Triangle ABC: AB=1862; BC=2753; AC=1943. When placed around the perimeter as shown the numbers add up to 17 along each side. The vertices: A=1; B=2; C=3. The non-vertex numbers can be interchanged: example AB=1682; BC=2573; AC=1493. So there are 8 arrangements of these central pairs.
The sum of the numbers 1 to 9 = 45. If we label the perimeter numbers a to i, then we add together all the numbers on the sides we get 2a+b+c+2d+e+f+2g+h+i because the numbers on the vertices, a, d and g are each used for two sides. So we have 45+a+d+g as the sum total of each side, and dividing by 3 gives us the sum of the four numbers on each side separately. The minimum value of a+d+g=6 (when a=1, b=2 and c=3. See first paragraph). So we have 45+3=51 for the three sides and each side adds up to 51/3=17 (e.g., 2+5+7+3=17 in first paragraph).
The next number after 51 would need to be 54 so that the sides total 18 and a+d+g=9. This tells us that (a,d,g)=(1,2,6), (1,3,5), or (2,3,4). For (1,2,6), a+b+c+d=18; b+c=18-3=15. Also, e+f=10 and h+i=11 because we know the values of a, d and g. The two other alternatives are (1,3,5): b+c=14; e+f=10; h+i=11; (2,3,4): b+c=13; e+f=11; h+i=12.
The trick is now to find values for b, c, e, f, h and i amongst the remainimg digits: 3, 4, 5, 7, 8 and 9. 8+7=15; but we can't make 10 and 11 out of the remaining digits. So (1,2,6) has no solution. (1,3,5) is slightly better because we can get 14 and 10, but not 11. (2,3,4): 8+5=13; but no others for 11 and 12.
Having eliminated these, we move to 57 and a side total of 19. We have a larger group of possibilities for a, d and g: (1,2,9), (1,3,8), (1,4,7), (1,5,6), (2,3,7), (2,4,6), (3,4,5). This time the sums of the pairs are 16, 8 and 9; 15, 8 and 9; 14, 8 and 10;
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