[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 127
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The points of extremum cannot be CA because, say, there is nothing above the maximum of cos(x) + 1 (your CA) :)
Here, for sines, it's multiples of Pi.
P.S. No, that's not what I'm saying. You mean points on the x-axis, of course? OK, take point 0 and draw the line y=x through it. From above and below it will intersect your cosines differently. At the same time, if you take Pi/2, everything is tip-top.
Even simpler: the straight line x=0 is enough. The CS is (0;0) in your case? It will intersect the figure at y=0 and y=2.
at n=1 is trivial. Further, if it is true for some n (1), then
4^(n+1)+15(n+1)-1=4*(4^n+15n-1)-45*n+18. The bracket is divisible by 9 by (1), the last two terms are obviously also multiples of 9. By virtue of the method of matinduction the divisibility is proved.
Strong, alsu, strong. Didn't you go to physics school, by any chance?
Next: Construct triangle ABC given its two vertices A and B, and the line containing the bisector of angle C.
P.S. On some matte forum (not mehmatic one) I bumped into a very famous trader and MQL4 programmer, who is a Fermatian. I have no doubts that it is him, because not only his nickname but his avatar as well matched. It happens, doesn't it?
Easy :).
Spit it out.
I get it, but you tell me.
All right, let's wait a bit.
The next one should be more complicated: There are 2000 points marked in the plane, no three of which lie on the same line. Prove that it is possible to draw a line (not passing through any of the marked points) on both sides of which there are 1000 points.
Mathemat писал(а) >>
I've got it, but you tell me.
Construct the symmetry of any of the points with respect to the bisector. The rest, I think, is clear.
I think it is more interesting to build a triangle knowing the lengths of two sides and the bisector between them.
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Somehow I don't know what to hang on to myself.
In geometry, the raw data is without lengths. "Knowing the lengths of the sides" is the same as "knowing all the sides". Well then you don't need bisectors either.
But to build a triangle by three bisectors (three segments) without knowing any angles between them is the problem.
OK, that's fine. We'll solve the "three bisectors" problem later.
ОК, можно и такую. Задачку "по трем биссектрисам" решим потом.
I have a vague suspicion that it is unsolvable...
I think there's also a two-sided and median problem, but I'm not sure.
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ZS, yes, there is and it seems to be much easier to solve than the bisector.