Henry Ernest Dudeney/Modern Puzzles/176 - The Five-Pointed Star/Solution

by : $176$

 * The Five-Pointed Star

Solution
's unsatisfyingly unstructured and prolix answer is given below.

A full and coherent analysis is welcomed.


 * Referring to Figure $\text I$, we will call $A$, $B$, $C$, $D$, $E$ the "pentagon", and $F$, $G$, $H$, $J$, $K$ the "points".
 * Write in the numbers $1$, $2$, $3$, $4$, $5$ in the pentagon in the order shown in Figure $\text {II}$, where you go round in a clockwise direction,
 * starting with $1$ and jumping over a disc to the place for $2$, jumping over another for $3$, and so on.
 * Now to complete the star for the constant summation of $24$, as required, use this simple rule.
 * To find $H$ subtract the sum of $B$ and $C$ from half the constant plus $E$.
 * That is, subtract $6$ from $15$.
 * We thus get $9$ as the required number for $H$.
 * Now you are able to write in successively $10$ at $F$ (to make $24$), $6$ at $J$, $12$ at $G$, and $8$ at $K$.
 * There is your solution.


 * Dudeney-Modern-Puzzles-176-solution.png


 * You can write any five numbers you like in the pentagon, in any order, and with any constant summation you wish,
 * and you will always get, by the rule shown, the only possible solution for that pentagon and constant.
 * But that solution may require the use of repeated numbers and even negative numbers.
 * Suppose, for example, I make the pentagon $1$, $3$, $11$, $7$, $4$, and the constant $26$, as in Figure $\text {III}$,
 * then I shall find the $3$ is repeated, and the repeated $4$ is negative and must be deducted instead of added.
 * You will also find that if we had written our pentagon numbers in Figure $\text {II}$ in any other order we should always get repeated numbers.


 * Let us confine our attention to solutions with ten different positive whole numbers.
 * Then $24$ is the smallest possible constant.
 * A solution for any higher constant can be derived from it.
 * Thus, if we want $26$, add $1$ at each of the points;
 * if we want $28$, add $2$ at every point or $1$ at every place in both points and pentagon.
 * Odd constants are impossible unless we use fractions.
 * Every solution can be "turned inside out".
 * Thus Figure $\text {IV}$ is simply a different arrangement of Figure $\text {II}$.
 * Also the four numbers in $G$, $K$, $D$, $J$ may always be changed, if repetitions do not occur.
 * For example, in Figure $\text {II}$ substitute $13$, $7$, $6$, $5$ for $12$, $8$, $5$, $6$ respectively.
 * Finally, in any solution the constant will be two-fifths of the sum of all the ten numbers.
 * So, if we are given a particular set of numbers we at once know the constant,
 * and for any constant we can determine the sum of the numbers to be used.