4. Solve for $x$ the equation

$$
\frac{2 p-5}{x+2}=\frac{3 x+4}{x^{2}+3 x+2}(p-1)+\frac{3}{x+1}(p-1)
$$

For which values of the real parameter $p$ is the solution of the equation less than -2?
## Solution.

For the denominators to be different from zero, it must be $x \neq -2$ and $x \neq -1$.

Multiply the equation by the common denominator $(x+1)(x+2)$, and we get:

$$
(2 p-5)(x+1)=(3 x+4)(p-1)+3(x+2)(p-1)
$$

After rearranging the equation, we can write it in the form:

$$
x(1-4 p)=8 p-5
$$

For $p \neq \frac{1}{4}$, the solution is $x=\frac{8 p-5}{1-4 p}$.

For $p=\frac{1}{4}$, the equation has no solution.

However, we still need to check that $x \neq -2$ and $x \neq -1$.

$$
\begin{gathered}
x=-2 \Leftrightarrow \frac{8 p-5}{1-4 p}=-2 \Leftrightarrow 8 p-5=-2(1-4 p) \Leftrightarrow -5=-2 \\
x=-1 \Leftrightarrow \frac{8 p-5}{1-4 p}=-1 \Leftrightarrow 8 p-5=-(1-4 p) \Leftrightarrow 4 p=4 \Leftrightarrow p=1
\end{gathered}
$$

This means that $x \neq -2$ for any $p$, and $x = -1$ only for $p=1$. Therefore, the equation has no solution for $p=1$ as well.

(5 points)

Now, let's answer the posed question: for which $p$ is $x > 0$.

The answer is: for all $p < \frac{1}{4}$.

(7 points)

![](https://cdn.mathpix.com/cropped/2024_05_30_3430c4665f45242670cag-4.jpg?height=71&width=1469&top_left_y=240&top_left_x=309)

Given that the vertices of all trapezoids are also the vertices of a regular dodecagon, what is the perimeter of that dodecagon?

## Solution.

The internal angle of a regular dodecagon is $150^{\circ}$,

![](https://cdn.mathpix.com/cropped/2024_05_30_3430c4665f45242670cag-4.jpg?height=524&width=503&top_left_y=562&top_left_x=806)

so we have

$$
2 \alpha + 90^{\circ} = 150^{\circ}, \alpha = 30^{\circ}
$$

Let's isolate one trapezoid:

![](https://cdn.mathpix.com/cropped/2024_05_30_3430c4665f45242670cag-4.jpg?height=174&width=529&top_left_y=1328&top_left_x=782)

If the lengths $|A F| = |F E| = |E B| = x$ and $|A B| = 1$, then

$$
|A G| = \frac{1 - x}{2}
$$

Triangle $A G F$ is half of an equilateral triangle with side length $x$, so its height $|A G|$ is $\frac{x \sqrt{3}}{2}$.

Therefore,

$$
\frac{1 - x}{2} = \frac{x \sqrt{3}}{2}
$$

or $x(1 + \sqrt{3}) = 1$, and finally

$$
x = \frac{1}{1 + \sqrt{3}} = \frac{\sqrt{3} - 1}{2}
$$

Thus, the perimeter of the dodecagon is $O = 12 x = 6(\sqrt{3} - 1)$.

Note. Parts of this problem can be solved in different ways. In any case, determining the angle between the side of the square and the dodecagon is worth 5 points, setting up the equation whose solution is the side of the dodecagon is worth 10 points, and solving that equation and the final result is worth another 5 points.