Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 140
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About the ants. By all accounts, they need 10 seconds at the most. How to prove it - I don't know yet. The solution must be beautiful.
Scan
Heehee
The solution is very beautiful and understandable even to a child) Literally in a couple of lines)
It's about ants again. It's a lot of boo-boo, it could probably be simpler and prettier, but still:
To find out the maximal time of "fermentation" it is enough to calculate the length of maximal ant's mileage. Take N, which is the number of ants that is large enough (ideally tending to infinity) and arranged evenly. The initial motion is opposite in one. Then the ant that is closest to the centre of the stick will oscillate while those on the edge, gradually, one from each edge, falls off outwards. The amplitude of the oscillations is half the initial distance between the neighbouring ants 10/(2N). The number of such oscillations until the space to leave to one of the edges is N/2. An ant will have moved (10/(2N))(N/2)=5 cmin that time. Now it will have to pass from the centre to the edge - another 5 cm. Total - 10 cm, i.e. 10 sec.
Yes, there is a really simple, geometric one. Almost no numbers in the calculations (apart from having to divide 10 by 1). That just counted :)
Also, your assumptions rely on the hypothesis of "maximality" of the solution for evenly spaced ants.
Try even simpler somehow. Most of the problems on braingames.ru have a very brief and elementary solution. Even the ones that don't seem to have such a solution.
2 Mischek: zadachka is good!
It's about ants again. It's a lot of boo-boo, it could probably be simpler and prettier, but still:
To find out the maximal time of "fermentation" it is enough to calculate the length of maximal ant's mileage. Take N, which is the number of ants that is large enough (ideally tending to infinity) and arranged evenly. The initial motion is opposite one through one. Then the ant that is closest to the centre of the stick will oscillate while those on the edge will gradually, one from each edge, fall off the stick. The amplitude of the oscillations is half the initial distance between the neighbouring ants 10/(2N). The number of such oscillations until the space to leave to one of the edges is N/2. An ant will have moved (10/(2N))(N/2)=5 cmin that time. Now it will have to pass from the centre to the edge - another 5 cm. Total - 10 cm, i.e. 10 sec.
Scan
Heehee
Notebook costs 26 roubles. 50 kopecks. Now try to prove otherwise.
Huh
(4) While looking at the relief map of Brainland, Megamozg suddenly noticed an interesting feature: the average height of any four points lying in the vertices of one square is zero. Is it true that Brainiac is perfectly flat?
Comment: no considerations of relief continuity apply. Brainland may well turn out to be extremely rugged in height - like a Dirichlet function, for example (this function is not continuous at any point).
The country is known to have no boundaries.
First class))
Let us draw Brainiac with Cartesian coordinate system and choose some point (x,y). We have for any a<>0 four squares from the given point:
h(x,y)+h(x+a,y)+h(x,y+a)+h(x+a,y+a)=0
h(x,y)+h(x-a,y)+h(x,y+a)+h(x-a,y+a)=0
h(x,y)+h(x+a,y)+h(x,y-a)+h(x+a,y-a)=0
h(x,y)+h(x-a,y)+h(x,y-a)+h(x-a,y-a)=0
Adding up, we get
4*h(x,y) + 2*[h(x+a,y)+h(x-a,y)+h(x,y+a)+h(x,y-a)] + [h(x+a,y+a)+h(x-a,y+a)+h(x+a,y-a)+h(x-a,y-a)] = 0
The second term in the parenthesis contains the sum of the heights of the vertices of the square and the third term also, hence they are both zero. So the first summand is zero too, i.e. Brainiac is in fact perfectly flat.Perfect. I have exactly the same solution, but on the third try :)
P.S. I also have a drawing; the solution is clearer:
P.S. The first "solution" was this:
DEFINITION:
Relief is a [real] function of the complex variable f(z) satisfying the following condition (w is an arbitrary complex number, see figure):
1/4 * ( f( z + w ) + f( z - w ) + f( z + w*i ) + f( z - w*i ) ) = 0
Since no one forbids us to take w = 0 in the relation, we obtain that f(z) = 0.
Brainiac is perfectly flat. No consideration of the continuity of the function is required.
Where is the error here?
Preliminary comments from the moderators included the fact that the function is defined at every point. However, to this my "solution" the moderator replied that there should be a square, not a point. Did I violate the possibility of discontinuity of the function, or what?