Find maximum value of number $a$ such that for any arrangement of numbers $1,2,\ldots ,10$ on a circle, we can find three consecutive numbers such their sum bigger or equal than $a$.
1. Define \( x_1, x_2, \ldots, x_{10} \) to be the numbers on the circle consecutively, with \( x_{11} = x_1 \) and \( x_{12} = x_2 \). Let \( y_k = x_k + x_{k+1} + x_{k+2} \) for \( 1 \le k \le 10 \).

2. Note that the sum of all numbers on the circle is:
   \[
   \sum_{i=1}^{10} x_i = 1 + 2 + \cdots + 10 = 55
   \]
   Since each \( y_k \) is the sum of three consecutive \( x_i \)'s, the sum of all \( y_k \)'s is:
   \[
   \sum_{k=1}^{10} y_k = 3 \sum_{i=1}^{10} x_i = 3 \times 55 = 165
   \]

3. By the pigeonhole principle, there must be at least one \( y_k \) such that:
   \[
   y_k \ge \frac{165}{10} = 16.5
   \]
   Since \( y_k \) must be an integer, it follows that:
   \[
   y_k \ge 17
   \]
   Therefore, \( a \ge 17 \).

4. Assume for contradiction that \( a \le 17 \). This implies \( a = 17 \) since we already established \( a \ge 17 \). Thus, all \( y_k \) must be less than or equal to 17.

5. Note that if \( y_i = y_{i+1} = 17 \) for any \( 1 \le i \le 10 \), then \( x_i \) and \( x_{i+3} \) must be equal. This is because:
   \[
   y_i = x_i + x_{i+1} + x_{i+2} = 17
   \]
   \[
   y_{i+1} = x_{i+1} + x_{i+2} + x_{i+3} = 17
   \]
   Subtracting these equations, we get:
   \[
   x_i = x_{i+3}
   \]
   This implies that there can be at most five \( y_i \)'s equal to 17. Without loss of generality, assume \( y_1 = y_3 = y_5 = y_7 = y_9 = 17 \). This forces \( y_2, y_4, y_6, y_8, y_{10} \le 16 \).

6. Since the sum of all \( y_i \)'s is 165, we have:
   \[
   y_2 + y_4 + y_6 + y_8 + y_{10} = 165 - 5 \times 17 = 165 - 85 = 80
   \]
   Therefore, each of these \( y_i \)'s must be exactly 16:
   \[
   y_2 = y_4 = y_6 = y_8 = y_{10} = 16
   \]

7. Now, consider the relationships between the \( x_i \)'s. From \( y_1 = 17 \) and \( y_2 = 16 \), we have:
   \[
   x_1 + x_2 + x_3 = 17
   \]
   \[
   x_2 + x_3 + x_4 = 16
   \]
   Subtracting these equations, we get:
   \[
   x_1 - x_4 = 1 \quad \text{or} \quad x_1 = x_4 + 1
   \]
   Similarly, from \( y_4 = 16 \) and \( y_5 = 17 \), we have:
   \[
   x_4 + x_5 + x_6 = 16
   \]
   \[
   x_5 + x_6 + x_7 = 17
   \]
   Subtracting these equations, we get:
   \[
   x_4 - x_7 = -1 \quad \text{or} \quad x_4 = x_7 - 1
   \]
   Combining these, we get:
   \[
   x_1 = x_7
   \]
   This is a contradiction since all \( x_i \)'s are distinct. Therefore, \( a \) cannot be equal to 17.

8. To show that \( a \le 18 \), consider the arrangement:
   \[
   10, 1, 7, 9, 2, 6, 4, 8, 3, 5
   \]
   The sums of any three consecutive numbers are:
   \[
   10 + 1 + 7 = 18, \quad 1 + 7 + 9 = 17, \quad 7 + 9 + 2 = 18, \quad 9 + 2 + 6 = 17, \quad 2 + 6 + 4 = 12
   \]
   \[
   6 + 4 + 8 = 18, \quad 4 + 8 + 3 = 15, \quad 8 + 3 + 5 = 16, \quad 3 + 5 + 10 = 18, \quad 5 + 10 + 1 = 16
   \]
   No three consecutive numbers have a sum larger than 18, so \( a \le 18 \).

Therefore, the maximum value of \( a \) is \( \boxed{18} \).