6
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Take an equilateral triangle with dots at the vertices and at the midpoints of each side: Triangle game template

There are a total of six dots.

In the triangle game, you have the numbers 1, 2, 3, 4, 5, 6. Place one number at each dot such that the numbers on each side adds up to the same total.

Here's a solution for side length 9:

Triangle game

Find all the distinct solutions of the triangle game, not including reflections and rotations.

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  • $\begingroup$ Do you count rotations of the same triangle as two different solutions, or as the same solution? $\endgroup$ – Bailey M May 26 '15 at 19:44
  • $\begingroup$ @BaileyM I mentioned in the last sentence: rotations and reflections count as the same solution. $\endgroup$ – mmking May 26 '15 at 19:45
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    $\begingroup$ Perhaps reading the full question would benefit me in the future, eh? $\endgroup$ – Bailey M May 26 '15 at 19:45
  • $\begingroup$ Would both the triangles described here be considered rotations of the example (starting from the top working clockwise): 5,1,6,2,4,3 and 3,5,1,6,2,4? $\endgroup$ – tfitzger May 26 '15 at 21:18
  • $\begingroup$ @tfitzger 3,5,1,6,2,4 would be considered a rotation, but 5,1,6,2,4,6 would not, since it has different side length (side length 12) $\endgroup$ – mmking May 27 '15 at 3:20
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Let $a,b,c$ be the numbers at the corners and $d,e,f$ be the numbers at the midpoints (say $d$ opposite $a$, $e$ opposite $b$, $f$ opposite $c$). To fix the rotation and reflection of a given solution, we can assume $a\leq b\leq c$. Clearly the largest of the three midpoint-numbers must be opposite the largest of the three corner-numbers and so on, so we also have $d\leq e\leq f$.

Let $x=a+b+c$ and $y=d+e+f$, and let $S$ be the sum from each side. We have two equations in $x$ and $y$:

  • $x+y=21$ (the sum of $1,\dots,6$)
  • $2x+y=3S$ (adding together all three copies of $S$ results in counting each corner twice and each midpoint once).

So $x$ and $y$ must both be multiples of $3$. Thus the three numbers $a,b,c$ summing to $x$ must consist of one that is a multiple of 3, one that is one more than a multiple of 3, and one that is one less that a multiple of 3; similarly for $d,e,f$. So the possibilities are:

  • $a=1,b=2,c=3,d=4,e=5,f=6$, giving $S=9$
  • a=1,b=2,c=6,d=3,e=4,f=5, required condition not satisfied
  • $a=1,b=3,c=5,d=2,e=4,f=6$, giving $S=10$
  • a=1,b=5,c=6,d=2,e=3,f=4, required condition not satisfied
  • $a=4,b=5,c=6,d=1,e=2,f=3$, giving $S=12$
  • a=3,b=4,c=5,d=1,e=2,f=6, required condition not satisfied
  • $a=2,b=4,c=6,d=1,e=3,f=5$, giving $S=11$
  • a=2,b=3,c=4,d=1,e=5,f=6, required condition not satisfied
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    $\begingroup$ nice and thorough answer +1 $\endgroup$ – A.D. May 26 '15 at 20:21
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    $\begingroup$ Funny how much your thought process resembles that of my math professor ;) $\endgroup$ – mmking May 27 '15 at 2:20

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