4. For APPLICANTS IN $\left\{ \begin{array} { l } \text { MATHEMATICS } \\ \text { MATHEMATICS \& STATISTICS } \\ \text { MATHEMATICS \& PHILOSOPHY } \end{array} \right\}$ ONLY.
Mathematics \& Computer Science, Computer Science and Computer Science \& Philosophy applicants should turn to page 14. A horse is attached by a rope to the corner of a square field of side length 1 . (i) What length of rope allows the horse to reach precisely half the area of the field? Another horse is placed in the field, attached to the corner diagonally opposite from the first horse. Each horse has a length of rope such that each can reach half the field. (ii) Explain why the area that both can reach is the same as the area neither can reach. [Figure] (iii) The angle $\alpha$ is marked in the diagram above. Show that $\alpha = \cos ^ { - 1 } \left( \frac { \sqrt { \pi } } { 2 } \right)$ and hence show that the area neither can reach is $\frac { 4 } { \pi } \cos ^ { - 1 } \left( \frac { \sqrt { \pi } } { 2 } \right) - \sqrt { \frac { 4 - \pi } { \pi } }$. Note that $\cos ^ { - 1 }$ can also be written as arccos. A third horse is placed in the field, and the three horses are rearranged. One horse is now attached to the midpoint of the bottom side of the field, and another horse is now attached to the midpoint of the left side of the field. The third horse is attached to the upper right corner. (iv) Given each horse can access an equal area of the field and that none of the areas overlap, what length of rope must each horse have to minimise the area that no horse can reach? The horses on the bottom and left midpoints of the field are each replaced by a goat; each goat is attached by a rope of length $g$ to the same midpoint as in part (iii). The remaining horse is attached to the upper right corner with rope length $h$. (v) Given that $0 \leqslant h \leqslant 1$, and that none of the animals' areas can overlap, show that $\frac { \sqrt { 5 } - 2 } { 2 } \leqslant g \leqslant \frac { 1 } { 2 \sqrt { 2 } }$ holds if the area that the animals can reach is maximised.
(i) [2 marks] We require that the area of the sector is $\frac { r ^ { 2 } \theta } { 2 } = \frac { 1 } { 2 } \cdot \theta = \frac { \pi } { 2 }$ as a quarter circle is formed. Hence the length of rope required is $\sqrt { \frac { 2 } { \pi } }$. [0pt]
\section*{4. For APPLICANTS IN $\left\{ \begin{array} { l } \text { MATHEMATICS } \\ \text { MATHEMATICS \& STATISTICS } \\ \text { MATHEMATICS \& PHILOSOPHY } \end{array} \right\}$ ONLY.}
Mathematics \& Computer Science, Computer Science and Computer Science \& Philosophy applicants should turn to page 14.
A horse is attached by a rope to the corner of a square field of side length 1 .\\
(i) What length of rope allows the horse to reach precisely half the area of the field?
Another horse is placed in the field, attached to the corner diagonally opposite from the first horse. Each horse has a length of rope such that each can reach half the field.\\
(ii) Explain why the area that both can reach is the same as the area neither can reach.\\
\includegraphics[max width=\textwidth, alt={}, center]{4f9b4c7b-cd2d-4862-9649-23a1574ede43-14_411_412_1078_820}\\
(iii) The angle $\alpha$ is marked in the diagram above. Show that $\alpha = \cos ^ { - 1 } \left( \frac { \sqrt { \pi } } { 2 } \right)$ and hence show that the area neither can reach is $\frac { 4 } { \pi } \cos ^ { - 1 } \left( \frac { \sqrt { \pi } } { 2 } \right) - \sqrt { \frac { 4 - \pi } { \pi } }$. Note that $\cos ^ { - 1 }$ can also be written as arccos.
A third horse is placed in the field, and the three horses are rearranged. One horse is now attached to the midpoint of the bottom side of the field, and another horse is now attached to the midpoint of the left side of the field. The third horse is attached to the upper right corner.\\
(iv) Given each horse can access an equal area of the field and that none of the areas overlap, what length of rope must each horse have to minimise the area that no horse can reach?
The horses on the bottom and left midpoints of the field are each replaced by a goat; each goat is attached by a rope of length $g$ to the same midpoint as in part (iii). The remaining horse is attached to the upper right corner with rope length $h$.\\
(v) Given that $0 \leqslant h \leqslant 1$, and that none of the animals' areas can overlap, show that $\frac { \sqrt { 5 } - 2 } { 2 } \leqslant g \leqslant \frac { 1 } { 2 \sqrt { 2 } }$ holds if the area that the animals can reach is maximised.