Star

[cinelli]

In this puzzle you are asked to place the twelve numbers 1, 2, …, 12 in the circles so that

every line of four numbers adds up to 26, and
the six numbers on the points of the star also add up to 26.

I believe there are six essentially different solutions.

Cinelli

[modellingman]

I believe there are six essentially different solutions.

Cinelli

I think you are right.

Very long spoiler...


The solution below first identifies possibilities for the six numbers on the points of the star. There are 7 such possibilities meeting the puzzle's requirement that the six numbers on the points of the star add up to 26. Six numbers for the points of a star can be split in 10 different way into the two sets of three numbers on the points of the two triangles making up the star. This results in 7*10=70 possible triangle-pairs. However, by developing a simple rule which identifies many of the individual triangles which do not meet the requirements of the puzzle, the 70 triangle-pairs are quickly reduced down to 22.

The simple rule is specified in terms of the sum of the three values on the points of the triangle. Development of the simple rule involved attempting to find, for a range of triangles, a solution to the triangle by identifying the 3 pairs of numbers along the sides of the triangle. Many triangles specified in terms of 3 numbers on the 3 points of the triangle have no solution that meets the requirement of the puzzle that the 4 numbers on each side of the triangle add up to 26.

Of the 22 remaining triangle-pairs, 10 contained triangles with no solution (and not previously identified by the simple rule) and 8 were eliminated because the solution(s) to one triangle of the pair utilised a number on a point of the other triangle. That left 4 triangle-pairs.

Of the 4 triangle-pairs, 2 each contain a triangle with 2 distinct solutions. Allowing for these different triangle solutions, increases the number of triangle-pair solution combinations to 6 and it is demonstrated that all 6 provide a solution to the puzzle.

In the discussion that follows, whenever a specific subset of n distinct numbers (2<=n<=6) numbers is drawn from the set {1,2,...,12}, the subset will be represented using round brackets and its elements will be listed in ascending order. In particular, a set of six numbers for the points of the star will be represented by eg (1,2,3,4,5,11), a triangle specified in terms of the 3 numbers on its points by eg (3,6,8) and the pair of additional numbers along the sides of a triangle by eg (5,12).

Part 1 - Points-of-Star Possibilities

The first part of the solution is to find the possibilities for the numbers associated with the points (external vertices) of the puzzle's 6 pointed star.

The puzzle's star has 6 points and the numbers associated with these 6 points are required to add up to 26. A points-of-star possibility, therefore, amounts to 6 distinct numbers drawn from {1,2,...,12} which sum to 26.

A points-of-star possibility must include the number 1. This follows from the observation that smallest sum of 6 distinct numbers without including the number 1 is 2+3+4+5+6+7 = 27.

Next, note that 1+2+3+4+5+6 = 21, so replacing the final 6 with 11 provides the first points-of-star possibility as (1,2,3,4,5,11).

The final pair, (5,11), can be replaced by the pairs (6,10) and (7,9) to give the second and third points-of-star possibilities: (1,2,3,4,6,10) and (1,2,3,4,7,9)

The first points-of-star possibility begins with the sequence 1,2,3,4,5 and is the only points-of-star possibility with this property. The second and third pair of points-of-star possibilities noted with the sequence 1,2,3,4 and have the 5th number not equal to 5.

Continuing this approach, the next set of points-of-star possibilities to be considered should begin with the sequence 1,2,3 and have the 4th, 5th and 6th numbers all greater than 4. Since the six numbers in a possibility sum to 26, these latter 3 numbers must sum to 20. There are only two such triples satisfying this requirement (5,6,9) and (5,7,8) leading to the fourth and fifth points-of-star possibilities as (1,2,3,5,6,9) and (1,2,3,5,7,8).

No points-of-star possibility can begin with the sequence 1,2,3,6 as this would require such a possibility's final pair of numbers to each be 7 or more, distinct and sum to 14 - an impossibility. Similar arguments apply to points-of-star possibilities beginning with a sequence 1,2,3,n where n>6 and, as a result, there are no other points-of-star possibilities beginning with the sequence 1,2,3. Therefore, the fourth and fifth points-of-star possibilities are the only points-of-star possibilities beginning with the sequence 1,2,3.

