Sum in Magic star puzzle

Question

I have the following problem:

Place the first 11 natural numbers in the circles so that the sum of the four numbers at the tops of each of the five sectors-beams of the star equals 25.

I came up with the fact that $6$ should be central number as the sum of numbers from 1 to 11 is 66. But how should I distribute all threes of sum=19 - i don't know. Would appreciate any help.

Source of the contest link

Answer 1

Let the central number be $C$, sum of five numbers in inner (inverted) pentagon be $P$ and sum of five numbers in outer pentagon be $Q$. Each sector has sum $25$, we can sum all five sectors to get a sum of $5 \times 25 =125$. In this total sum, numbers in outer pentagon are present only once, numbers in inner pentagon twice and the central number five times. $$5C+2P+Q=125$$ Subtracting $C+P+Q=66$ from this gives $4C+P=59$.

Choosing $C=11$ gives $P=15, Q=40$. Since $15=1+2+3+4+5$ and $40=6+7+8+9+10$, very first choice looks promising.

It makes sense to fill outer pentagon first since numbers appear only once.

One sector can be

   6 
 /   \ 

 \   / 
   11

Remaining two numbers on this sector must be $5,3$. Continuing on and playing around a bit, I got

              6
         3        5     
  10         11         7
        1          2
             4
    9                8

which satisfies given conditions.