[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 125
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Proof in the non-trivial case - by a figure we mean a part of the plane bounded by a closed line (without self-intersections).
Suppose the figure X has symmetry centre O, i.e. for every point of the figure the condition of the problem holds. Suppose there is at least one other symmetry centre O', not coincident to O. Draw the line O'. Obviously it intersects the boundary in finitely even number of points (at least in two). Choose one of such points A among those which are on the same side of O' such that the distance from A to O is maximal (1). Let B be also a point of the figure symmetric to A with respect to O.
Note that every point on the line O' that is at a greater distance from O than A does not belong to the figure X according to (1). (2)
Let B' be a point symmetric to B with respect to O', then by definition of symmetry B' belongs to X. However, OA=OB<O'B=O'B'= OB'-OB'<OB', then it follows from (2) that point B' does not belong to X. We obtain a contradiction, which proves that the assumption about the second symmetry centre is incorrect. The theorem is proved.
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Well done, alsu!
At first I went the other way (found the image of one CA relative to another and started to prove that I got a third CA), but then I found the solution you described.
In principle we can consider the case of infinite figure (strip or something similar), there it turns out that symmetry centres can be generated an infinite number (by "walking excavator" method) :) But I think, in fact, a finite figure is enough.
Нашел монетку написано: HALF PENNY это сколько будет в %% от GBP? :)
Yes, I forgot to say! The coin is from 1958, that's important!
And I have a neat solution to a geometric problem, if anyone remembers it ("There are two circles and a point. Construct a segment whose ends lie on the given circles, and whose middle is at the given point"). Here it is half an hour ago.
Yurixx, and it is extremely characteristic that the construction itself determines when the solution to the problem exists. I.e. writing out constraints in the problem condition is almost the same as solving it.
Hint: the solution popped into my head just after I saw alsu's solution.
You have a strange centre of symmetry.
If by the centre of symmetry we mean the point about which a 180-degree rotation results in an exact match, then 2 centres are hard to come by. But infinitely many of them are welcome.
Let us take the graphs of functions F1(x) = cos(x)+1 and F2(x) = cos(x)-1 in the plane. The part of the plane between these graphs is our figure. Its symmetry centres are all points x divisible by pi.