For points-of-star possibilities beginning with the sequence 1,2 with subsequent numbers all greater than 3, first consider points-of-star possibilities beginning with the sequence 1,2,4. The 4th, 5th and 6th numbers are required to sum to 19 and must each be 5 or more. There is only one triple satisfying this requirement (5,6,8) leading to a sixth points-of-star possibility of (1,2,4,5,6,8). No points-of-star possibility can begin with the sequence 1,2,5 since this would require the possibility's final 3 numbers each to be 6 or more, distinct and sum to 18 - an impossibility. Again, similar arguments apply to points-of-star possibilities beginning with a sequence 1,2,n where n>5 and, as a result, there are no other points-of-star possibilities beginning with the sequence 1,2.

Finally for Part 1, consider the points-of-star possibilities beginning with 1 with subsequent numbers being greater than 2. The smallest sum of six such numbers is 1+3+4+5+6+7=26. This meets the requirement of a points-of-star possibility. Since any different set of six numbers with the required properties (first number is 1 and subsequent numbers are distinct and greater than 2) will necessarily sum to more than 26, the only points-of-star possibility beginning with 1 with subsequent numbers greater than 2 is the seventh points-of-star possibility of (1,3,4,5,6,7).

Summarising, there are seven (and only seven) points-of-star possibilities for the star in the puzzle. These are:

1. (1,2,3,4,5,11)
2. (1,2,3,4,6,10)
3. (1,2,3,4,7,9)
4. (1,2,3,5,6,9)
5. (1,2,3,5,7,8)
6. (1,2,4,5,6,8)
7. (1,3,4,5,6,7)

The puzzle's star comprises two triangles. Therefore, one way of identifying solutions is to allocate the six numbers within a points-of-star possibility to the points (vertices) of the two triangles, creating a triangle-pair. However, since there are 10 ways (see Part 3) of doing this for each points-of-star possibility, this results in 7*10 = 70 possible triangle-pairs. However, not all of these triangle-pairs will provide a valid solution to the puzzle. Therefore, a way of reducing this number of triangle-pairs has been sought. The approach involved identifying triangles, specified in terms of the three numbers at the points of the triangle, which do not meet the requirements of the puzzle that the 4 numbers on each side of the triangle must add up to 26.

Part 2 - Infeasible triangles

Each of the two large triangles in the puzzle has 3 numbers at its points and a further pair of numbers alongside each side making 9 numbers in total. All are required to be distinct. The 4 numbers along each side must sum to 26.

A large triangle can be denoted by (a,b,c) where a,b,c are 3 distinct numbers from a points-of-star possibility with a<b<c. Define Side 1 of the triangle as being the side (edge) of the triangle being between points (vertices) corresponding to the numbers a and b. Similarly define Side 2 in terms of numbers a and c and Side 3 in terms of b and c.

Consider now the triangle (1,2,3). The additional pair of numbers on Side 1 must sum to 23 (since all 4 numbers on the side are required to sum to 26). The only possibility for this pair is (11,12). For Side 2, the additional pair of numbers must sum to 22 and again there is only a single possibility of (10,12). However, this pair cannot be used for Side 2 since the number 12 is already taken by Side 1. Therefore, triangle (1,2,3) has no solutions which satisfy the requirements of the puzzle. Triangle (1,2,3) is an infeasible triangle.

Consider a slightly less simple case, triangle (1,2,6). Again, the additional pair of numbers on Side 1 are (11,12). For Side 2 the required sum is 19 and now there are 3 possibilities for the additional pair on Side 2: (7,12), (8,11) and (9,10). However, the first two of these pairs can be ruled out since they include numbers which are on Side 1, so Side 2's additional pair must be (9,10). For Side 3 the required sum is 18 and the possible additional pairs are therefore (6,12), (7,11) and (8,10). However, none of these pairs can be used on Side 3 since each contains a number taken by either Side 1 or Side 2. Triangle (1,2,6) has no solutions which satisfy the requirements of the puzzle and so is an infeasible triangle.

