Henry Ernest Dudeney/Puzzles and Curious Problems/238 - A Maypole Puzzle/Solution

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Puzzles and Curious Problems by Henry Ernest Dudeney: $238$

A Maypole Puzzle
During a gale a maypole was broken in such a manner that it struck the level ground at a distance of $20$ feet from the base of the pole,
where it entered the earth.
It was repaired, and broken by the wind a second time at a point $5$ feet lower down,
and struck the ground at a distance of $30$ feet from the base.
What was the original height of the pole?


Solution

$50$ feet.


Proof

Let $L$ feet be the length of the maypole.

Let $d$ feet be the distance above the ground of the point where first it snapped.

The maypole, the snapped-off part, and the ground form a right triangle.

Hence we can use Pythagoras's Theorem to obtain:

\(\text {(1)}: \quad\) \(\ds \paren {L - d}^2\) \(=\) \(\ds d^2 + 20^2\) ... it struck the level ground at a distance of $20$ feet from the base of the pole,
\(\text {(2)}: \quad\) \(\ds \paren {L - \paren {d - 5} }^2\) \(=\) \(\ds \paren {d - 5}^2 + 30^2\) ... second time at a point $5$ feet lower down, and struck the ground at a distance of $30$ feet from the base.
\(\ds \leadsto \ \ \) \(\ds L^2 - 2 L d + d^2\) \(=\) \(\ds d^2 + 400\) expanding
\(\ds L^2 - 2 L \paren {d - 5} + \paren {d - 5}^2\) \(=\) \(\ds \paren {d - 5}^2 + 900\)
\(\text {(3)}: \quad\) \(\ds \leadsto \ \ \) \(\ds L^2 - 2 L d\) \(=\) \(\ds 400\) simplifying
\(\text {(4)}: \quad\) \(\ds L^2 - 2 L d + 10 L\) \(=\) \(\ds 900\)
\(\ds \leadsto \ \ \) \(\ds 10 L\) \(=\) \(\ds 500\) $(4) - (3)$
\(\ds \leadsto \ \ \) \(\ds L\) \(=\) \(\ds 50\)

$\blacksquare$


Sources

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