פתרון ארבע חיפושיות


Each beetle walks 10 meters before all the beetles unite. To see that, imagine , that time is measured at discrete steps at time units, and in each such steps a beetle moves D meters towards the beetle it "hunts" (the "huntee", opposite of "hunter"). All beetles move that distance, as they all have the same speed. Now, if these movements are drawn as dots, a square is formed, in an inclination to the original square of the locations of the beetles before the current time step. Now, let's consider the distance between a beetle and its huntee. By the Pythagorean theorem, it is sqrt((d-D)*(d-D)+D*D), where "d" is the distance a time unit before. Now, the D*D factor can be neglected if D is small enough, and that equal sqrt((d-D)*(d-D)) or d-D. That . is, if a beetle moves D meters, it gets closer to the its huntee by D meters Now, let's consider the case of a move followed by another move: the beetle moves 2*D meters, and gets closer 2*D meters. Continuing this way, the beetles meet when they get closer "d" meters, where "d" is their . initial distance. Now, that equals 10 meters - the length of a side of the square

This solution by Yuval Filmus of Tet4.

N.B. I'm aware it is a bit (and more) problematic.

However, I'm quite sure the answer by itself is correct .

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אחת לשניה. מסלול כל אחת מהן לכן

מאונך כל הזמן למסלול החיפושית

הנרדפת. לכן אורך כל ספירלה יהיה

בדיוק אורך צלע הריבוע, כלומר 10 מטר.


פותר : יובל פילמוס ט' 4


Document URL: http://ohel-shem.com/subjects/math/
Date: 27/2/98