In a (very) shorthand notation, the analysis of these two triangles can be written as

(1,2,3) Side 1 23 (11,12), Side 2 22 (10,12) Infeasible
(1,2,6) Side 1 23 (11,12), Side 2 19 (7,12) or (8,11) or (9,10), Side 3 18 (6,12) or (7,11) or (8,10) Infeasible

In this notation, strikethrough of a pair ie (p,q), indicates that the pair (p,q) cannot be used on a triangle's Side because one of the numbers p or q (or both) has been used on a lower numbered Side or at a point of the triangle.

It can be shown (and is shown at the end of this post) that for all triangles (a,b,c) where a+b+c<=11 the triangle is an infeasible triangle. This provides a convenient way of quickly ruling out some of the 10 triangle-pairs from each points-of-star possibility. A triangle-pair containing an infeasible triangle cannot form the basis of a solution to the puzzle.

There is just one exception to this rule of 11-or-less:

Triangle (2,4,5) is not infeasible and meets the requirements of the puzzle with Side 1 as (8,12), Side 2 as (9,10) and Side 3 as (6,11)

The converse of the rule does not hold since there are triangles (a,b,c) where a+b+c>11 which are also infeasible eg triangle (1,2,11) is an infeasible triangle because the only possible pair for Side 1 is (11,12) but this cannot be used as the number 11 is associated with a point of the triangle.

Part 3 - Assessment of Points-of-Star Possibilities

The seven points-of-star possibilities identified in Part 1 all contain the number 1. Therefore, a points-of-star possibility can be generally written as (1,a,b,c,d,e) (where, of course, a<b<c<d<e). There are ten different ways that the six numbers within a points-of-star possibility can be allocated to the two triangles within the puzzle to form triangle-pairs. These are listed below.

1. (1,a,b) & (c,d,e)
2. (1,a,c) & (b,d,e)
3. (1,a,d) & (b,c,e)
4. (1,a,e) & (b,c,d)
5. (1,b,c) & (a,d,e)
6. (1,b,d) & (a,c,e)
7. (1,b,e) & (a,c,d)
8. (1,c,d) & (a,b,e)
9. (1,c,e) & (a,b,d)
10. (1,d,e) & (a,c,e)

Points-of-Star Possibility 1 - (1,2,3,4,5,11)

The rule about infeasible triangles in Part 2 can be used to rule out all but 2 of the 10 possible triangle-pairs from this points-of-star possibility. These 2 remaining triangle-pairs are:

(1,2,11) & (3,4,5)
(1,3,11) & (2,4,5)

The first of these triangle-pairs can be easily ruled out. Its first triangle is (1,2,11) and, as noted at the end of Part 2, triangle (1,2,11) is an infeasible triangle.

The second triangle-pair can also be ruled out. Its second triangle is (2,4,5) and is the exception noted for the rule in Part 2, where it is also noted that its sides comprise the three pairs of (8,12), (9,10) and (6,11). However, the number 11 is one of the points of the first triangle of (1,3,11) in the triangle-pair. In the puzzle a number cannot simultaneously be at a point of a triangle and part of a side pair, so it follows that the triangle-pair (1,3,1) & (2,4,5) cannot be part of a solution to the puzzle. This triangle-pair is an infeasible triangle-pair.

The points-of-star possibility 1 of (1,2,3,4,5,11) has no triangle-pairs that can form a solution to the puzzle.

Points-of-Star Possibility 2 - (1,2,3,4,6,10)

The rule in Part 2 can be used to eliminate 8 of the 10 possible triangle-pairs from this points-of-star possibility. The 2 triangle-pairs remaining are:

(1,2,10) & (3,4,6)
(1,3,10) & (2,4,6)

The second of these can be ruled out by considering its first triangle (1,3,10). The pair on Side 1 is required to sum to 22 meaning that it can only be the pair (10,12). However, the number 10 is associated with a point of the triangle so (1,3,10) is an infeasible triangle.

