Henry Ernest Dudeney/Puzzles and Curious Problems/231 - The Rose Garden/Solution

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Puzzles and Curious Problems by Henry Ernest Dudeney: $231$

The Rose Garden
A man has a rectangular garden, and wants to make exactly half of it into a large bed of roses,
with a gravel path of uniform width round it.
Can you find a general rule that will apply equally to any rectangular garden, whatever its proportions?
All the measurements must be made in the garden.
A plain ribbon, no shorter than the length of the garden, is all the material required.


Solution

Dudeney-Puzzles-and-Curious-Problems-231-solution.png

Construct $AD$ one quarter the length of $AB$.

Construct $AF$ and $DE$ one quarter the length of $BC$.

Construct $EG = DF$.

Then $AG$ is the required width of the path.


Proof

Let the length and breadth of the garden be $a$ and $b$.

Let $c$ be the width of the path.

We have:

\(\ds \paren {a - 2 c} \paren {b - 2 c}\) \(=\) \(\ds \dfrac {a b} 2\)
\(\ds \leadsto \ \ \) \(\ds 2 \paren {a b - 2 a c - 2 b c + 4 c^2}\) \(=\) \(\ds a b\) multiplying out and clearing fractions
\(\ds \leadsto \ \ \) \(\ds 8 c^2 - 4 \paren {a + b} c + a b\) \(=\) \(\ds 0\) simplifying
\(\ds \leadsto \ \ \) \(\ds c\) \(=\) \(\ds \dfrac {4 \paren {a + b} \pm \sqrt {\paren {4 \paren {a + b} }^2 - 4 \times 8 \times a b} } {2 \times 8}\) Quadratic Formula
\(\ds \) \(=\) \(\ds \dfrac {a + b \pm \sqrt {a^2 + b^2} } 4\) simplifying

It is apparent that it is the negative square root of $\sqrt {a^2 + b^2}$ we need in the above, as the positive one would result in the path being wider than the garden.

Hence:

$c = \dfrac {a + b - \sqrt {a^2 + b^2} } 4$

$\Box$


Now we investigate the geometry.

Let $a = AB$ and $b = BC$.

We have that:

\(\ds AD\) \(=\) \(\ds \dfrac a 4\)
\(\ds AF\) \(=\) \(\ds \dfrac b 4\)
\(\ds AE\) \(=\) \(\ds \dfrac a 4 + \dfrac b 4 = \dfrac {a + b} 4\)
\(\ds \leadsto \ \ \) \(\ds AG\) \(=\) \(\ds AE - EG\)
\(\ds \leadsto \ \ \) \(\ds EG = DF\) \(=\) \(\ds \sqrt {\paren {\dfrac a 4}^2 + \paren {\dfrac b 4}^2}\) Pythagoras's Theorem
\(\ds \leadsto \ \ \) \(\ds EG = DF\) \(=\) \(\ds \dfrac {\sqrt {a^2 + b^2} } 4\) simplifying
\(\ds \leadsto \ \ \) \(\ds AG\) \(=\) \(\ds \dfrac {\sqrt {a^2 + b^2} } 4\) simplifying
\(\ds \leadsto \ \ \) \(\ds AG\) \(=\) \(\ds AE - EG\)
\(\ds \) \(=\) \(\ds \dfrac {a + b} 4 - \dfrac {\sqrt {a^2 + b^2} } 4\)

and it is seen that $AG$ is the same as what was calculated algebraically above.

$\blacksquare$


Sources

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