Triangle Game

Label the circles, starting at the top circle and going clockwise, $(a, b, c, d, e, f).$

We wish to find all integer sums $n$ such that

$n = a + b + c = c + d + e = e + f + a.$

Now $3n = (a + c + e) + (a + b + c + d + e + f) = (a + c + e) + 21.$

Thus we are looking for $n$ such that $n = 7 + \dfrac{a + c + e}{3}.$

Now $a + c + e$ can take on any value from $1 + 2 + 3 = 6$ to $4 + 5 + 6 = 15$ but what we require is that $3 | (a + c + e)$ in order for $n$ to be an integer. So at most we can have $4$ possible sums, corresponding to $a + c + e$ equalling $6, 9, 12$ and $15$ , giving us possible values for $n$ of $9, 10, 11$ and $12.$

We then need to confirm that these values are indeed possible, which they are using the following configurations for $(a,b,c,d,e,f)$ :

• $(1,5,3,4,2,6) \Longrightarrow n = 9$ ,

• $(1,4,5,2,3,6) \Longrightarrow n = 10$ ,

• $(2,3,6,1,4,5) \Longrightarrow n = 11$ ,

• $(6,2,4,3,5,1) \Longrightarrow n = 12$ .

This confirms that there are indeed $\boxed{4}$ possible sums.

Python 2.7:

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9

from itertools import permutations
found_sums = set()
for combo in permutations(xrange(1,7)):
    side_1 = sum(combo[0:3])
    side_2 = sum(combo[2:5])
    side_3 = sum(combo[4:6]+(combo[0],))
    if side_1 == side_2 and side_1 == side_3:
        found_sums.add(side_1)
print "Answer:", len(found_sums)

the sequences are: 1 5 6 3 4 2 sum equals to 9 1 4 6 5 2 3 sum equals to 10 2 3 5 6 1 4 sum equals to 11 4 2 3 6 1 5 sum equals to 12