A clueless ant makes the following route: starting at point $ A $ goes $ 1$ cm north, then $ 2$ cm east, then $ 3$ cm south, then $ 4$ cm west, immediately $ 5$ cm north, continues $ 6$ cm east, and so on, finally $ 41$ cm north and ends in point $ B $. Calculate the distance between $ A $ and $ B $ (in a straight line).

1. **Understanding the Path:**
   The ant starts at point \( A \) and follows a spiral path with increasing distances in the north, east, south, and west directions. The distances increase by 1 cm each time. The sequence of movements is:
   - 1 cm north
   - 2 cm east
   - 3 cm south
   - 4 cm west
   - 5 cm north
   - 6 cm east
   - 7 cm south
   - 8 cm west
   - ...
   - 41 cm north

2. **Breaking Down the Movements:**
   Let's break down the movements into horizontal (east-west) and vertical (north-south) components:
   - North-South movements: \(1, -3, 5, -7, 9, -11, \ldots, 41\)
   - East-West movements: \(2, -4, 6, -8, 10, -12, \ldots\)

3. **Summing the Movements:**
   - **Vertical (North-South) Sum:**
     The vertical movements form an arithmetic sequence with the first term \(a = 1\) and common difference \(d = 2\). The sequence is:
     \[
     1, -3, 5, -7, 9, -11, \ldots, 41
     \]
     We need to find the sum of this sequence. The positive terms are \(1, 5, 9, \ldots, 41\) and the negative terms are \(-3, -7, -11, \ldots\).

     The number of terms in the sequence can be found by solving:
     \[
     1 + (n-1) \cdot 4 = 41 \implies n = 11
     \]
     So, there are 11 terms in the sequence.

     The sum of the positive terms:
     \[
     S_+ = 1 + 5 + 9 + \ldots + 41
     \]
     This is an arithmetic series with \(a = 1\), \(d = 4\), and \(n = 6\):
     \[
     S_+ = \frac{n}{2} \cdot (2a + (n-1)d) = \frac{6}{2} \cdot (2 \cdot 1 + 5 \cdot 4) = 3 \cdot (2 + 20) = 3 \cdot 22 = 66
     \]

     The sum of the negative terms:
     \[
     S_- = -3 - 7 - 11 - \ldots - 39
     \]
     This is an arithmetic series with \(a = -3\), \(d = -4\), and \(n = 5\):
     \[
     S_- = \frac{n}{2} \cdot (2a + (n-1)d) = \frac{5}{2} \cdot (2 \cdot -3 + 4 \cdot -4) = \frac{5}{2} \cdot (-6 - 16) = \frac{5}{2} \cdot -22 = -55
     \]

     The total vertical displacement:
     \[
     S_{\text{vertical}} = S_+ + S_- = 66 - 55 = 11
     \]

   - **Horizontal (East-West) Sum:**
     The horizontal movements form an arithmetic sequence with the first term \(a = 2\) and common difference \(d = 4\). The sequence is:
     \[
     2, -4, 6, -8, 10, -12, \ldots
     \]
     The number of terms in the sequence is 10 (since the last term is 40).

     The sum of the positive terms:
     \[
     S_+ = 2 + 6 + 10 + \ldots + 38
     \]
     This is an arithmetic series with \(a = 2\), \(d = 4\), and \(n = 5\):
     \[
     S_+ = \frac{n}{2} \cdot (2a + (n-1)d) = \frac{5}{2} \cdot (2 \cdot 2 + 4 \cdot 4) = \frac{5}{2} \cdot (4 + 16) = \frac{5}{2} \cdot 20 = 50
     \]

     The sum of the negative terms:
     \[
     S_- = -4 - 8 - 12 - \ldots - 40
     \]
     This is an arithmetic series with \(a = -4\), \(d = -4\), and \(n = 5\):
     \[
     S_- = \frac{n}{2} \cdot (2a + (n-1)d) = \frac{5}{2} \cdot (2 \cdot -4 + 4 \cdot -4) = \frac{5}{2} \cdot (-8 - 16) = \frac{5}{2} \cdot -24 = -60
     \]

     The total horizontal displacement:
     \[
     S_{\text{horizontal}} = S_+ + S_- = 50 - 60 = -10
     \]

4. **Calculating the Distance:**
   The final position of the ant relative to the starting point \( A \) is:
   \[
   (x, y) = (-10, 11)
   \]
   The distance between \( A \) and \( B \) is given by the Euclidean distance formula:
   \[
   \text{Distance} = \sqrt{(-10)^2 + 11^2} = \sqrt{100 + 121} = \sqrt{221} = \boxed{29}
   \]