Within the first triangle-pair, for triangle (1,2,10) the pair on Side 1 is required to sum to 23 and can only be (11,12). The pair on Side 2 is required to sum to 15 yielding possible pairs as (3,12), (4,11), (5,10), (6,9) and (7,8). However, all except (7,8) can be eliminated: pairs (3,12), (4,11) and (5,10) are eliminated by use of the numbers 10, 11, 12 either as a point of the triangle or on Side 1; pair (6,9) is eliminated by the use of number 6 as a point of the second triangle in the triangle-pair, triangle (3,4,6). The pair on Side 3 of triangle (1,2,10) is required to sum to 14 yielding possible pairs as (2,12), (3,11), (4,10), (5,9) and (6,8). All can be eliminated except (5,9).

Therefore, the first triangle of the first triangle-pair, triangle (1,2,10), has (11,12) on Side 1, (7,8) on Side 2 and (5,9) on Side 3.

The second triangle of the first triangle-pair, triangle (3,4,6), can be similarly evaluated. In this case, though there are two solutions for the triangle, these being in the shorthand notation developed earlier:

(3,4,6) Side 1 19 (7,12) Side 2 17 (8,9) Side 3 (5,11)
(3,4,6) Side 1 19 (8,11) Side 2 17 (5,12) Side 3 (7,9)

The first triangle-pair of (1,2,10) & (3,4,6) therefore contains two feasible triangles and one of its triangles has two solutions. This triangle-pair and its solutions are considered further in Part 4.

Points-of-Star Possibility 3 - (1,2,3,4,7,9)

The rule in Part 2 can be used to eliminate 6 of the 10 possible triangle-pairs from this points-of-star possibility. The 4 triangle-pairs remaining are:

(1,2,9) & (3,4,7)
(1,3,9) & (2,4,7)
(1,4,7) & (2,3,9)
(1,4,9) & (2,3,7)

Within the first triangle-pair, the first triangle of (1,2,9) has a sole solution of Side 1 as (11,12), Side 2 as (6,10) and Side 3 as (7,8). However the Side 3 pair uses the number 7 which is a point of the second triangle, triangle (3,4,7). The triangle-pair of (1,2,9) & (3,4,7) is an infeasible triangle-pair.

Within the third triangle-pair, triangle (1,4,7) is an infeasible triangle.

Within the fourth triangle-pair, the first triangle of (1,4,9) has a sole solution of Side 1 as (10,11), Side 2 as (7,9) and Side 3 as (5,8). However, the Side 2 pair has the number 7 which is a point of the second triangle (2,3,7). The triangle-pair of (1,4,9) & (2,3,7) is an infeasible triangle-pair.

For the second triangle-pair of (1,3,9) & (2,4,7) each of the two triangles has a sole solution. These solutions are:

(1,3,9) Side 1 22 (10,12), Side 2 16 (5,11), Side 3 14 (6,8)
(2,4,7) Side 1 20 (8,12), Side 2 17 (6,11), Side 3 15 (5,10)

This triangle-pair is considered further in Part 4.

Points-of-Star Possibility 4 - (1,2,3,5,6,9)

The rule in Part 2 can be used to eliminate 7 of the 10 possible triangle-pairs for this points-of-star possibility. The 3 triangle-pairs remaining are listed below.

(1,2,9) & (3,5,6)
(1,3,9) & (2,5,6)
(1,5,6) & (2,3,9)

All 3 triangle-pairs are infeasible triangle-pairs. The first triangle in each triangle-pair has a unique solution but, in each case one of the numbers in the Side pairs is used as a point of the second triangle. For triangle (1,2,9) it is the number 6 with pair (6,10) being used on Side 2; for triangle (1,3,9) both 5 and 6 cause infeasibility with pair (5,11) on Side 2 and (6,8) on Side 3; and, for triangle (1,5,6) it is 9 with pair (9,10) on Side 2.

Points-of-Star Possibility 5 - (1,2,3,5,7,8)

The rule in Part 2 can be used to eliminate 7 of the 10 possible triangle-pairs from this points-of-star possibility. The 3 triangle-pairs remaining are listed below.

(1,3,8) & (2,5,7)
(1,5,7) & (2,3,8)
(1,5,8) & (2,3,7)

Triangles (1,3,8) and (1,5,8) are both infeasible triangles, leaving just triangle-pair (1,5,7) & (2,3,8) from this points-of-star possibility. Triangle (1,5,7) has one solution and triangle (2,3,8) has two solutions. The solutions are listed below.

