5. In triangle $\mathrm{ABC}$, the sine of angle $\mathrm{A}$ is equal to $\frac{\sqrt{3}}{2}$. On side $\mathrm{AC}$, point $\mathrm{M}$ is taken such that $\mathrm{CM}=b$, on side $\mathrm{AB}$, point $\mathrm{N}$ is taken such that $\mathrm{BN}=a, \mathrm{~T}$ is the midpoint of $\mathrm{NC}$, and $\mathrm{P}$ is the midpoint of $\mathrm{BM}$. Find $\mathrm{PT}$.
Solution. Method 1. Since we know the lengths of vectors $\overrightarrow{M C}, \overrightarrow{N C}$ and the angle between them, it is obvious that we need to express the vector $\overrightarrow{P T}$ in terms of vectors $\overrightarrow{M C}$ and $\overrightarrow{N C}$. This can be done in different ways. We will show one of the possible methods. Note that $\overrightarrow{A P}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A M})$ and $\overrightarrow{A T}=\frac{1}{2}(\overrightarrow{A N}+\overrightarrow{A C})$. Since $\overrightarrow{A M}=\overrightarrow{A C}-\overrightarrow{M C}$ and $\overrightarrow{A N}=\overrightarrow{A B}-\overrightarrow{N B}$, then $\overrightarrow{A P}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C}-\overrightarrow{M C})$ and $\overrightarrow{A T}=\frac{1}{2}(\overrightarrow{A B}-\overrightarrow{N B}+\overrightarrow{A C})$. Now $\overrightarrow{P T}=\overrightarrow{A T}-\overrightarrow{A P}=\frac{1}{2}(\overrightarrow{M C}-\overrightarrow{N B})$. It remains to find the length of the vector $\overrightarrow{P T}$

$$
\begin{array}{r}
|P T|^{2}=\left|\frac{1}{2}(\overrightarrow{M C}-\overrightarrow{N C})\right|^{2}=\frac{1}{4}\left(M C^{2}+N C^{2}-2 \cdot M C \cdot N C \cdot \cos (A)\right)= \\
=\frac{1}{4}\left(b^{2}+a^{2} \pm 2 \cdot b \cdot a \cdot \frac{1}{2}\right)=\frac{1}{4}\left(b^{2}+a^{2} \pm b a\right)
\end{array}
$$

In the last formula, the sign "plus" is taken when angle A is $120^{\circ}$, and "minus" when angle A is $60^{\circ}$.

Answer. $P T=\frac{1}{2} \sqrt{b^{2}+a^{2} \pm b a}$.

Method 2. Let $\mathrm{K}$ be the midpoint of BC. Then PK and TK are the midlines of triangles MBC and NBC, respectively. From the properties of the midlines of a triangle, it follows that

$T K\|A B, \quad P K\| A C$ and $\angle P K T=\angle A$, and also $P K=\frac{b}{2}, \quad T K=\frac{a}{2}$. Now, by the cosine theorem for triangle PKT, we get:

$$
\begin{aligned}
& P T^{2}=P K^{2}+T K^{2}-2 P K \cdot T K \cdot \cos (A)= \\
& =\frac{1}{4}\left(b^{2}+a^{2} \pm 2 \cdot b \cdot a \cdot \frac{1}{2}\right)=\frac{1}{4}\left(b^{2}+a^{2} \pm b a\right)
\end{aligned}
$$

Answer. $P T=\frac{1}{2} \sqrt{b^{2}+a^{2} \pm b a}$

Recommendations for evaluating solutions. Correctly found a representation of the form $\overrightarrow{P T}=$ $\frac{1}{2}(\overrightarrow{M C}-\overrightarrow{N C})$ or an analog for method 2 - 3 points. Correctly analyzed only one of the two cases - 4 points.