Problem 1285

1. Identify the sides of the triangle and the given angle:
The sides of the triangle are $x$, $2x + 1$, and $x + 2$. One of the angles is $60^\circ$.

2. Apply the Law of Cosines:
The Law of Cosines states that for any triangle with sides $a$, $b$, and $c$ and an angle $\gamma$ opposite side $c$:
$$c^2 = a^2 + b^2 - 2ab \cos(\gamma)$$
Here, we assume the angle $60^\circ$ is opposite the side $x + 2$. Thus, we have:
$$(x + 2)^2 = x^2 + (2x + 1)^2 - 2 \cdot x \cdot (2x + 1) \cdot \cos(60^\circ)$$

3. **Substitute $\cos(60^\circ) = \frac{1}{2}$:**
$$(x + 2)^2 = x^2 + (2x + 1)^2 - 2 \cdot x \cdot (2x + 1) \cdot \frac{1}{2}$$
Simplify the equation:
$$(x + 2)^2 = x^2 + (2x + 1)^2 - x(2x + 1)$$

4. Expand and simplify the equation:
$$(x + 2)^2 = x^2 + 4x^2 + 4x + 1 - 2x^2 - x$$
$$x^2 + 4x + 4 = 3x^2 + 3x + 1$$

5. Rearrange the equation to form a quadratic equation:
$$x^2 + 4x + 4 = 3x^2 + 3x + 1$$
$$0 = 3x^2 + 3x + 1 - x^2 - 4x - 4$$
$$0 = 2x^2 - x - 3$$

6. **Solve the quadratic equation $2x^2 - x - 3 = 0$:**
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 2$, $b = -1$, and $c = -3$:
$$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2}$$
$$x = \frac{1 \pm \sqrt{1 + 24}}{4}$$
$$x = \frac{1 \pm \sqrt{25}}{4}$$
$$x = \frac{1 \pm 5}{4}$$
$$x = \frac{6}{4} \quad \text{or} \quad x = \frac{-4}{4}$$
$$x = \frac{3}{2} \quad \text{or} \quad x = -1$$
Since $x$ must be positive, we have $x = \frac{3}{2}$.

7. Calculate the perimeter of the triangle:
$$\text{Perimeter} = x + (2x + 1) + (x + 2)$$
$$\text{Perimeter} = \frac{3}{2} + 2 \cdot \frac{3}{2} + 1 + \frac{3}{2} + 2$$
$$\text{Perimeter} = \frac{3}{2} + 3 + 1 + \frac{3}{2} + 2$$
$$\text{Perimeter} = \frac{3}{2} + \frac{3}{2} + 3 + 1 + 2$$
$$\text{Perimeter} = 3 + 3 + 3$$
$$\text{Perimeter} = 9$$

The final answer is $\boxed{9}$.