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.
Best Answer
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 $$(x+y+z)\left(a+b+c\right)=60(a+b+c)\\ x+y+z=60$$