23. Four children are in the four corners of a $10 \mathrm{~m} \times 25 \mathrm{~m}$ pool. Their coach is standing somewhere on one side of the pool. When he calls them, three children get out and walk as short a distance as possible round the pool to meet him. They walk $50 \mathrm{~m}$ in total. What is the shortest distance the coach needs to walk to get to the fourth child's corner?
A $10 \mathrm{~m}$
B $12 \mathrm{~m}$
C $15 \mathrm{~m}$
D $20 \mathrm{~m}$
E $25 \mathrm{~m}$
Solution
D

Consider two children in opposite corners of the pool. Wherever their trainer stands, the total distance these two pupils would need to walk to meet him is half the perimeter of the pool, as illustrated in the first diagram. Therefore, if all four children walked to meet their trainer, the total distance they would walk is $(2 \times 10+2 \times 25) \mathrm{m}=70 \mathrm{~m}$. Since we are given that three of the children walked $50 \mathrm{~m}$ in total, the distance the trainer would have to walk to get to the fourth child is $(70-50) \mathrm{m}=20 \mathrm{~m}$. The second diagram shows one possible position of the trainer and child which satisfies this situation although there are others.