(1,5,7) Side 1 20 (9,11), Side 2 18 (6,12), Side 3 14 (4,10)
(2,3,8) Side 1 21 (9,12), Side 2 16 (6,10), Side 3 15 (4,11)
(2,3,8) Side 1 21 (10,11), Side 2 16 (4,12), Side 3 15 (6,9)

The triangle-pair of (1,3,8) & (2,5,7) and its triangle solutions are considered further in Part 4.

Points-of-Star Possibility 6 - (1,2,4,5,6,8)

The rule in Part 2 can be used to eliminate 6 of the 10 possible triangle-pairs for this points-of-star possibility. The 4 triangle-pairs remaining are listed below.

(1,4,8) & (2,5,6)
(1,5,6) & (2,4,8)
(1,5,8) & (2,4,6)
(1,6,8) & (2,4,5)

The first triangle-pair has a solution. The second triangle-pair is an infeasible triangle-pair: triangle (1,5,6) has a solution but this involves pair (8,12) on Side 1 and pair (4,11) on Side 3 making the triangle-pair (1,5,6) & (2,4,8) an infeasible triangle-pair through numbers 4 and 8. The third and fourth triangle-pairs are both eliminated because triangles (1,5,8) and (1,6,8) are both infeasible triangles.

For the triangle-pair (1,4,8) & (2,5,6), triangle (1,4,8) has a sole solution and triangle (2,5,6) has two solutions, though one of these would result an infeasible triangle-pair and so can be ignored. The solutions that provide a valid triangle-pair are

(1,4,8) Side 1 21 (9,12), Side 2 17 (7,10), Side 3 14 (3,11)
(2,5,6) Side 1 19 (9,10), Side 2 18 (7,11), Side 3 15 (3,12)

The triangle-pair of (1,4,8) & (2,5,6) is considered further in Part 4.

Points-of-Star Possibility 7 - (1,3,4,5,6,7)

The rule in Part 2 can be used to eliminate 6 of the 10 possible triangle-pairs from this points-of-star possibility. The 4 triangle-pairs remaining are listed below.

(1,4,7) & (3,5,6)
(1,5,6) & (3,4,7)
(1,5,7) & (3,4,6)
(1,6,7) & (3,4,5)

Triangles (1,4,7) and (1,6,7) are both infeasible triangles, ruling out the first and fourth triangle-pairs as part of potential solutions to the puzzle. Triangle (1,5,6) has a single solution but the pair for Side 3 is (4,11) making the second triangle-pair of (1,5,6) & (3,4,7) an infeasible triangle-pair through the number 4. Both triangles in the third triangle-pair have two solutions. However the two solutions for triangle (3,4,6) both utilise the numbers 5 and 7 as part the solution pairs for the triangle's sides meaning that the third triangle-pair of (1,5,7) & (3,4,6) will be an infeasible triangle-pair irrespective of the solution chosen for triangle (3,4,6).

Part 4 - Solutions to the Puzzle

The triangle-pairs not eliminated in Part 3 are listed below. For two of the triangle-pairs, the first and third below, there are two distinct solutions for one of the triangles. In both cases it is the second triangle of the triangle-pair (triangles (3,4,6) and (2,3,8), respectively) that have two solutions and this means that in these cases there are two combinations of solutions applicable to the triangle-pair. For the other two triangle-pairs there is just a single triangle-pair solution combination.

(1,2,10) & (3,4,6) - 2 triangle-pair solution combinations
(1,3,9) & (2,4,7) - 1 triangle-pair solution combination
(1,5,7) & (2,3,8) - 2 triangle-pair solution combinations
(1,4,8) & (2,5,6) - 1 triangle-pair solution combination

The six triangle-pair solution combinations are listed below using the shorthand notation.

