The Mystic Triangle

The sum of the digits from $0$ to $9$ is $45$ .

The three corner triangle sums + center number = $45$ . Since each triangle sum is equal and $45$ is divisible by $3$ , center number can only be $0, 3, 6, 9$

$(A)$ If center vertex is $0$ the sum of each triangle is $15$ . Now the middle three triangles that share the center number each must sum up to $15$ . But we don't have three sets of $2$ numbers to sum up to $15$ in the middle triangles. Only $(7,8)$ and $(9,6)$ . So center vertex cant be $0$ .

$(B)$ If center vertex is $3$ , the sum of each triangle is $14$ . The middle three triangles that share the center number can have three sets of two numbers that sum up to $11$ viz $(5, 6)$ , $(4, 7)$ and $(9, 2)$ . So $3$ can be center vertex. The numbers not used $0, 1, 8,$ are our corner vertices.

$(C)$ If center vertex is $6$ , the sum of each triangle is $13$ . The middle three triangles that share the center number can have three sets of two numbers that sum up to $7$ viz $(0, 7)$ , $(2, 5)$ and $(3, 4)$ . So $6$ can be center vertex. The numbers not used $1, 8, 9$ are our corner vertices.

$(D)$ If center vertex is $9$ the sum of each triangle is $12$ . Now the middle three triangles that share the center number each must sum up to $12$ . But we don't have three sets of $2$ numbers to sum up to $3$ in the middle triangles. Only $(0,3)$ and $(2,1)$ . So center vertex cant be $9$ .

Our answer is $18$ .

Here is a minizinc model to solve this with a legendary solver

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include "globals.mzn";

int: n = 9;

array[1..n+1] of var 0..n: A;

solve satisfy;

constraint
  A[1] + A[2] + A[5] = A[2] + A[3] + A[6]
  /\
  A[2] + A[3] + A[6] = A[3] + A[4] + A[7]
  /\
  A[3] + A[4] + A[7] = A[5] + A[6] + A[8]
  /\
  A[5] + A[6] + A[8] = A[6] + A[7] + A[9]
  /\
  A[6] + A[7] + A[9] = A[8] + A[9] + A[10]
;

constraint
  alldifferent ([A[i]|i in 1..n+1])
;

output [show(A)];

Run with ./mzn-g12fd --all-solutions /tmp/mystic.mzn

Refer to my (in progress) wiki on discrete probability distribution - uniform distribution , to know what this means

Let $x,y,z,T$ be the center value, the sum of the hexagon ring, the sum of the triangle corners, and the mutual triangle sum, respectively.

Note the following relationships:

$0+1+\dots + 4 + 5 = 15 \leq y \leq 39 = 4 + 5 + \dots + 8 + 9$

$3x + 2y + z = 6T$ and $3x + y = 3T$ imply that $z = 3x$

$x + y + z = 45$ then implies that $4x + y = 45$

From the equations above, derive that $x \equiv y \equiv z \equiv 0\mod 3$ and $y \equiv 1\mod 4$ .

Therefore, the possibilities for $x,y,z$ are $6,21, 18$ and $3, 33,9$ , respectively. A quick check indicates that both are indeed possible.

It is easy to see that total (T) of all numbers of vertices of the graph is divisible by 6. But this total in the range [65..96]. Try some values: 66, 72, 78, 84, 90, 96. But this total minus total of number of the range: [0..9] is: total of number of three inner triangles-number in the center C. $T-45=3 \times \frac{T}{6} - C = \frac{T}{2}-C$ $C=45-\frac{T}{2}$ Try more conditions to get C=6 when T=78 or C=3 by T=84. $Result = \boxed{18}$