4
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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.

enter image description here

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

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  • $\begingroup$ FYI: sum 25 seems a bit arbitrary (except for 25=5*5), but, for example, sum 18 and sum 30 have less solutions, and may therefore be harder to spot. $\endgroup$ Nov 3 '21 at 16:52
6
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Assuming "sector beams" are the five kite-shaped things joined in the centre one solution would be

                  7


     8        2       5        6

                  11
            4           3
                  1

          9              10
 

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  • $\begingroup$ Any step-by-step solution? $\endgroup$
    – justhalf
    Nov 6 '21 at 6:51
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Complement to loopy walt answer:

There are 8 'distinct' solutions apparently (if I coded well in Julia), each solution having 10 variants (rotation and mirror -wise).

The program runs for any n=numVertices and only provides solutions for n=4 and n=5.

Perhaps, for n=3, 3^2=9 is too small, and, for n=6, 6^2=36 is too big.

code:

using Printf
using Combinatorics

# number of vertices of polygon
numVertices = 5 # e.g. 5 = pentagon

# number of points
# vertices of two polygons and one center point
numPoints = 2 * numVertices + 1

# array of all points involved
Points = collect(1 : numPoints)

# try every point as center
for center in Points

 # total of four values in vertices of each kite-shape
 # taken to be square of number of vertices in the exercise
 totKiteShape = numVertices * numVertices # e.g. 5 * 5 = 25
 @printf("center = %d\n", center)

 # total of three values in vertices of each triangle
 # (all kite-shapes share same center)
 totTriangle = totKiteShape - center

 # split Points minus center in two equal parts
 # which are potential vertices of two polygons
 # inner and outer polygon

 # make indices of polygons go clockwise

 # smaller inner polygon
 
 #     1 2
 #    5   3
 #      4


 # larger outer polygon

 #      1
 #      
 #  5       2
 #  
 #    4   3


 # take out center from copy of Points
 allVertices = filter!(x -> x ≠ center, copy(Points))
 for outerPolygon in combinations(allVertices, numVertices)
  innerPolygon = filter!(y -> y ∉ outerPolygon, copy(allVertices))

  # if we sum up all vertices of all kite-shapes
  # inner polygon are counted twice
  # outer polygon are counted once
  # these two must be equal
  if (2 * sum(innerPolygon) + sum(outerPolygon)) == (numVertices * totTriangle)
   innerPermutations = permutations(innerPolygon)
   outerPermutations = permutations(outerPolygon)

   # go over all permutations of inner polygon and outer polygon
   for ip in innerPermutations
    for op in outerPermutations

     # check if each triangle adds up to requested total
     ok = 1
     for vertex in (1 : numVertices)
      nextvertex = vertex + 1
      # largest and smallest indices are next to one another
      if vertex == numVertices
       nextvertex = 1
      end

      # add up the values and compare with total
      if ((op[vertex] + (ip[vertex] + ip[nextvertex])) ≠ totTriangle)
       ok = 0
       break
      end
     end

     # print the solution
     if ok == 1
      @printf("inner = ")
      for elem in ip
       @printf("%02d ", elem)
      end
      @printf(" ")
      @printf("outer = ")
      for elem in op
       @printf("%02d ", elem)
      end
      @printf("\n")
     end

    end
   end

  end
 end

end

Here is relevant part of output (I marked loopy walt solution with <<--), of course, all 9 others are variants of that solution with center 11 :

center = 4
inner = 06 07 11 09 10  outer = 08 03 01 02 05 
inner = 06 10 09 11 07  outer = 05 02 01 03 08 
inner = 07 06 10 09 11  outer = 08 05 02 01 03 
inner = 07 11 09 10 06  outer = 03 01 02 05 08 
inner = 09 10 06 07 11  outer = 02 05 08 03 01 
inner = 09 11 07 06 10  outer = 01 03 08 05 02 
inner = 10 06 07 11 09  outer = 05 08 03 01 02 
inner = 10 09 11 07 06  outer = 02 01 03 08 05 
inner = 11 07 06 10 09  outer = 03 08 05 02 01 
inner = 11 09 10 06 07  outer = 01 02 05 08 03

inner = 05 09 11 08 10  outer = 07 01 02 03 06 
inner = 05 10 08 11 09  outer = 06 03 02 01 07 
inner = 08 10 05 09 11  outer = 03 06 07 01 02 
inner = 08 11 09 05 10  outer = 02 01 07 06 03 
inner = 09 05 10 08 11  outer = 07 06 03 02 01 
inner = 09 11 08 10 05  outer = 01 02 03 06 07 
inner = 10 05 09 11 08  outer = 06 07 01 02 03 
inner = 10 08 11 09 05  outer = 03 02 01 07 06 
inner = 11 08 10 05 09  outer = 02 03 06 07 01 
inner = 11 09 05 10 08  outer = 01 07 06 03 02 

