The numbers 1 through 10 can be arranged along the vertices and sides of a pentagon so that the sum of the three numbers along each side is the same. The diagram below shows an arrangement with sum 16. Find, with proof, the smallest possible value for a sum and give an example of an arrangement with that sum.

[asy]
int i;
pair[] A={dir(18+72*0), dir(18+72*1), dir(18+72*2),dir(18+72*3), dir(18+72*4), dir(18+72*0)};
pair O=origin;
int[] v = {7,1,10,4,3};
int[] s = {8, 5, 2, 9, 6};
for(i=0; i<5; i=i+1) {
label(string(v[i]), A[i], dir(O--A[i]));
label(string(s[i]), A[i]--A[i+1], dir(-90)*dir(A[i]--A[i+1]));
}
draw(rotate(0)*polygon(5));[/asy]
1. **Labeling the vertices and sides:**
   We label the vertices and sides of the pentagon as follows:
   - Vertices: \(a, c, e, g, i\)
   - Sides: \(b, d, f, h, j\)
   The sum of the numbers along each side of the pentagon is given by:
   \[
   a + b + c = c + d + e = e + f + g = g + h + i = i + j + a
   \]
   Let this common sum be \(S\).

2. **Expressing the sum \(S\):**
   We can express \(S\) in terms of the total sum of the numbers from 1 to 10. The total sum of the numbers from 1 to 10 is:
   \[
   1 + 2 + 3 + \cdots + 10 = 55
   \]
   Since each number appears exactly once in the arrangement, the sum of all the numbers along the sides of the pentagon is:
   \[
   (a + b + c) + (c + d + e) + (e + f + g) + (g + h + i) + (i + j + a)
   \]
   Simplifying, we get:
   \[
   2(a + c + e + g + i) + (b + d + f + h + j) = 55
   \]
   Since each side sum \(S\) appears twice in the above expression, we have:
   \[
   5S = 2(a + c + e + g + i) + (b + d + f + h + j)
   \]

3. **Finding the minimum value of \(S\):**
   To find the minimum value of \(S\), we need to minimize \(a + c + e + g + i\). The smallest possible values for \(a, c, e, g, i\) are 1, 2, 3, 4, and 5. Thus:
   \[
   a + c + e + g + i = 1 + 2 + 3 + 4 + 5 = 15
   \]
   Substituting this into the equation for \(S\), we get:
   \[
   5S = 2 \times 15 + (b + d + f + h + j)
   \]
   Since \(b + d + f + h + j\) must be the remaining numbers 6, 7, 8, 9, and 10, we have:
   \[
   b + d + f + h + j = 6 + 7 + 8 + 9 + 10 = 40
   \]
   Therefore:
   \[
   5S = 2 \times 15 + 40 = 30 + 40 = 70
   \]
   Solving for \(S\), we get:
   \[
   S = \frac{70}{5} = 14
   \]

4. **Example of an arrangement with sum 14:**
   We need to find an arrangement of the numbers such that the sum along each side is 14. One possible arrangement is:
   - Vertices: \(1, 4, 2, 5, 3\)
   - Sides: \(9, 8, 7, 6, 10\)
   This arrangement satisfies:
   \[
   1 + 9 + 4 = 14, \quad 4 + 8 + 2 = 14, \quad 2 + 7 + 5 = 14, \quad 5 + 6 + 3 = 14, \quad 3 + 10 + 1 = 14
   \]

The final answer is \(\boxed{14}\).