1. [(1,2,10) Side 1 (11,12), Side 2 (7,8), Side 3 (5,9)] & [(3,4,6) Side 1 (7,12), Side 2 (8,9), Side 3 (5,11)]
2. [(1,2,10) Side 1 (11,12), Side 2 (7,8), Side 3 (5,9)] & [(3,4,6) Side 1 (8,11), Side 2 (5,12), Side 3 (7,9)]
3. [(1,3,9) Side 1 (10,12), Side 2 (5,11), Side 3 (6,8)] & [(2,4,7) & [(2,4,7) Side 1 (8,12), Side 2 (6,11), Side 3 (5,10)]
4. [(1,5,7) Side 1 (9,11), Side 2 (6,12), Side 3 (4,10,)] & [(2,3,8) Side 1 (9,12), Side 2 (6,10), Side 3 (4,11)]
5. [(1,5,7) Side 1 (9,11), Side 2 (6,12,), Side 3 (4,10)] & [(2,3,8) Side 1 (10,11), Side 2 (4,12), Side 3 (6,9)]
6. [(1,4,8) Side 1 (9,12), Side 2 (7,10), Side 3 (3,11)] & [(2,5,6) Side 1 (9,10), Side 2 (7,11,), Side 3 (3,12)]

Consider the first of these. It can be represented as the two triangles shown below.

..........................................
.........1.................Side 3.........
......../.\.............4---5,11----6.....
Side 1./...\Side 2.......\........./......
...11,12...7,8............\......./.......
...../.......\............7,12..8,9.......
..../.........\......Side 1.\.../.Side 2..
...2----5,9---10.............\./..........
.......Side 3.................3...........
..........................................

If these two triangles are to provide a solution to the puzzle then in both triangles the 6 pairs of numbers showing on the sides of the two triangles must be arranged along the sides and the second triangle must then be placed over the first so that the numbers along the sides coincide. The second triangle may need to be rotated and/or reflected to achieve the required coincidence.

The solution method involves using the two pairs on Sides 1 and 2 of the first triangle. These are (11,12) and (7,8). Taking a number from the first pair and one from the second pair generates four possible pairs these are (7,11), (7,12), (8,11) and (8,12). Of these 4, only 1 matches to a pair on a Side of the second triangle, this is the pair (7,12). In terms of a puzzle solution this means that Side 1 of the second triangle must be the parallel to Side 3 of of the first triangle. It also means that numbers 12 and 7 on Sides 1 and 2 of the first triangle must lie closer to the point with the number 1. This can be shown as

....................
.........1..........
......../.\.........
...A---12--7---B....
....\./.....\./.....
.....11......8......
..../.\...../.\.....
...2---C---D---10...
........\./.........
.........6..........
....................

A and B must each take values of 3 or 4 (and must not take the same value) and C and D values of 5 or 9 (and, again, not take the same value). The side of the second triangle between points 6 and A contains the number 11. This implies A is 4 and C is 5 because the only side pair in the second triangle containing the number 11 is (5,11) on Side 3 between the points with the numbers 4 and 6. Assigning B as 3 and D as 9 completes the solution.

....................
.........1..........
......../.\.........
...4---12--7---3....
....\./.....\./.....
.....11......8......
..../.\...../.\.....
...2---5---9---10...
........\./.........
.........6..........
....................

Note that compared to its initial representation, the second triangle has been transformed by reflection in the axis of symmetry passing through the point with the number 4.

The same solution method can been applied to the remaining 5 triangle-pair solution combinations to get the puzzle solutions shown below.

....................
.........1..........
......../.\.........
...3---11--8---4....
....\./.....\./.....
.....12......7......
..../.\...../.\.....
...2---5---9---10...
........\./.........
.........6..........
....................
....................
.........1..........
......../.\.........
...4---10--5---7....
....\./.....\./.....
.....12......7......
..../.\...../.\.....
...3---8---6---9....
........\./.........
.........2..........
....................
....................
.........1..........
......../.\.........
...3---9---12--2....
....\./.....\./.....
.....11......6......
..../.\...../.\.....
...5---4---10--7....
........\./.........
.........8..........
....................
....................
.........1..........
......../.\.........
...3---9---6---8....
....\./.....\./.....
.....11......12.....
..../.\...../.\.....
...5---10--4---7....
........\./.........
.........2..........
....................
....................
.........1..........
......../.\.........
...5---9---10--2....
....\./.....\./.....
.....12......7......
..../.\...../.\.....
...4---3---11--8....
........\./.........
.........6..........

