Henry Ernest Dudeney/Puzzles and Curious Problems/157 - Counting the Matches/Solution

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Puzzles and Curious Problems by Henry Ernest Dudeney: $157$

Counting the Matches
A friend bought a box of midget matches, each one inch in length.
He found he could arrange them all in the form of a triangle whose area was just as many square inches as there were matches.
He then used up $6$ of the matches,
and found that with the remainder he could again construct a triangle whose area was just as many square inches as there were matches.
And using another $6$ matches he could again do precisely the same.
How many matches were there in the box originally?
The number is less than $40$.


Solution

First we recall Heron's Formula for the area of a triangle with sides equal to $a$, $b$ and $c$:

$\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$

where $s = \dfrac {a + b + c} 2$ is the semiperimeter of $\triangle ABC$.


With $36$ matches you can make a triangle with sides $17$, $10$, $9$ whose area $\AA$ is calculated by Heron's Formula to be:

\(\ds \AA\) \(=\) \(\ds \sqrt {18 \paren {18 - 17} \paren {18 - 10} \paren {18 - 9} }\)
\(\ds \) \(=\) \(\ds \sqrt {18 \times 1 \times 8 \times 9}\)
\(\ds \) \(=\) \(\ds \sqrt {1296}\)
\(\ds \) \(=\) \(\ds 36\)


With $30$ matches you can make a triangle with sides $13$, $12$, $5$, which is the $5$-$12$-$13$ Pythagorean Triangle whose area $\AA$ is calculated by Area of Triangle in Terms of Side and Altitude to be:

\(\ds \AA\) \(=\) \(\ds \dfrac 1 2 \times 5 \times 12\)
\(\ds \) \(=\) \(\ds 30\)


With $24$ matches you can make a triangle with sides $10$, $8$, $6$, which is the $3$-$4$-$5$ Pythagorean Triangle scaled up by a factor of $2$, whose area $\AA$ is calculated by Area of Triangle in Terms of Side and Altitude to be:

\(\ds \AA\) \(=\) \(\ds \dfrac 1 2 \times 6 \times 8\)
\(\ds \) \(=\) \(\ds 24\)

$\blacksquare$


Sources

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