Let $N$ be the foot of the perpendicular from $T$ to $P B$.

Let $|P N|=x$. Then $|B N|=|B P|-|P N|=1-x$.

![](https://cdn.mathpix.com/cropped/2024_05_30_a630fb9dab34a5225850g-11.jpg?height=462&width=913&top_left_y=1194&top_left_x=503)

Triangle $P T B$ is a right triangle because $\overline{P B}$ is the diameter of the circle. Applying the Euclidean theorem to triangle $P T B$, we get

$$
|N T|^{2}=|P N| \cdot|N B|=x(1-x)
$$

By the Pythagorean theorem for triangle $A T N$, we have $(1+x)^{2}+x(1-x)=2$.

This simplifies to $1+3 x=2$, i.e., $x=\frac{1}{3}$.

From the previous equalities, we directly get

$$
|N B|=1-x=\frac{2}{3} \quad \text{and} \quad|N T|=\sqrt{x(1-x)}=\frac{\sqrt{2}}{3}
$$

Using the Pythagorean theorem in triangle $P T N$, we get $|P N|^{2}+|N T|^{2}=|P T|^{2}$. Substituting $|P N|$ and $|N T|$, we obtain

$$
|P T|=\frac{1}{\sqrt{3}}
$$

1 point

Note: The statement $|A T|^{2}=|A P| \cdot|A B|$ can be justified by considering the power of point $A$ or by using the similarity of triangles $A P T$ and $A T B$, as in the first solution.

Stewart's theorem can be directly applied to triangle ATO. This gives us

$$
|A P| \cdot|T O|^{2}+|O P| \cdot|A T|^{2}=|A O| \cdot\left(|P T|^{2}+|A P| \cdot|P O|\right)
$$

If we previously calculate $|A T|=\sqrt{2}$, we can easily find $|P T|$.