(related to Problem: A New Match Puzzle)
- The easiest way is to arrange the eighteen matches as in Diagrams $1$ and $2,$ making the length of the perpendicular $AB$ equal to a match and a half. Then, if the matches are an inch in length, Fig. $1$ contains two square inches and Fig. $2$ contains six square inches — $4 \times 1\frac 12.$
- The second case is a little more difficult to solve. The solution is given in Figs. $3$ and $4.$ For the purpose of construction, place matches temporarily on the dotted lines. Then it will be seen that as $3$ contains five equal equilateral triangles and $4$ contains fifteen similar triangles, one figure is three times as large as the other, and exactly eighteen matches are used.
Thank you to the contributors under CC BY-SA 4.0!
- Dudeney, H. E.: "Amusements in Mathematics", The Authors' Club, 1917
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