This completes the puzzle. There are 6 solutions and (sticking my neck out) these are the only solutions.

List of infeasible triangles as noted in Part 2

The list of triangles analysed below is the list of all possible triangles (a,b,c) where the sum a+b+c is between 6 and 11.

a+b+c = 6
(1,2,3) Side 1 23 (11,12), Side 2 22 (10,12) Infeasible

a+b+c = 7
(1,2,4) Side 1 23 (11,12), Side 2 21 (9,12) or (10,11) Infeasible

a+b+c = 8
(1,2,5) Side 1 23 (11,12), Side 2 20 (8,12) or (9,11) Infeasible
(1,3,4) Side 1 22 (10,12), Side 2 21 (9,12) or (10,11) Infeasible

a+b+c = 9
(1,2,6) Side 1 23 (11,12), Side 2 19 (7,12) or (8,11) or (9,10), Side 3 18 (6,12) or (7,11) or (8,10) Infeasible
(1,3,5) Side 1 22 (10,12), Side 2 20 (8,12) or (9,11), Side 3 18 (6,12) or (7,11) or (9,10) Infeasible
(2,3,4) Side 1 21 (9,12) or (10,11), Side 2 20 (8,12) or (9,11) implies
(2,3,4) Side 1 21 (9,12) or (10,11), Side 2 (8,12) or (9,11), Side 3 19 (7,12) or (8,11) or (9,10) Infeasible

a+b+c = 10
(1,2,7) Side 1 23 (11,12), Side 2 18 (6,12) or (7,11) or (8,10), Side 3 17 (5,12) or (6,11) or (7,10) or (8,9) Infeasible
(1,3,6) Side 1 22 (10,12), Side 2 19 (7,12) or (8,11) or (9,10), Side 3 17 (5,12) or (6,11) or (7,10) or (8,9) Infeasible
(1,4,5) Side 1 21 (9,12) or (10,11), Side 2 20 (8,12) or (9,11) implies
(1,4,5) Side 1 21 (9,12) or (10,11), Side 2 20 (8,12) or (9,11), Side 3 17 (5,12) or (6,11) or (7,10) or (8,9) Infeasible
(2,3,5) Side 1 21 (9,12) or (10,11), Side 2 19 (7,12) or (8,11) or (9,10) implies
(2,3,5) [Case 1] Side 1 21 (9,12), Side 2 19 (7,12) or (8,11) or (9,10), Side 3 18 (6,12) or (7,11) or (8,10) Infeasible
(2,3,5) [Case 2] Side 1 21 (10,11), Side 2 19 (7,12) or (8,11) or (9,10), Side 3 18 (6,12) or (7,11) or (8,10) Infeasible

a+b+c = 11
(1,2,8) Side 1 23 (11,12), Side 2 17 (5,12) or (6,11) or (7,10) or (8,9), Side 3 16 (4,12) or (5,11) or [/s](6,10) or (7,9) Infeasible
(1,3,7) Side 1 22 (10,12), Side 2 18 (6,12) or (7,11) or (8,10) Infeasible
(1,4,6) Side 1 21 (9,12) or (10,11), Side 2 19 (7,12) or (8,11) or (9,10) implies
(1,4,6) [Case 1] Side 1 21 (9,12), Side 2 19 (7,12) or (8,11) or (9,10), Side 3 16 (4,12) or (5,11) or (6,10) or (7,9) Infeasible
(1,4,6) [Case 2] Side 1 21 (10,11), Side 2 19 (7,12) or (8,11) or (9,10), Side 3 16 (4,12) or (5,11) or (6,10) or (7,9) Infeasible
(2,3,6) Side 1 21 (9,12) or (10,11), Side 2 18 (6,12) or (7,11) or (8,10) implies
(2,3,6) [Case 1] Side 1 21 (9,12), Side 2 18 (6,12) or (7,11), Side 3 17 (5,12) or (6,11) or (7,10) or (8,9) Infeasible
(2,3,6) [Case 2] Side 1 21 (9,12), Side 2 18 (6,12) or (8,10), Side 3 17 (5,12) or (6,11) or (7,10) or (8,9) Infeasible
(2,3,6) [Case 3] Side 1 21 (10,11), Side 2 18 (6,12) or (7,11) or (8,10) Infeasible
(2,4,5) Side 1 20 (8,12) or (9,11), Side 2 19 (7,12) or (8,11) or (9,10) implies
(2,4,5) [Case 1] Side 1 20 (8,12), Side 2 19 (7,12) or (8,11) or (9,10), Side 3 17 (5,12) or (6,11) or (7,10) or (8,9) Feasible
(2,4,5) [Case 2] Side 1 20 (9,11), Side 2 19 (7,12) or (8,11) or (9,10), Side 3 17 (5,12) or (6,11) or (7,10) or (8,9) Infeasible

