4. $\quad$ in triangle $ABC$ the sine of angle $A$ is equal to $3/5$. On side $AC$ a point $M$ is taken such that $CM=15$, on side $AB$ a point $N$ is taken such that $BN=7, AN=AM, T$ is the midpoint of $NC$, $P$ is the midpoint of $B$. Find the length of segment $PT$.
Answer: $\sqrt{26.5}, \sqrt{110.5}$

Solution. Let the length $\mathrm{AN}=\mathrm{AM}=\mathrm{x}$. Introduce a system of two unit vectors: let vector $\mathbf{~collinear~with~vector~} \mathrm{AB}$, and vector $\mathbf{c}$ be collinear with vector $\mathrm{AC}$. Then the following vector relations hold:

$$
\begin{gathered}
\overrightarrow{P T}=\overrightarrow{A T}-\overrightarrow{A P}, \overrightarrow{A P}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A M}), \overrightarrow{A T}=\frac{1}{2}(\overrightarrow{A C}+\overrightarrow{A N}) \\
\overrightarrow{A B}=(x+7) \mathbf{b}, \overrightarrow{A C}=(x+15) \mathbf{c}, \quad \overrightarrow{A M}=x \mathbf{c}, \quad \overrightarrow{A N}=x \mathbf{b} \\
\overrightarrow{P T}=\frac{1}{2}(\overrightarrow{A C}+\overrightarrow{A N}-\overrightarrow{A B}-\overrightarrow{A M})=\frac{1}{2}(15 \mathbf{c}-7 \mathbf{b})
\end{gathered}
$$

Calculating the scalar square of vector $\overrightarrow{P T}$, and considering that the cosine of angle A can be equal to $4 / 5$ for an acute angle and $-4 / 5$ for an obtuse angle, we get:

$$
\begin{aligned}
& \overrightarrow{P T} \cdot \overrightarrow{P T}=\frac{1}{4}(49 \mathbf{b} \cdot \mathbf{b}+225 \mathbf{c} \cdot \mathbf{c}-210 \mathbf{b} \cdot \mathbf{c})=26.5 \\
& \overrightarrow{P T} \cdot \overrightarrow{P T}=\frac{1}{4}(49 \mathbf{b} \cdot \mathbf{b}+225 \mathbf{c} \cdot \mathbf{c}-210 \mathbf{b} \cdot \mathbf{c})=110.5
\end{aligned}
$$

Criteria. A correct answer for both cases is obtained with proper justification - 7 points. The solution can be planimetric and not rely on the vector method. One case is addressed - 3 points. Arithmetic error - minus 1 point.