Triangle Sudoku

The Setup

Someone recently shared a fun math puzzle with me. It doesn’t require immense math skills, but its quite satisfying when everything works out. Credit for the puzzle goes to Joachim Laukenmann, who regularly posts “Zahlendreher” (German for “number twister”), math puzzles from ETH Zürich. They are regularly posted in the Tages Anzeiger, a Swiss newspaper. Click here to see the original post on this puzzle.

The puzzle goes as follows. You have this setup of triangles:

The task is simple:

1. Distribute all the numbers from $1$ to $8$ in the circles, using each number exactly once.
2. When adding the three numbers around a gray triangle, the sum should be $14$.
3. Adding the numbers around a white triangle should result in $13$.

Curious to try it youself? Should be solveable in 15 minutes.

The solution(s)

The first important insight stems from looking at two of the triangles next to each other. We will use arbitrary numbers $a,b,c$ and $d$ (yes, yes, I know, technically they are letters) and assume they satisfy the requirements number 2 and 3.

Requirement 2 and 3 tell us that the sum in the grey triangle is $14$, the sum in the white triangle $13$. We write this down in equation form below.

$$\begin{align} a + b + c &= 13 \\ a + b + d &= 14 \end{align}$$

Subtracting the first from the second gives us $d - c = 1$, or equivalently $d = c + 1$. This simplifies things a lot. We have four corners to fill, with just three unkowns $a,b$ and $c$. Lets start applying this pattern. We do this by setting the corners of the upper middle triangle to $a,b,c$ and $c+1$, as depicted below.

From there on it is very easy to continue. For each triangle pair, if one side is complete, we fill in the opposite, missing corner. Doing this we get the following.

Now we apply the requirement number 1. All of the numbers need to be distinct, thus the numbers from $1$ to $8$ need to be turned into three non-overlapping and consecutive groups.

$$\begin{array}{c} \{ a-1, a, a+1 \} \\ \{ b, b+1 \} \\ \{ c-1, c, c+1 \} \end{array}$$

Further, we use the fact that since $a,b$ and $c$ sit on a white triangle, their sum must be $13$. There is 6 combinations on how to arrange the groups on the range from $1$ to $8$, only one of which satisfies $a+b+c = 13$. Either by manually checking, or by some more reasoning about parity and the group sizes, one arrives at the solution $a=2$, $b = 4$ and $c = 7$. Plugging in the numbers into our triangle, we can check that this is indeed a solution.