## Task B-3.3.

Determine the integer solutions of the inequality $\frac{x^{2}-16 x+39}{\tan^{2} \frac{\pi x}{4}-1} \geqslant 0$.
## Solution.

Let's write down the conditions for the given expression.

Due to the domain of the tangent function, $\frac{\pi x}{4} \neq \frac{\pi}{2} + k \pi$ so $x \neq 4k + 2, k \in \mathbb{Z}$.

Furthermore, from $\operatorname{tg}^{2}\left(\frac{\pi x}{4}\right) - 1 \neq 0$ it follows that $\operatorname{tg}^{2}\left(\frac{\pi x}{4}\right) \neq 1$ or $\operatorname{tg}\left(\frac{\pi x}{4}\right) \neq \pm 1$.

Thus, $\frac{\pi x}{4} \neq \pm \frac{\pi}{4} + k \pi$ so $x \neq 4k \pm 1$.

From these conditions, it follows that $x = 4k, k \in \mathbb{Z}$, since we are looking for only integer solutions.

Now let's solve the inequality. We have two cases:

1. Case:

$$
\begin{aligned}
& x^{2} - 16x + 39 \geqslant 0 \\
& \operatorname{tg}^{2}\left(\frac{\pi x}{4}\right) - 1 > 0
\end{aligned}
$$

From $(x-3)(x-13) \geqslant 0$ it follows that $x \in (-\infty, 3] \cup [13, \infty)$.

$$
\begin{gathered}
\text{From } \operatorname{tg}^{2}\left(\frac{\pi x}{4}\right) > 1 \text{ it follows that} \\
\operatorname{tg}\left(\frac{\pi x}{4}\right) > 1 \text{ or } \operatorname{tg}\left(\frac{\pi x}{4}\right) < -1 \\
-\frac{\pi}{2} + k \pi < \frac{\pi x}{4} < -\frac{\pi}{4} + k \pi \text{ or } \frac{\pi}{4} + k \pi < \frac{\pi x}{4} < \frac{\pi}{2} + k \pi \\
-2 + 4k < x < -1 + 4k \text{ or } 1 + 4k < x < 2 + 4k, k \in \mathbb{Z}
\end{gathered}
$$

Thus, these are the intervals $\ldots (-2, -1) \cup (1, 2) \cup (2, 3) \cup (5, 6) \cup (6, 7) \cup (9, 10) \ldots$

Due to the condition $x = 4k, k \in \mathbb{Z}$, this case has no solutions.

2. Case:

$$
\begin{aligned}
& x^{2} - 16x + 39 \leqslant 0 \\
& \operatorname{tg}^{2}\left(\frac{\pi x}{4}\right) - 1 < 0
\end{aligned}
$$

From $(x-3)(x-13) \leqslant 0$ it follows that $x \in [3, 13]$.

$$
\begin{gathered}
\text{From } \operatorname{tg}^{2}\left(\frac{\pi x}{4}\right) < 1 \text{ it follows that} \\
-1 < \operatorname{tg}\left(\frac{\pi x}{4}\right) < 1 \\
-\frac{\pi}{4} + k \pi < \frac{\pi x}{4} < \frac{\pi}{4} + k \pi \\
-\frac{1}{4} + k < \frac{x}{4} < \frac{1}{4} + k \\
4k - 1 < x < 4k + 1, k \in \mathbb{Z}
\end{gathered}
$$

Thus, $x = 4k, k \in \mathbb{Z}$ and $x \in [3, 13]$ so $x \in \{4, 8, 12\}$, which is the final solution.

## Problem B-3.4

In triangle $ABC$, point $D$ is on side $\overline{AB}$, and point $E$ is on side $\overline{BC}$ such that $|AD| : |DB| = |CE| : |EB| = 3 : 4$. Segments $\overline{AE}$ and $\overline{CD}$ intersect at point $P$.

Express the vector $\overrightarrow{AP}$ as a linear combination of vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$.

## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_ea3ef29009412d1e0d2bg-15.jpg?height=432&width=574&top_left_y=338&top_left_x=752)

Express the vector $\overrightarrow{AE} = \overrightarrow{AD} + \overrightarrow{DE}$, where $\overrightarrow{AD} = \frac{3}{7} \overrightarrow{AB}$ and $\overrightarrow{DE} = \frac{4}{7} (\overrightarrow{AB} + \overrightarrow{BC}) = \frac{4}{7} \overrightarrow{AC}$.

Substituting, we get $\overrightarrow{AE} = \frac{3}{7} \overrightarrow{AB} + \frac{4}{7} \overrightarrow{AC}$.

Vectors $\overrightarrow{AE}$ and $\overrightarrow{AP}$ are collinear, so $\overrightarrow{AP} = \alpha \overrightarrow{AE} = \alpha \left( \frac{3}{7} \overrightarrow{AB} + \frac{4}{7} \overrightarrow{AC} \right)$

or

$$
\overrightarrow{AP} = \frac{3}{7} \alpha \overrightarrow{AB} + \frac{4}{7} \alpha \overrightarrow{AC}
$$

Express the vector $\overrightarrow{AP}$ in another way.

$\overrightarrow{AP} = \overrightarrow{AC} + \overrightarrow{CP}$, where $\overrightarrow{CP} = \beta \overrightarrow{CD}$ and

$\overrightarrow{CD} = \overrightarrow{AD} - \overrightarrow{AC} = \frac{3}{7} \overrightarrow{AB} - \overrightarrow{AC}$.

Substituting, we get $\overrightarrow{AP} = \overrightarrow{AC} + \beta \left( \frac{3}{7} \overrightarrow{AB} - \overrightarrow{AC} \right)$,

or

$$
\overrightarrow{AP} = \frac{3}{7} \beta \overrightarrow{AB} + (1 - \beta) \overrightarrow{AC}
$$

From the two expressions for vector $\overrightarrow{AP}$ (1) and (2), we get the system:

$$
\begin{aligned}
\alpha & = \beta \\
\frac{4}{7} \alpha & = 1 - \beta
\end{aligned}
$$

The solution to the system is $\alpha = \beta = \frac{7}{11}$.

Finally, $\overrightarrow{AP} = \frac{3}{11} \overrightarrow{AB} + \frac{4}{11} \overrightarrow{AC}$.