Problem 686

We label the unknown numbers in the circles as shown:

[[IMAGE0]]

Since the sum of the numbers along each side of the triangle is $S$ then $$S = a+v+w+b \qquad S = a+x+y+c \qquad S = b+z+c$$ When we add the numbers along each of the three sides of the triangles, we include each of $a$, $b$ and $c$ twice and obtain $$S+S+S = (a+v+w+b)+(a+x+y+c)+(b+z+c) = (a+v+w+b+z+c+y+x) + a+b+c$$ Now the numbers $a,v,w,b,z,c,y,x$ are the numbers $1,2,3,4,5,6,7,8$ in some order.

This means that $a+v+w+b+z+c+y+x=1+2+3+4+5+6+7+8=36$.

Therefore, $$3S = 36 + a+b+c$$ Since $3S$ is a multiple of 3 and $36$ is a multiple of 3, then $a+b+c$ (which equals $3S-36$) must also be a multiple of 3.

Looking at the possible numbers that can go in the circles, the smallest that $a+b+c$ can be is $1+2+3$ or 6, which would make $3S = 36 + 6 = 42$ or $S=14$. $S$ cannot be any smaller than 14 because $a+b+c$ cannot be any smaller than 6 and so $3S$ cannot be any smaller than 42.

Looking at the possible numbers that can go in the circles, the largest that $a+b+c$ can be is $6+7+8$ or 21, which would make $3S = 36 + 21 = 57$ or $S=19$. $S$ cannot be any larger than 19 because $a+b+c$ cannot be any larger than 21 and so $3S$ cannot be any larger than 57.

So which of the values $S=14,15,16,17,18,19$ is actually possible?

The following diagrams show ways of completing the triangle with $S=15,16,17,19$:

[[IMAGE1]]    [[IMAGE2]]    [[IMAGE3]]    [[IMAGE4]]

Coming up with these examples requires a combination of reasoning and fiddling.

For example, consider the case when $S=15$.

Since $3S=36+a+b+c$ and $S=15$, then $a+b+c=3\times 15 - 36 = 9$.

In the given example, we have $a=1$, $b=2$ and $c=6$.

Since the bottom row ($b+z+c$) has the smallest number of circles, we put the largest of $a,b,c$ here ($b=2$ and $c=6$) and then set $z = 15 - b- c = 7$. A bit of fiddling allows us to choose $u,v,x,y$ appropriately to get the desired sums on the two other sides.

We note that there are other possible combinations of $a,b,c$ with $a+b+c=9$ (namely, $1,3,5$ and $2,3,4$). It turns out that neither of these possibilities can produce a triangle with $S=15$.

In a similar way, we can determine examples like those shown with $S=16, 17,19$.

To complete the solution, we show that $S=14$ and $S=18$ are not possible.

Suppose that $S=14$.

In this case, $a+b+c=3S-36 = 3\times 14 - 36 = 6$.

The only integers from the list $1,2,3,4,5,6,7,8$ which give this sum are $1,2,3$.

Consider the bottom row, which should have $b+z+c=14$.

Since $a,b,c$ are $1,2,3$ in some order, then $b+c$ is at most $2+3=5$.

Since the maximum number in the triangle is 8, then $z$ is at most 8.

This makes $b+z+c$ at most $5+8=13$, which means that $b+z+c$ cannot equal 14.

This means that we cannot build a triangle with $S=14$.

Suppose that $S=18$.

In this case, $a+b+c=3S-36 = 3\times 18 - 36 = 18$.

There are several possible sets of values for $a,b,c$: $3,7,8$ and $4,6,8$ and $5,6,7$.

Consider the bottom row again, which should have $b+z+c=18$.

Here we have $a+b+c=18$ and $b+z+c=18$.

Since $b$ and $c$ are common to these sums and the total is the same in each case, then $a=z$, which is not allowed.

This means that we cannot build a triangle with $S=18$.

In summary, the possible values of $S$ are $15,16,17,19$.

The sum of these values is $67$.