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

by : $177$

 * The Six-Pointed Star

Solution
's solution is as follows:


 * Dudeney-Modern-Puzzles-177-solution-1.png


 * $(1): \quad$ 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 $6 + 13 = 8 + 11$.


 * Dudeney-Modern-Puzzles-177-solution-2.png


 * $(5): \quad$ 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.


 * Dudeney-Modern-Puzzles-177-solution-3.png $\qquad$ Dudeney-Modern-Puzzles-177-solution-4.png


 * 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 did not find.