To swim or to run?
Problem
The famous film star Birkhoff Maclane is sunning herself by the side of her enormous circular swimming pool (with centre $O$) at a point $A$ on its circumference. She wants a drink from a small jug of iced tea placed at the diametrically opposite point $B$. She has three choices:
(i) to swim directly to $B$.
(ii) to choose $\theta$ with $0<\theta<\pi,$ to run round the pool to a point $X$ with $\angle AOX=\theta$
and then to swim directly from $X$ to $B$.
(iii) to run round the pool from $A$ to $B$.
She can run $k$ times as fast as she can swim and she wishes to reach her tea as fast as possible. Explain, with reasons, which of (i), (ii) and (iii) she should choose for each value of $k$. Is there one choice from (i), (ii) and (iii) she will never take whatever the value of $k$?
There are some hints and suggestions for how to approach this problem in the Getting Started section.
Garret Birkhoff and Saunders Mac Lane were two American mathematicians, who together in 1941 wrote "A Survey of Modern Algebra" one of the first modern algebra textbooks for undergraduates.
STEP Mathematics II, 1995, Q5. Question reproduced by kind permission of Cambridge Assessment Group Archives. The question remains Copyright University of Cambridge Local Examinations Syndicate ("UCLES"), All rights reserved.
Getting Started
It is important to note that $\theta$ is measured in radians.
The first step is to draw a diagram! You might also want to pick a variable to represent the speed at which Birkhoff Maclane can swim (such as $v$), and then you can write her running speed as $kv$. You will also want to pick a variable to represent the diameter or radius of the pool (Radius might be easier).
It might be easiest to start by considering just cases (i) and (iii). Can you find expressions for the length of time it will take her to swim across the pool, and the time it will take her to run around? What is the optimum strategy if $k=1$? What if $k=1000$? Is there a value of $k$ where the times are the same?
When considering the case where she runs a bit and then swims, it is a good idea to check that when you substitute in $\theta = 0$ (when she swims straight across) or $\theta = {\pi} $ (when she runs around) you get the same answers as in cases (i) and (iii).
You might be able to simplify your expression for the time in case (ii) by considering double-angle trig formulae.
Can you use differentiation to find the minimum time it takes if she uses method (ii)? How can you check that this is a minimum?
Student Solutions
Well done to Mahdi from Mahatma Gandhi International School in India who sent in this thorough solution:
Swimming Speed: 1
Running Speed: $k$
Radius of pool: $1$
(i) Case 1 (swim directly to B) is covered in this diagram when $\theta = 0$
(iii) Case 3 is covered in this diagram when $\theta = \pi$
Running Distance: arc AX which has length $\dfrac{\theta}{2\pi}\times 2\pi r = \theta r$
Running Time: $\dfrac{\theta r}k$ because the running speed is $k$
Swimming Distance: Chord $\text{XB}$ which has length $\text{2XM.}$ $\angle\text{XOM}=90-\frac{\theta}2$
Why $90-\frac{\theta}2$? $\angle\text{BOX}$ and $\theta$ lie on a straight line. $\angle\text{BOX}$ is in an isosceles triangle with $\angle\text{OBX}$ and $\angle\text{OXB,}$ therefore $\angle\text{OBX}$ and $\angle\text{OXB} = \angle\text{OXM}$ are both equal to $\dfrac{\theta}2.$ Triangle $\text{OMX}$ is right-angled at $\text{M.}$
$$\begin{align}\sin{\left(90-\frac{\theta}2\right)}&=\frac{\text{XM}}{\text{OX}}\\ \text{XM}&=r\cos{\left(\frac \theta 2 \right)}\\ \text{XB} &= 2r\cos{\left(\frac \theta 2 \right)}\end{align}$$
Swimming Time: $\dfrac{2r\cos{\left(\dfrac \theta 2 \right)}}1 = 2r\cos{\left(\dfrac \theta 2 \right)}$
So, total time, $T$ = Running Time + Swimming Time $$ T = \dfrac{\theta r}k + 2r\cos{\left(\dfrac \theta 2 \right)}$$
To find the minimum time, $\dfrac{\text d T}{\text d \theta}=0$ and $\dfrac{\text d^2 T}{\text d \theta^2}\gt0$ $$\begin{align}\dfrac{\text d T}{\text d \theta}&=\frac r k - 2r\sin{\left(\frac\theta2\right)}\times\frac12\\ &=r\left[\frac1k-\sin{\left(\frac\theta2\right)}\right]\end{align}$$
Now the double derivative has to be positive for there to be a minimum $$\begin{align}\dfrac{\text d^2 T}{\text d \theta^2} &= r\left[0-\cos{\left(\frac\theta2\right)}\times\frac12\right]\\ &=\frac{-r}2\cos{\left(\frac\theta2\right)}\end{align}$$
For $0\le\theta\le\pi,$ $\cos{\left(\frac\theta2\right)}\ge0$ and so $\dfrac{\text d^2 T}{\text d \theta^2}\le0$ which implies there is no minimum time between $0\le\theta\le\pi.$ Hence, the minimum is at either of the end points. There are two cases for that, as shown in the two graphs below. $r,$ the radius of pool, just adds vertical stretch.
There is a value of $k$ for which the time taken at $\theta = 0$ is equal to the time taken at $\theta = \pi.$ And the time taken is $2$ for both [times $r$]. Let $r=1$ since it just adds vertical stretch to the graph.
For this value of $k,$ running all the way ($\theta = \pi$) takes time $2$ (the same amount of time that swimming all the way takes). $$2=\frac{\theta}{k}+2r\cos{\left(\frac{\theta}2\right)}\\ 2=\frac{\pi}{k}+2r\cos{\left(\frac{\pi}2\right)} \\ 2=\frac{\pi}{k}$$ So we get $k = \dfrac \pi2$. We can thus conclude that:
For $0\lt\theta\lt\dfrac\pi2,$ the minimum time is achieved at $\theta = 0.$ So option (i) swim directly.
For $\dfrac\pi2\lt\theta\lt\pi,$ the minimum time is achieved at $\theta = \pi.$ So option (iii) run around.
Teachers' Resources
Why do this problem?
This problem uses ideas of arc length and chord lengths, as well as possibly differentiation or graph sketching to justify the final results (which might be surprising!).
Possible Approach
This problem featured in an NRICH Secondary webinar in March 2022.
You may want to ask students to consider Where to Land before introducing them to this problem.
Start by just considering methods (i) and (iii).
Ask students for expressions for the time it takes Birkhoff to swim directly across and to run all the way around. They will need to choose some variables for speed and either radius or diameter. Can they work out the values of $k$ for which running is the better option?
Method (ii) - if Birkhoff decides to run part of the way, and then swim, can they find expressions for the length that she runs and swims? What is the total time taken for this method?
How could we check our expression for the time taken in method (ii)?
Substituting $\theta = 0$ and $\theta = \pi$ should give the same answers as for methods (i) and (iii).
Can we find the minimum time taken with method (ii)? How can we be sure it is a minimum?
Possible Extension
What happens if the drink is not diametrically opposite? What if it is $\frac 1 3$ or $\frac 1 4$ of the way around the circle?
Note
Garrett Birkhoff and Saunders Mac Lane published their Survey of Modern Algebra in 1941. The book was written because the authors could find no adequate text to use with their students at Harvard.