[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 128
You are missing trading opportunities:
- Free trading apps
- Over 8,000 signals for copying
- Economic news for exploring financial markets
Registration
Log in
You agree to website policy and terms of use
If you do not have an account, please register
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.
Yeah, man, you're right as always. Fucked up. The functions F1(x) = 1+cos(x) and F2(x) = -1-cos(x). In short, raise one cosine by 1, and get the other one by its mirror reflection relative to Oh.
Sorry for the sloppiness. :-)
Yurixx, we are not boys any more, mistakes are forgivable :)
2 TheXpert: Once again, let's clarify the problem. Given two sides of a triangle (two segments) and a line containing the bisector. Construct the triangle. Right?
Mathemat писал(а) >>
2 TheXpert: Clarify the problem again. Given two sides of a triangle (two segments) and a line containing the bisector. Construct a triangle. Right?
No. There are three segments
1. the lengths of the two sides and the length of the bisector between them
2. the lengths of two sides and the length of the median between them
3. The lengths of three medians (this problem seems to have a geometric solution).
4. lengths of three bisectors (this one does not seem to have a solution)
Ой, об этом не подумал. У меня было другое решение.
Следующая: Докажите, что число 4n + 15n – 1 делится на 9.
It is easily proved that it is divisible by 3:
4 mod 3 =1 mod 3,
15 mod 3= 0 mod 3 => (4n + 15n - 1) mod 3 ≡ (1n + 0*n - 1) mod 3 ≡ (1 + 0*n - 1) mod 3 ≡ 0 mod 3.
But the divisibility by 9 is a bit harder to prove, because I have forgotten and I can't remember the property right now.
Hi Rosh. Well alsu has already solved it by matinduction here.
Concerning the triangle problems:
2. длины двух сторон и длина медианы между ними
Let the sides a, b, median m. Obviously, m is strictly between the remaining two numbers. Consider that a is the minimum, b the maximum.
Draw three circles from the common centre with radii a, b, m. It remains to draw a segment between points on the outer (b) and inner circles (a) so that it is divided by the middle circle (m) in half. There's probably a neat solution here by the inversion method.
P.S. By the way, problem 3 (on three medians) easily reduces to problem 2. That is, if we can solve 2, we can solve 3.
P.P.S. And vice versa too! In other words, if we know how to solve one problem, we can easily solve the other.
P.P.P.S. The problem (any of these two medians) is reduced to this: reconstruct a parallelogram by its adjacent sides and the diagonal coming from their common vertex.
I'm tired of writing afterwords. The "on three medians" problem is solved like this:
We divide the medians so that we build 2/3 of each. I hope this won't be a problem, it's not a trisection of the angle :)
We construct a triangle by these three pieces of medians, complete it to a parallelogram, taking any of the sides of the triangle as its diagonal. Then the second diagonal of the parallelogram will be one of the sides of the desired triangle. Further it is easily constructed.
The problem "by two sides and the median between" reduces to the same one.
To be sure of all this, just plot the triangle and its medians and remember that the medians at the point of intersection divide 1:2.
I remember from school that the solution is simple.
Similar bisector problems should be more difficult.
Mathemat писал(а) >>
We divide the medians so that we build 2/3 of each. I hope this won't be a problem, it's not a trisection of the angle :)
We construct a triangle by these three pieces of medians, complete it to a parallelogram, taking any of the sides of the triangle as its diagonal. Then the second diagonal of the parallelogram will be one of the sides of the desired triangle. Further it is easily constructed.
The problem "by two sides and the median between" reduces to the same one.
Yes. But I solved it in a different way and vice versa.
The problem "on two sides (1) (2) and the median between (3)":
Draw one of the sides of (1), divide it in two. From the middle of the segment we draw a circle of radius (2)/2 .
From the original vertex a circle of radius (3). the intersection of the circles -- the other end of the median.
Further it is easy.
And the median problem is reduced to plotting the sides and median with 2/3(1) 2/3(2) 1/3(3) by the above property of medians.
Аналогичные задачи о биссектрисах должны быть сложнее.
With the bisector, you should apparently use the fact that the third side is divided by it in the ratio a:b
С биссектрисой, видимо, следует использовать тот факт, что третья сторона делится ей в соотношении a:b
Yes, this is the first step.