1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I am seeing equations of the form $\frac {BC}a+\frac {CA}b+\frac {AB}c=a+c-b$ . (ignoring minutes to hours conversion). Adding these three equations gives a fairly simple expression. $\endgroup$ – lulu Jul 21 '18 at 22:30
  • $\begingroup$ @RossMillikan I got exactly what you got (well, I persist in confusing $60$ and $\frac 1{60}$, but never mind). I don't think you can go further. $\endgroup$ – lulu Jul 21 '18 at 22:34
  • $\begingroup$ @lulu: amd pointed out the rates are minutes per mile, not miles per minute. I'll fix my answer. $\endgroup$ – Ross Millikan Jul 21 '18 at 22:35
  • $\begingroup$ @RossMillikan Ah, entirely missed that. Certainly simplifies things. $\endgroup$ – lulu Jul 21 '18 at 22:38
3
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @RossMillikan They’re trying to be tricky. The rates are given in minutes per mile. $\endgroup$ – amd Jul 21 '18 at 22:34
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Though I do like this solution, I think the one by Ross Millikan is more elegant. $\endgroup$ – S. Sharma Jul 21 '18 at 22:44
0
$\begingroup$

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 ... $$ P(a+b+c) =60(a+b+c) \\P=60$$

$\endgroup$
  • $\begingroup$ The rates are given in minutes per mile, not miles per minute. $\endgroup$ – amd Jul 21 '18 at 22:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.