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
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
- 1932: Henry Ernest Dudeney: Puzzles and Curious Problems ... (previous) ... (next): Solutions: $231$. -- The Rose Garden
- 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $279$. The Rose Garden