Henry Ernest Dudeney/Modern Puzzles/177 - The Six-Pointed Star/Solution

Modern Puzzles by Henry Ernest Dudeney: $177$

The Six-Pointed Star

In this case we can always use the twelve consecutive numbers $1$ to $12$ and the sum of the four numbers in every line will always be $26$.

The numbers at the six points of the star may add up to any even number from $24$ to $54$ inclusive, except $28$ and $50$, which are impossible.

It will be seen in the example that the six points add up to $24$.

If for every number in its present position you substitute its difference from $13$ you will get another solution, its complementary,

with the points adding up to $54$, which is $78$ less $24$.

The two complementary totals will always sum to $78$.

I will give the total number of different solutions and point out some of the pretty laws which govern the problem,

but I will leave the reader this puzzle to solve.

There are six arrangements, and six only, in which all the lines of four and the six points also add up to $26$.

Can you find one or all of them?

Solution

Dudeney's solution is as follows:

$(1): \quad$ In every solution, the sum of the numbers in triangle $ABC$ of Figure $\text I$ must equal the sum of the numbers in the triangle $DEF$.

This can be anywhere from $12$ to $27$ inclusive, except $14$ and $25$, which are impossible.

We need only obtain solutions for $12$, $13$, $15$, $16$, $17$, $18$ and $19$, because those of the complementaries may be derived by substituting for every number its difference from $13$.

$(2): \quad$ Every arrangement is composed of three independent diamonds $AGHF$, $DKBL$ and $EMCI$, each of which must add up to $26$.

$(3): \quad$ The sum of the numbers in opposite external triangles will always be equal, for example, $AIK$ equals $LMF$.

$(4): \quad$ If the difference between $26$ and its triangle sum $ABC$ is added to any number at a point, say $A$, it will give the sum of the two numbers in the relative positions of $L$ and $M$.

Thus for example in Figure $\text {II}$: $10 + 13 = 11 + 12$, and $6 + 13 = 8 + 11$.

$(5): \quad$ There are $6$ pairs summing the $13$, which are: $12 + 1$, $11 + 2$, $10 + 3$, $9 + 4$, $8 + 5$, $7 + 6$.

One or two such pairs may occur among the numbers at the points, but never three.

The relative positions of these pairs determine the type of solution.

In the regular type, as in Figure $\text {II}$, $A$ and $F$ and also $G$ and $H$, as indicated by the dotted lines, always sum to $13$, but this class can be subdivided.

Figures $\text {III}$ and $\text {IV}$ are examples of the two irregular types.

$\qquad$

There are $37$ solutions in all, or $74$ if we count complementaries.

$32$ of these are regular, and $5$ are irregular.

Of the $37$ solutions, $6$ have their points summing to $26$. These are as follows:

$\begin {array} {rrrrrrrrrrrr}

10 & 6 & 2 & 3 & 1 & 4 & 7 & 9 & 5 & 12 & 11 & 8 \\ 9 & 7 & 1 & 4 & 3 & 2 & 6 & 11 & 5 & 10 & 12 & 8 \\ 5 & 4 & 6 & 8 & 2 & 1 & 9 & 12 & 3 & 11 & 7 & 10 \\ 5 & 2 & 7 & 8 & 1 & 3 & 11 & 10 & 4 & 12 & 6 & 9 \\ 10 & 3 & 1 & 4 & 2 & 6 & 9 & 8 & 7 & 12 & 11 & 5 \\ 8 & 5 & 3 & 1 & 2 & 7 & 10 & 4 & 11 & 9 & 12 & 6 \\ \end {array}$

The first of these is Figure $\text {II}$, and the last but one is Figure $\text {III}$.

A reference to those diagrams shows how to write out the star.

The first four are of the regular type and the last two are irregular.

Also note that where the $6$ points add to $24$, $26$, $30$, $32$, $34$, $36$ or $38$, the respective number of solutions is $3$, $6$, $2$, $4$, $7$, $6$ and $9$, making $37$ in all.

There are also another $6$ solutions which Dudeney did not find.

Historical Note

Martin Gardner reported this problem in his Mathematical Games column in Scientific American in the December $1965$ issue.

In correspondence, he received confirmation from E.J. Ulrich and A. Domergue that there are in fact $80$ patterns for this six-pointed star, rather than the $74$ offered by Dudeney.

Sources

• 1926: Henry Ernest Dudeney: Modern Puzzles ... (previous) ... (next): Solutions: $177$. -- The Six-Pointed Star
• 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $394$. The Six-Pointed Star