[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 170
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Окружности расположены именно так и не иначе?
Yes. I.e. each one touches the other two, no circle lies in the other circle.
ZS: I don't remember the solution myself any more.
Кто решит задачку и докажет правильность своего решения, может считать себя крутым математиком.
Для трех окружностей произвольного радиуса найти треугольник максимальной площади, вписанный в заштрихованную фигуру.
Но это так -- если будет куча свободного времени и амбиций и желание сломать мозг.
It doesn't work like that.
Draw a line connecting the two points
one where the left circle touches the upper right.
the other where the upper right circle touches the lower right circle
parallel to this line, draw a line inside the hatched area, so that it touches the upper right circle
one side is ready
the others in the same way
no proof (
alsu, большая просьба, не выкладывай решение. Думаю, ты ее давно решил.
Richie, хочешь почувствовать радость решения скучной математической задачки - пусть даже с небольшими подсказками?
P.S. Ладно, Richie уже спит, наверно. Будем решать, кому интересно и кто не спит еще.
then I'll post about 2,000 points
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) which has 1000 points on each side
.Consider some Cartesian coordinate system xOy, in which points have coordinates (xi,yi), i=1...2000.
If xi!=xj for any i!=j, then it is obviously sufficient to order the set of points by arranging them in ascending abscissa and dividing it in half. If a is the largest abscissa in group 1 (with smaller xi), and b is the smallest in group 2 (with larger xi), then by choosing some a<x0<b and drawing the line x=x0 we obtain the solution.
If we still find xi=xj for some pair(s) i!=j, then we apply the following method. Introduce a coordinate system x'Oy' with the same centre, but rotated about it by the angle alpha. The abscissas of the points are transformed by the law xi'=xi*cos(alpha). Gradually changing the angle alpha from 0 to 2pi, we will from time to time obtain coincident abscissa in the new coordinate system. The set of all non-empty subsets of points with power greater than 1 (i.e. the set of variants of their abscissa xi') is finite, hence finite is the mapping to the set of all angles alpha corresponding to the given matches. However, since the set of all rotation angles is known to have the power of a continuum, we can say that there exists an alpha=alpha0 such that at no pair of points the abscissas coincide. In this case the construction described in the first part of the solution is possible.
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I would add that the condition that no three points must lie on the same line is not used in the proof, and therefore it is not essential. In fact, it is sufficient that the points are simply pairwise different.
Shit. I didn't think too much about the finiteness of the set of lines...
А так не прокатит -
It might work... I'll have to rebuild the solution. I'll have time to draw.
___________________
It'll work :) But it won't be easy to prove :) . But... It's worth a shot.
Then the problem is transformed to this: prove that this triangle has the maximal area of all those inscribed to this figure.
Окружности расположены именно так и не иначе?
the radius is arbitrary, so it may be different
I've already written the solution, see earlier. Richie doesn't want to feel happy about it, so be it.
2 TheXpert: In the three circles problem, is the geometric solution necessary? Or is an analytical one enough?
Да я уже написал решение, смотри чуть раньше. Richie не хочет ощущать радость, ну и ладно.
2 TheXpert: в задаче о трех окружностях - геометрическое решение обязательно? Или достаточно аналитического?
It is unlikely that the analytical one exists. The geometrical one doesn't have to, it's easy there, you just need a proof.
Вот это задачка, так задачка - Польский ученый доказал, что Бог существует. Цитата - "Геллер разработал сложную формулу, которая позволяет объяснить все, даже случайность, путем математических подсчетов".
The formula in the studio,
we don't accept in ex4.
although ...for sure the fit