With one exception all of these triangles are all infeasible triangles. The exception is triangle (2,4,5) which has solution

(2,4,5) Side 1 20 (8,12), Side 2 19 (9,10), Side 3 17 (6,11)


modellingman

[9873210]

I believe you can shorten your solution significantly by observing that the sums of the vertices of both big triangles must be 13. This is a stronger theorem than the one you use that the sum must be greater than 11. Also the proof is simpler.

Divide the star into three parts,
U - the three corners of the up pointing triangle ;
D - the corners of the down pointing triangle ; and
I - the six remaining interior points.

Then, with loose notation, sum (2*U+I) = 3*26 = sum( 2*D+I) so sum(U) = sum(D) and it's given that sum(D + U) = 26

You could use this to proceed as you did. But shorten many of your cases.

[modellingman]

I believe you can shorten your solution significantly by observing that the sums of the vertices of both big triangles must be 13. This is a stronger theorem than the one you use that the sum must be greater than 11. Also the proof is simpler.

Divide the star into three parts,
U - the three corners of the up pointing triangle ;
D - the corners of the down pointing triangle ; and
I - the six remaining interior points.

Then, with loose notation, sum (2*U+I) = 3*26 = sum( 2*D+I) so sum(U) = sum(D) and it's given that sum(D + U) = 26

You could use this to proceed as you did. But shorten many of your cases.

Your are correct.

I did think after some initial playing around with the points-of-star possibilities that a rule of 13 (which you elegantly derive) might be applicable. However, by that point I became rather blinkered by numerical exploration of triangles and triangle-pairs and failed to step back and look at the whole puzzle algebraically as you have done. The cost, as you observe, is a solution that is much longer than it could have been. A second factor adding to my blinkers was that I found 13 is not a sufficient condition for the triangle point sum when I found that triangle (1,2,10) was infeasible.

The rule of 11 was simply a pragmatic way of eliminating some of the 70 possibilities I knew existed in the framework I was using. It started by looking at the triangle sum of 6 and building up to 7, 8 and so on. I stopped at 11 when I discovered that for the first sum of 12 for triangle (1,2,9) was a soluble triangle.

modellingman

[cinelli]

Congratulations to Modellingman for his explanation of real scholarship.
This problem has some interesting properties.

.        A          
        / \         
   D---I---K---E    
    \ /     \ /     
     G       H      
    / \     / \     
   C---L---J---B
        \ /         
         F

The six rows of four must add to 26 but the sum of the points of the
start, A E B F C D, can be any number from 12 to 27, except 14 and 25,
which are impossible. A+B+C must be the same as D+E+F and the three
diamonds, AGFH, DLBK, CJEI must sum to 26.

Cinelli

[9873210]

The six rows of four must add to 26 but the sum of the points of the
start, A E B F C D, can be any number from 12 to 27, except 14 and 25,
which are impossible. A+B+C must be the same as D+E+F

Cinelli

I may have misunderstood, but
A+B+C = D+E+F would suggest that A+B+C+D+E+F must be even. How can it be say 13 or 27?

[cinelli]

I may have misunderstood, but
A+B+C = D+E+F would suggest that A+B+C+D+E+F must be even. How can it be say 13 or 27?

You are right. Thanks for the correction. I referred to the wrong triangles. It should have been that

A+I+K = F+J+L
D+G+I = B+H+J
C+L+G = E+K+H

Cinelli