3 Towns and a Man problem 
A, B, C are three towns forming a triangle. A man has to walk from one to the next, ride thence to the next, and drive thence to his starting point. He can walk, ride, and drive a mile in a, b, c minutes respectively. If he starts from B he takes $a + c - b$ hours, if he starts from C, he takes $b + a - c$ hours, and if he starts from A he takes $c + b - a$ hours. Find the length of the circuit.

I did the following in an attempt to solve it:
Let $d = AB + BC + CA$, or the perimeter of the triangle.
Then, we have the rates in miles per minute:
$$\frac{d}{60(a + c - b)}$$
$$\frac{d}{60(b + a - c)}$$
$$\frac{d}{60(c + b - a)}$$
At this point, I wasn't sure what else I could do, because in order to find the distance, I need to know the average rate from each starting point.
The only other thing I noticed is that the sum of the hours it takes from each starting point is $a + b + c$. However, I don't think this property seems relevant right now.
If anyone can help me deduce the rates or find another approach, then I can carry out the problem from there.
Thanks.
 A: Let $AB=x, BC=y, CA=z$ be the distances in miles between the towns.  Then if he starts from $A$ he takes $xa+yb+ zc$ minutes, which we are told is $(c+b-a)$  hours.  Using the other two as well we get $$xa+yb+ zc=60(c+b-a)\\
xb+\ yc+za=60(a+c-b)\\
xc+ ya+ zb=60(b+a-c)$$
and we are asked to find $x+y+z$.  If we add these together we get
$$(x+y+z)\left(a+b+c\right)=60(a+b+c)\\
x+y+z=60$$
A: You’re on the right track. I would suggest keeping the three distances separate at first, so that your equations are $aBC+bCA+cAB = 60(a+c-b)$ and so on. If you add up the three equations, the left-hand side will be some multiple of $BC+CA+AB$ and the right-hand side looks like it will cancel nicely.  
Looking at it from a different angle, the man effectively travels the circuit thrice: once walking, once riding, once driving. The total time for this is (a+b+c)d, and this is equal to the sum of the given times.
A: Imagine taking three trips , one from each vertex, the total time for the three trips is $60(a+b+c)$ minutes
In the process of doing this you will end up travelling the entire perimeter using each mode of transportation
Let $P$ represent the perimeter of the triangle in miles, so the time spent driving is $\frac Pc $ minutes
Equating the total time for the three trips ...
$$\frac Pa +\frac Pb +\frac Pc =60(a+b+c)
\\P\left(\frac{bc+ac+ab}{abc}\right) =60(a+b+c)
\\P=\frac{60abc(a+b+c)}{bc+ac+ab}$$
*EDIT (rates given in minutes per mile - thanks @amd  )
Let $P$ represent the perimeter of the triangle in miles, so the time spent driving is $Pc$ minutes
Equating the total time for the three trips ...
$$ P(a+b+c) =60(a+b+c)
\\P=60$$
A: Alright, so I think I have an answer based upon what everyone has said until now:
In $a + b + c$ minutes, the man covers 3 miles.
Now, if he starts from each of those points and completes a circuit, of $d$ length, then he will cover $3d$ miles. His average time completing the circuit 3 times is equivalent to $60(a + c - b) + 60(b + a - c) + 60(c + b - a)$ minutes which is $60(a + b + c)$ minutes.
Therefore, the following ratios can be stated equal:
$$\frac{3}{a + b + c} = \frac{3d}{60(a + b + c)} = \frac{d}{20(a + b + c)}$$
Through basic algebra, we can then see that $d = 60$ miles.
