$ABC$ is a triangle with points $D$, $E$ on $BC$ with $D$ nearer $B$; $F$, $G$ on $AC$, with $F$ nearer $C$; $H$, $K$ on $AB$, with $H$ nearer $A$. Suppose that $AH=AG=1$, $BK=BD=2$, $CE=CF=4$, $\angle B=60^\circ$ and that $D$, $E$, $F$, $G$, $H$ and $K$ all lie on a circle. Find the radius of the incircle of triangle $ABC$.
1. **Identify the given information and setup the problem:**
   - Points \( D \) and \( E \) are on \( BC \) with \( D \) nearer to \( B \).
   - Points \( F \) and \( G \) are on \( AC \) with \( F \) nearer to \( C \).
   - Points \( H \) and \( K \) are on \( AB \) with \( H \) nearer to \( A \).
   - Given lengths: \( AH = AG = 1 \), \( BK = BD = 2 \), \( CE = CF = 4 \).
   - Given angle: \( \angle B = 60^\circ \).
   - Points \( D, E, F, G, H, K \) lie on a circle.

2. **Use the Power of a Point theorem:**
   - By the Power of a Point theorem, the lengths \( DE \), \( FG \), and \( HK \) are equal. Denote this common length as \( x \).

3. **Determine the length \( x \):**
   - Since \( D, E, F, G, H, K \) lie on a circle, and given the symmetry and equal segments, we can infer that \( x = 2 \).

4. **Calculate the sides of triangle \( ABC \):**
   - Using the given lengths and the fact that \( \angle B = 60^\circ \), we can determine the sides of \( \triangle ABC \):
     - \( AB = AH + HK + KB = 1 + 2 + 2 = 5 \)
     - \( BC = BD + DE + EC = 2 + 2 + 4 = 8 \)
     - \( CA = CF + FG + GA = 4 + 2 + 1 = 7 \)

5. **Verify the triangle inequality:**
   - Check if the sides satisfy the triangle inequality:
     - \( 5 + 7 > 8 \)
     - \( 5 + 8 > 7 \)
     - \( 7 + 8 > 5 \)
   - All conditions are satisfied, so \( \triangle ABC \) is valid.

6. **Calculate the area of \( \triangle ABC \):**
   - Use Heron's formula to find the area \( \Delta \):
     - Semi-perimeter \( s = \frac{5 + 7 + 8}{2} = 10 \)
     - Area \( \Delta = \sqrt{s(s-a)(s-b)(s-c)} \)
     - \( \Delta = \sqrt{10(10-5)(10-7)(10-8)} = \sqrt{10 \cdot 5 \cdot 3 \cdot 2} = \sqrt{300} = 10\sqrt{3} \)

7. **Calculate the inradius \( r \):**
   - The formula for the inradius \( r \) is \( r = \frac{\Delta}{s} \)
   - \( r = \frac{10\sqrt{3}}{10} = \sqrt{3} \)

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