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You have to go from a point $A$ (start) to a point $B$ (arrival) by crossing a river $(d)$ and traveling as little distance as possible.

enter image description here

Pupils first do a search by trying several paths $1,2,3,4$ and completing a table with the different perimeters $p_1, p_2, p_3, p_4$. I use perimeter as on geobra for the length of a broken line. Out of two classes to which I proposed the exercise, none of the pupils thought of using symmetry on their own, although the exercise is part of the chapter Symmetry; On the other hand, they all very naturally thought of the straight line to reach the river from $A$. Once the indication was given: "perhaps symmetry could be used", the solution was quickly found by some of the students. On the other hand, doubts about this solution have arisen, with some students thinking that they have measured a shorter path than the red path. In addition, with a geogebra animation, because of the rounding, students pointed out that there were perhaps several possibilities... From then on, I was forced to give a demonstration.

The knowledge that "the shortest path from one point to another is the straight line" having been implicitly accepted, the triangular inequality $\color{red}{(1)}$ flowed from it.

In addition, the students knew that "$B'$ symmetrical of $B$ with respect to $d$" was equivalent to writing "$d$ mediatrix of $BB'$"$\color{green}{(2)}$;

that if a point is on the mediatrix of a segment, then it is equidistant from the endpoints"$\color{orange}{(3)}$;

that "if $a=b$, then $a\color{red}{+c}=b\color{red}{+c}$"$ (4)$, e.g. through the example "$5=2+3$ therefore $5\color{red}{+7}=2+3\color{red}{+7}$";

that if $M$ is on $[AB]$, then $AB=AM+MB$ $\color{Magenta}{(5)}$

We then had to formalize the problem: we are looking for $H$ as for any point $M$ on $d,AM+MB>AH+HB$.

Then prove that $H=I$ is the only point that fits.

Here is the demonstration I provided, with obvious notations that I obviously spared the pupils: let $M$ any point other than $I$ on $[AB]$, $B'$ the symmetric of $B$ with respect to $d$ and $I:=[AB']\cap d $ $$\color{red}{AM+}MB=\color{red}{AM+}MB'>AB'=\color{red}{AI+}IB'=\color{red}{AI+}IB$$ according to, in order of use, $\color{green }{(2)}$,$\color{orange}{(3)}$, $ (4)$, $\color{red}{(1)}$, $\color{Magenta} {(5)}$ and finally again $\color{green }{(2)}$,$\color{orange}{(3)}$ et $(4)$.

Aware that the demonstration was not envisaged by the authors of the exercise and that only the first research was, giving the demonstration to twelve-year-old students, some of whom have very sharp minds, did not seem to me useless or too problematic for the more difficult students.

Is the proof mathematically correct?

What about its relevance to twelve-year-olds?

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    $\begingroup$ I think this should be used as motivation for the topic of symmetry by showing that, in addition to how symmetry can be found in nature (e.g. symmetry in plants and animals), symmetry can sometimes be a useful mathematical tool. And rather than make a big production out of who can "see the idea" without prompting, the initial trial-and-error process mainly serves the purpose of indicating that the answer (for nearly everyone) isn't obvious. And for this age it's probably better to just give a physical demonstration using paper in which segment $IB$ is rotated at point $P$ about line $MI.$ $\endgroup$ Feb 10 at 17:30
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    $\begingroup$ Given your solution it looks like the river does not get crossed. The problem description should be corrected. Maybe “You have to go from point $A$ to the bank of the river $d$ and then to point $B$ while traveling the least possible distance.” $\endgroup$ Feb 11 at 22:38
  • $\begingroup$ Yes @WillOrrick, even something like "Minimize AM+MB" to avoid uninteresting ambiguities. $\endgroup$ Feb 12 at 7:07
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    $\begingroup$ @WillOrrick: ... or something like "You need to go from $A$ to $B$ with a bucket of water you get from river $d$. At $A$ you have a bucket, you go to the river $d$, you fill the bucket and you go with your full bucket to $B$. Beware: you don't care about minimising the effort of carrying the bucket, you need to minimise the total distance.". $\endgroup$
    – Dominique
    Feb 12 at 10:08
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    $\begingroup$ A and B are on the same side of the river, so this question would confuse 12-year-old me. $\endgroup$
    – shoover
    Feb 12 at 22:31

1 Answer 1

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The exercise is part of the chapter Symmetry... Is the proof mathematically correct? What about its relevance to twelve-year-olds?

The proof seems fine, but I think there is a more intuitive way to explain the relevance of symmetry to twelve-year-olds using this problem as an example.

If I were to explain the solution to this problem to a twelve-year-old, I'd do the following:

  1. First establish baseline intuition about the relevance of symmetry to optimization problems. In my experience, kids will typically pick up on this pretty quickly if you give them a few simple yet illustrative problems. For instance, if you have $20$ meters of fencing and you want to fence off the largest possible area for a rectangular garden, then what should the dimensions of the garden be? The kids might need to try a few configurations but they'll see the trend and arrive at $5 \times 5$ pretty quickly.

  2. Then, connect that intuition to the problem at hand. In the problem you've given, the core idea is that you want to minimize the sum of two diagonal distances, and each distance is the hypotenuse of a right triangle (the legs of the triangle are parallel and perpendicular to the river). Just as no side of the rectangular garden is any more special than the other, neither triangle in this problem is any more special than the other. So the triangles should have the same relative dimensions (one triangle might be bigger than the other, but their angles should all be the same).

Of course, it remains to actually construct the two triangles that have the same relative dimensions. As you show in your diagram, we can use symmetry here too: reflect $B$ over the river to $B',$ draw $AIB',$ and then draw $IB.$

But the relevance of symmetry here, and especially its relevance to twelve-year-olds, is not so much this specific construction but rather the general idea that an "optimal configuration of things," when no particular thing is substantively different from the others, will exhibit symmetry.

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  • $\begingroup$ Vagn L. Hansen had written a whole short text to make young people aware of optimization problems, here (www2.mat.dtu.dk/people/V.L.Hansen/square.html) the $5\times5$ has the largest area. $\endgroup$ Feb 12 at 7:24
  • $\begingroup$ @StéphaneJaouen the link is broken; is there another place the text can be found? $\endgroup$ Feb 12 at 20:58
  • $\begingroup$ I AM THE GREATEST Vagn Lundsgaard Hansen $\endgroup$ Feb 12 at 21:26

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