In the triangle, each of the numbers $1,2,3,4,5,6,7,8$ is placed into a different circle. The sums of the numbers on each of the three sides of the triangle are equal to the same number, $S$. The sum of all of the different possible values of $S$ is
(A) 85
(B) 99
(C) 66
(D) 81
(E) 67

![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-026.jpg?height=393&width=347&top_left_y=1798&top_left_x=1323)
We label the unknown numbers in the circles as shown:

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-113.jpg?height=452&width=393&top_left_y=1346&top_left_x=926)

Since the sum of the numbers along each side of the triangle is $S$ then

$$
S=a+v+w+b \quad S=a+x+y+c \quad S=b+z+c
$$

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,

$$
3 S=36+a+b+c
$$

Since $3 S$ is a multiple of 3 and 36 is a multiple of 3 , then $a+b+c$ (which equals $3 S-36$ ) 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 $3 S=36+6=42$ or $S=14 . S$ cannot be any smaller than 14 because $a+b+c$ cannnot be any smaller than 6 and so $3 S$ 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 $3 S=36+21=57$ or $S=19$. $S$ cannot be any larger than 19 because $a+b+c$ cannot be any larger than 21 and so $3 S$ 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 $S=15,16,17,19$ :
![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-114.jpg?height=280&width=1042&top_left_y=558&top_left_x=602)

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

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

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

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

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 .

ANswer: (E)