center = 6
inner = 03 05 10 08 09  outer = 11 04 01 02 07 
inner = 03 09 08 10 05  outer = 07 02 01 04 11 
inner = 05 03 09 08 10  outer = 11 07 02 01 04 
inner = 05 10 08 09 03  outer = 04 01 02 07 11 
inner = 08 09 03 05 10  outer = 02 07 11 04 01 
inner = 08 10 05 03 09  outer = 01 04 11 07 02 
inner = 09 03 05 10 08  outer = 07 11 04 01 02 
inner = 09 08 10 05 03  outer = 02 01 04 11 07 
inner = 10 05 03 09 08  outer = 04 11 07 02 01 
inner = 10 08 09 03 05  outer = 01 02 07 11 04

center = 7
inner = 01 06 10 05 09  outer = 11 02 03 04 08 
inner = 01 09 05 10 06  outer = 08 04 03 02 11 
inner = 05 09 01 06 10  outer = 04 08 11 02 03 
inner = 05 10 06 01 09  outer = 03 02 11 08 04 
inner = 06 01 09 05 10  outer = 11 08 04 03 02 
inner = 06 10 05 09 01  outer = 02 03 04 08 11 
inner = 09 01 06 10 05  outer = 08 11 02 03 04 
inner = 09 05 10 06 01  outer = 04 03 02 11 08 
inner = 10 05 09 01 06  outer = 03 04 08 11 02 
inner = 10 06 01 09 05  outer = 02 11 08 04 03 

center = 8
inner = 02 04 07 09 05  outer = 11 06 01 03 10 
inner = 02 05 09 07 04  outer = 10 03 01 06 11 
inner = 04 02 05 09 07  outer = 11 10 03 01 06 
inner = 04 07 09 05 02  outer = 06 01 03 10 11 
inner = 05 02 04 07 09  outer = 10 11 06 01 03 
inner = 05 09 07 04 02  outer = 03 01 06 11 10 
inner = 07 04 02 05 09  outer = 06 11 10 03 01 
inner = 07 09 05 02 04  outer = 01 03 10 11 06 
inner = 09 05 02 04 07  outer = 03 10 11 06 01 
inner = 09 07 04 02 05  outer = 01 06 11 10 03

inner = 01 05 10 04 07  outer = 11 02 03 06 09 
inner = 01 07 04 10 05  outer = 09 06 03 02 11 
inner = 04 07 01 05 10  outer = 06 09 11 02 03 
inner = 04 10 05 01 07  outer = 03 02 11 09 06 
inner = 05 01 07 04 10  outer = 11 09 06 03 02 
inner = 05 10 04 07 01  outer = 02 03 06 09 11 
inner = 07 01 05 10 04  outer = 09 11 02 03 06 
inner = 07 04 10 05 01  outer = 06 03 02 11 09 
inner = 10 04 07 01 05  outer = 03 06 09 11 02 
inner = 10 05 01 07 04  outer = 02 11 09 06 03

center = 9
inner = 01 07 03 02 10  outer = 08 06 11 04 05 
inner = 01 10 02 03 07  outer = 05 04 11 06 08 
inner = 02 03 07 01 10  outer = 11 06 08 05 04 
inner = 02 10 01 07 03  outer = 04 05 08 06 11 
inner = 03 02 10 01 07  outer = 11 04 05 08 06 
inner = 03 07 01 10 02  outer = 06 08 05 04 11 
inner = 07 01 10 02 03  outer = 08 05 04 11 06 
inner = 07 03 02 10 01  outer = 06 11 04 05 08 
inner = 10 01 07 03 02  outer = 05 08 06 11 04 
inner = 10 02 03 07 01  outer = 04 11 06 08 05 

center = 11
inner = 01 03 05 02 04  outer = 10 06 07 08 09 
inner = 01 04 02 05 03  outer = 09 08 07 06 10 
inner = 02 04 01 03 05  outer = 08 09 10 06 07 
inner = 02 05 03 01 04  outer = 07 06 10 09 08 <<--
inner = 03 01 04 02 05  outer = 10 09 08 07 06 
inner = 03 05 02 04 01  outer = 06 07 08 09 10 
inner = 04 01 03 05 02  outer = 09 10 06 07 08 
inner = 04 02 05 03 01  outer = 08 07 06 10 09 
inner = 05 02 04 01 03  outer = 07 08 09 10 06 
inner = 05 03 01 04 02  outer = 06 10 09 08 07 

FYI Here are 2 'distinct' solutions for n=4 (and sum 16=4*4). (rotated 45 degrees)

6 2 9   8 1 9
7 1 4   5 2 4
5 3 8   6 3 7

Some particular extra equality about the solutions for n=4 is that sum of opposite corners is same (or, if you wish, sum along diagonals is same) as in

6+8=5+9=14 and 8+7=6+9=15

This equality is not too hard to prove.

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