RIDGE

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Let s be a line and A and B two points at the same side of s. For which point P in s is AP + PB the shortest way joining A and B? 

Der kürzeste Umweg

Gegeben sei eine Gerade g und zwei Punkte A und B außerhalb der Geraden. Sei C ein beliebiger Punkt auf g, d.h. C ist auf g frei wählbar bzw. kann auf g verschoben werden. Gesucht ist der kürzeste Umweg von A über C nach B. An welcher Stelle Cmin der Geraden g ist der Weg minimal?  Beschreiben Sie, wie man den Punkt Cmin exakt konstruieren kann! Warum ist der Weg über Cmin dann minimal?

[The shortest detour: Given a straight line g and two points A and B not on the straight line. C is any point on g, i.e. C can be chosen anywhere on g or can be moved on g. Find shortest path from A  to B via C. At what point Cmin on the straight line g is the path minimal? Describe how one can construct the point Cmin accurately! Why is the path via Cmin the shortest path?]

Note: This solution seems to be more likely to be found if the problem is presented in Knipping's form (see above). In that case, the possibility of exploring the situation where the points are on opposite sides of the line is present.  This may be a useful step in arriving at this solution, which otherwise seems to rely too heavily on insight.

Consider the diagram shown in Figure 4a. In this special case  the position of P is in the centre, which suggests Conjecture 2: 

P is the intersection between s and the perpendicular bisector of AB. (see Figure 4). 

[We have retained the numbering of the conjectures used by Figueiras & Deulofeu (2005) but we believe this conjecture to be the beginning of a discovery process.]

It is possible that those who made  conjecture 2 used their knowledge that the perpendicular  bisector is the locus of equidistance between two points. Why  this particular knowledge was active for these students is not clear (Perhaps they had solved a problem in which it was important recently??).  But it is reasonable to make use of this  knowledge.  They are considering a path made up  of two parts  AP and PB.  Not knowing how to minimise the total distance  they might have considered the easier problem of minimising  the longer part. If the total distance were fixed then the longer  part is minimal when the two arts are equal, which occurs  when P lies on the perpendicular bisector of AB (see Figure 4).   

Unfortunately, the total distance is not fixed (as we are trying  to minimise it) so the reasoning is flawed and a counter  example can be found, when the horizontal distance between A  and B is small compared to the vertical distance between them  (see Figure 5).  Conjectures 2 & 3 illustrate an interesting  confusion that occurs in solving such problems:  equal distances are often confused with shortest distances. Knipping  (2005) discusses a similar confusion in another minimum  distance problem, finding the point such that the total distance  to the sides is minimal (the Beach Problem).  Their initial conjecture was the centre of the incircle. This diagram (Figure 5) in the context of the reasoning that  has gone before, suggests the next conjecture:

Conjecture 3: P is the intersection between s and its  perpendicular through the mid-point of AB. (see Figure 5).

The problem it reveals is that the point P should stay on the  segment ED that is the projection of AB onto s.  This  constraint, combined with the earlier consideration of trying to  equalise distances, suggests that the midpoint of ED, or  alternately the projection of the midpoint of AB onto s, is the  point P. This new construction also accounts for the known  special case shown in Figure 4a.  

If A is on s, however, a new  counterexample is produced (Figure 5a). This diagram shows  clearly the correct solution in another specific case: When A is  on s then the shortest path is AB itself. The emphasis shifts to  defining P is such a way that it can be seen as a continuous  transformation from this initial situation. As A moves up, P  must move to the right.  

The segment EB provides a  mechanism to produce this motion: A is projected horizontally  onto EB to the point O and then O is projected vertically onto s  to the point P (Figure 6).   This gives rise to:

Conjecture 4: r is the perpendicular to BD through A; O the  intersection of r and BE; and P the intersection between s and  its perpendicular through O. (see Figure 6).

Again a counter-example is not hard  to find (Figure 6a), because this construction does not work in  the original special case, when EA=DB. 

Combining the two special cases (and the mirror image of the  second) produces the diagram shown in Figure 7a, which immediately suggests: 

Conjecture 5: O is the intersection  between AD and BE. P is the intersection between s and its  perpendicular through O (see Figure 7). 

Thus through a sequence of conjectures, each one based on a  single special case (except the last which is based on two  special cases), a conjecture is produced that turns out to be  correct.