On a circumference of a unit radius, take points $A$ and $B$ such that section $AB$ has length one. $C$ can be any point on the longer arc of the circle between $A$ and $B$. How do we take $C$ to make the perimeter of the triangle $ABC$ as large as possible?
1. **Identify the given information and the goal:**
   - We have a unit circle with points \( A \) and \( B \) such that the chord \( AB \) has length 1.
   - We need to find the point \( C \) on the longer arc \( AB \) that maximizes the perimeter of triangle \( ABC \).

2. **Calculate the central angle \( \theta \) subtended by chord \( AB \):**
   - Since \( AB \) is a chord of length 1 in a unit circle, we can use the chord length formula:
     \[
     AB = 2 \sin\left(\frac{\theta}{2}\right)
     \]
     Given \( AB = 1 \), we have:
     \[
     1 = 2 \sin\left(\frac{\theta}{2}\right) \implies \sin\left(\frac{\theta}{2}\right) = \frac{1}{2} \implies \frac{\theta}{2} = 30^\circ \implies \theta = 60^\circ
     \]

3. **Express the perimeter of \( \triangle ABC \):**
   - Let \( \alpha \) be the angle subtended by arc \( AC \) at the center of the circle.
   - The perimeter of \( \triangle ABC \) is given by:
     \[
     P = AB + AC + BC
     \]
   - Since \( AB = 1 \), we need to find \( AC \) and \( BC \).

4. **Calculate \( AC \) and \( BC \):**
   - The lengths \( AC \) and \( BC \) are the chord lengths corresponding to angles \( \alpha \) and \( 120^\circ - \alpha \) respectively.
   - Using the chord length formula:
     \[
     AC = 2 \sin\left(\frac{\alpha}{2}\right)
     \]
     \[
     BC = 2 \sin\left(\frac{120^\circ - \alpha}{2}\right)
     \]

5. **Formulate the expression for the perimeter:**
   - The perimeter \( P \) is:
     \[
     P = 1 + 2 \sin\left(\frac{\alpha}{2}\right) + 2 \sin\left(\frac{120^\circ - \alpha}{2}\right)
     \]

6. **Simplify the expression:**
   - Let \( x = \frac{\alpha}{2} \). Then the perimeter becomes:
     \[
     P = 1 + 2 \sin(x) + 2 \sin(60^\circ - x)
     \]
   - Using the angle subtraction formula for sine:
     \[
     \sin(60^\circ - x) = \sin 60^\circ \cos x - \cos 60^\circ \sin x = \frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x
     \]
   - Therefore:
     \[
     P = 1 + 2 \sin(x) + 2 \left( \frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x \right) = 1 + 2 \sin(x) + \sqrt{3} \cos x - \sin(x)
     \]
     \[
     P = 1 + \sin(x) + \sqrt{3} \cos x
     \]

7. **Maximize the expression:**
   - To maximize \( P = 1 + \sin(x) + \sqrt{3} \cos x \), we use the Cauchy-Schwarz inequality:
     \[
     (\sin x + \sqrt{3} \cos x)^2 \leq (1^2 + (\sqrt{3})^2)(\sin^2 x + \cos^2 x) = 4
     \]
     \[
     \sin x + \sqrt{3} \cos x \leq 2
     \]
   - Therefore, the maximum value of \( \sin x + \sqrt{3} \cos x \) is 2, and the maximum perimeter is:
     \[
     P = 1 + 2 = 3
     \]

The final answer is \( \boxed{3} \).