I proposed this problem to my calculus students. It turned out to be more interesting than I thought (thanks, Ignacio and Noema).

The government intends to build a superhighway without speed limit so, therefore, without curves. It will start from city A, and should pass as close as possible to cities B and C.

In order to solve it you should start by stating what you *mean* by *“as close as possible”*. An option is to minimize the sum of the distances. But then, you get the following funny configuration:

Cities *B* and* C* are at the same distance from *A*, and make up a right angle. Intuition dicatates that the best route for the highway would be the bisector. Let *d(A,C)=d(A,B)=1*. Then, the sum of the distances from the cities to the highway is . *But* there is a better highway! You can just break the symmetry between *B* and *C* and make the road pass through *C* exactly. Then, the sum of the distances is just… *1*.

If this was a real highway, the politician in charge would tell us that *symmetry* is also worth. Citizens of *B* may riot with un asymmetric solution… So, now let us change our *target function*. Let us minimize the sum of the squares of the distances. In that case, the bisector gives a square total distance of *1*, same as the asymmetric road… Can you explain it?

Which is the best choice? Of course, it depends on *the reason for which you’re fighting with the problem.* If it is just to pass an exam, any of them will do…:) [Of course, there are many other alternatives. A student minimized the sum of the squares of the distances from the points to the straight line *in the y direction*. This makes sense in some cases, e.g.: when you’re fitting experimental points to a line.]

So, choosing the right function to minimize is crucial in practice… Kadanoff once explained in a talk that the government of the city of London has ordered a huge global study of traffic, taking into account both public and private transport, energetic and economic issues, taxes and prices, eeeeeverything. They made an enormous computer program that was running for days and, finally, told them *the* answer, how to optimize traffic in London. *They had to remove all traffic lights*. Why???? The program had many things into account… also the fact that in street car accidents, it’s normally old people who die. And old people do not *pay* taxes, they *receive* their pension money from the government. So it was convenient to remove the traffic lights… So, you see: garbage-in, garbage-out. Yes, maths is a nice girlfriend, she gives you more than you put in the relationship… but she can do no miracles. If you’re stupid, she can’t fix that…

Puff, ya decía yo que me salían cosas muuuy raras haciendo este problema… 0/0 y inf/inf…

Normal, todas las soluciones eran válidas!!

PD:Huy, como te vea mi profesora de técnicas de búsqueda de la información utilizando fotos sin permiso del autor…

Vilxes, there are “degenerate cases”, where all solutions are valid… but that’s because the situation is pathological. In our case, why would anyone want a road passing “close” to B and C???? But for most cases, there is a single valid solution. Or it should! :)

About the permission… well, he’s a personal friend ;) (don’t believe it? I never lie on tuesdays!)

wtF??? i could be seen as an “isosceles” triangle? I was doing it for an “escaleno” triangle I am turning mad because im trying to apply relationships and proyections but without any positive result…

javi… when i came home that day i was thinking about this problem and i got a solution for it…though i dont really know if it is correct. I though about a linear algebra problem. What if you use the least squares line to solve it? Because you have three point with different coordinates….and the system Ax=b is inconsistent so… could it be an option?

@Ycs: when u say isosceles, I assume that you mean that both cities B and C are at the same distance to A, and scalene when they’re different. Insist, it is just some trigonometry.

@Viktor Yes, you can use least squares… if you can formulate it as a linear problem, Ax=b. In this case, as it is formulated, I think it that can’t be done. If you can, then just use the normal equation AtAx=Atb.

The problem has much more interest than I thought initially. In fact, the solution is surprising and there is a nice

apparentparadox.