Henry Ernest Dudeney/Modern Puzzles/137 - Hurdles and Sheep/Solution

From ProofWiki
Jump to navigation Jump to search

Modern Puzzles by Henry Ernest Dudeney: $137$

Hurdles and Sheep
A farmer says that four of his hurdles will form a square enclosure just sufficient for one sheep.
That being so, what is the smallest number of hurdles that he will require for enclosing ten sheep?


Solution

$12$ hurdles.


Proof

From the problem definition, an area of $1$ hurdle length squared holds $1$ sheep.

Hence the problem is to find the largest area enclosed by a given number of hurdles.

This is that enclosed by a regular polygon.


This article, or a section of it, needs explaining.
In particular: Prove the above.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Explain}} from the code.


Let $A_n$ be the area enclosed by a regular $n$-gon.

From Area of Regular Polygon:

$A_n = \dfrac n 4 \cot \dfrac \pi n$

Hence:

\(\ds A_{11}\) \(=\) \(\ds \dfrac {11} 4 \cot \dfrac \pi {11}\)
\(\ds \) \(\approx\) \(\ds 9.37\)
\(\ds A_{12}\) \(=\) \(\ds \dfrac {12} 4 \cot \dfrac \pi {12}\)
\(\ds \) \(\approx\) \(\ds 11.2\)

Hence the result.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Henry_Ernest_Dudeney/Modern_Puzzles/137_-_Hurdles_and_Sheep/Solution&oldid=560781"