Time-and-Work and Motorcycle Tyres A problem about motorcycle tyres, related to Time-and-Work or rate-of-work methods.
This is not a homework question, nor, as far as I know, a contest question. It is intended as a challenge for Year 10/11 or 15/16-year olds, but should require no knowledge of calculus.
Readers will no doubt be familiar with the type of rate-of-work problem, often seen at elementary school, which goes like this:
Alice can paint an $X$ metre-long fence in $A$ hours if she works on her own. Bob can paint the same fence in $B$ hours if he works on his own. If they work together, starting at opposite ends, how long will it take them to paint the fence?
Solution: Every hour Alice will paint $\frac{X}{A}$ metres of fencing, and likewise Bob will paint $\frac{X}{B}$ metres. Together, without interference, they will paint $\frac{X}{A}+\frac{X}{B}$ metres and therefore the hours they take complete the job is $$\frac{X}{\frac{X}{A}+\frac{X}{B}}=\frac{1}{\frac{1}{A}+\frac{1}{B}}=\frac{AB}{A+B}$$
So far so good. But the following three-part question is not quite so obvious.
QUESTION:
A motorcycle’s wheels wear out tyres at constant, but different, rates. A tyre fitted to the front wheel will wear out after $F$ miles, and a tyre fitted to the back wheel will wear out after $B$ miles.
a)  By swapping the tyres between front and back wheels, what is the maximum distance the owner can travel on just two tyres? When should she swap the tyres?
b)  What is the maximum distance she can travel if she also has a spare tyre? When should she swap the tyres around?
c)  The owner decides to fit a side-car to her motorcycle, and a tyre fitted to the side-car will wear out after $S$ miles. What is the maximum distance she can travel on the three tyres? When should she swap the tyres around?
We assume, no doubt, that all the tyres are identical and all are brand new to start with, and that all tyres will be worn out by the time she has driven the maximum distance. In part a) we assume that the tyres are swapped just once, and that there will be two change-overs in parts b) and c).
My Approach
a)  Let the maximum distance be $D$ and let the distance travelled to the swap-over time be $x$. Let $r$ be the depth of rubber on each tyre, measured in mm.
For each mile travelled, the back wheel loses $\frac{r}{B}$ mm of rubber. Therefore in travelling a distance $x$, the loss of rubber on this tyre is $\frac{rx}{B}$.
At the point of swap-over, the remaining rubber on this tyre is $r-\frac{rx}{B}$ and this tyre will be fitted to the front wheel to travel the remaining distance $D-x$.
When fitted to the front wheel, the loss of rubber for every mile is $$\frac{r}{F}=\frac{r-\frac{rx}{B}}{D-x}.$$
Rearranging this equation gives $$x=\frac{B(D-F)}{B-F}.$$
Since $x>0$, we have that either $B<D<F$ or $F<D<B$, which makes sense since we would expect the maximum journey to be less than $\max(B,F)$.
We can now write down the equivalent expression if we consider the front tyre in exactly the same way, and then equate the different expressions for $x.$ Thus we have:
$$\frac{B(D-F)}{B-F}=\frac{F(D-B)}{F-B}.$$
We can rearrange this and get an uncannily familiar-looking expression:
$$D=\frac{2}{\frac{1}{F}+\frac{1}{B}}$$
When this is substituted back to get $x$, we get: $$x=\frac {1}{\frac{1}{F}+\frac{1}{B}}$$
Obviously this is exactly the same type of expression obtained in the Alice & Bob rate-of-work problem. And this is what suggests to me that there must be an easier and more intuitive method to obtain the result.
Furthermore I don’t think it would be easy to apply my method to parts b) and c), so there must be an easier way. 
Can anyone give me a hint? I think I’m missing something obvious.
 A: (a)
Let the distance travelled be $FB$ mi.
Then front tyre wears out at B/mi, and back one wears out at F/mi.
For max. dist, each tyre must cover $0.5(B+F)$, thus $D_a = \frac{FB}{0.5(F+B)}$,
and the tyres need to be rotated at half this distance.
(b)
$3$ tyres are now sharing the load, so $D_b = 1.5D_a$
Changes should be made at $\frac13\;\;and\;\;\frac23$ of this distance
(c)
Here, let the distance travelled be $FBS$ mi, then by a similar procedure,
maximum distance $D_c = \frac{FBS}{\frac13\cdot{(BS+FS+BS)}}$
The rationale of the formula is that the $F,B,S$ positions of tyres must share the load in the ratio $F:B:S$  [Further explained in comments to a query on it]
Rotate tyres at every $\frac13$  of this maximum distance.
Added:Numerical example 
Suppose $F,B,S$ are respectively $40,50,60$.
Max. dist. travelled $= \frac{40*50*60}{\frac13\cdot(50*60 +40*60+40*50)}= 48.65$
Fractional usage of each tyre $= \frac13\cdot(48.65/40 +48.65/50 + 48.65/60) = 1$
A: OK I think the simple idea that had eluded me initially has now occurred to me, and in retrospect seems very obvious. I will therefore have to answer my own question.
The key work-rate idea that needs to be used is the number of tyres per mile that the different wheels are depleting.
a) The front wheel uses up $\frac 1F$ tyres per mile. The back wheel uses up $\frac 1B$ tyres per mile. Together, they use up $2$ tyres over a maximum distance $D$. Therefore, $$\frac 1F+\frac 1B=\frac 2D$$
Hence the result that I and others have found already, that $D$ is the harmonic mean of $B$ and $F$. Clearly by symmetry, change-over must occur at a distance $$\frac D2=\frac{BF}{B+F}$$
b) whatever is holding the spare tyre depletes it at a rate zero tyres per mile, so by the same argument, $$\frac 1F+\frac 1B+0=\frac 3D$$ since there are now three tyres to use up. Change-over must happen at distances $\frac D3$ and $\frac {2D}{3}$, where $$D=\frac{3BF}{B+F}$$
c) by similar argument, $$\frac 1F+\frac 1B+\frac 1S=\frac 3D$$ with change-over occurring again at distances $\frac D3$ and $\frac {2D}{3}$.
In the general case of an $n$-wheeled vehicle, where the $i$th wheel will deplete the tyre over a distance $R_i$, if the owner has $M\geq n$ new tyres to start with, then if she is prepared to do all these time-consuming change-overs at regular intervals, the maximum distance she can travel, ending up with all the tyres worn out, is $D$, given by $$\sum_{i=1}^{n}\frac {1}{R_i}=\frac MD$$
A: The thing you might be missing is the Harmonic Mean. 
For a) the solution is straightforward: maximum Distance is the harmonic mean, point of swap is half of harmonic mean.
In case of b), put some thoughts up front: the travelled distance is maximum as long as all three tyres are completely worn out at the end, independent of when exactly the swaps happen. So it is sufficient to find a solution with all tyres worn out. This means: change first worn out tyre for spare and proceed as in a)
In case of b), add a third mileage S, put up the harmonic mean and let $S\rightarrow\infty$, that's the maximum distance. Swaps happen at one third and two thirds. 
For c), maximum distance again should be harmonic mean, swaps should happen at one third and two thirds of harmonic mean.
