[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 92
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And I'm trying my ass off.
I've taken as a special case the points in the corners of the square (like aspire)
so far no luck.
There's something not right about this problem.
I'm getting confused...
Начинаю подсказывать?
Yes, you can. Just a little bit at a time.
OK. There is no need to draw circles as additional constructions.
(Correction -- not necessary. It's nicer and clearer to build vertices with, but that's for later... and it has little effect on the solution)
Я правильно понял, я рисую квадрат, ставлю точки на сторонах,стираю квадрат,даю Вам, Вы восстанавливаете именно мой квадрат?
Exactly, except when the dots form a square -- then it's unlikely that yours will work.
In principle, we are already well advanced and have drawn circles on the sides of the quadrilateral. It remains to find the right starting point of one circle to start drawing from in order to get an exact square.
It is very easy to obtain the analytic condition of such a square. It will involve the segments of the two adjacent sides of the quadrilateral and the corresponding angles. It is sufficient to equate the adjacent sides of the resulting rectangle with each other. I got it and analyzed it for the "degenerate case" specified by TheXpert. Yes, that's right: whichever point of the circle we start from (or at whatever angle to the side of the quadrilateral-square), the restored figure will always turn out to be a square too.
In principle, the analytic expression (equation) itself, which determines the angle at which the first side of the original square must be built, allows the construction of the original square directly. But the expression itself is very ugly. I would certainly like something more elegant.
Perhaps a hint is in order.
Очень легко получить аналитическое условие такого квадрата ....
So you were looking for a parameter for the graphical construction? I immediately wrote out equations for coordinates of vertices of the square (it is enough to find three, so the unknowns are 6), but I did not solve them, it is really not interesting. But if compare my "analytical" post with yours, it is clear that we used essentially the same elements: three circles, two sides and the condition of their equality :)
There is a simple graphical way of constructing rhombuses on four points, there will be an infinite number of them. Successive constructions of different rhombuses will give some trajectories of their vertices, the intersection of these trajectories with the same circles will just give the vertices of the square. But it can be considered a graphical solution only if a graphical way of drawing these trajectories is specified. Maybe there is one, I just did not do these constructions. Both due to the lack of a circular and, apparently, due to the lack of motivation.
Generally speaking, the problem boils down to this: there is a quadrilateral. You have to draw two parallel and perpendicular lines through both pairs of opposite vertices to make a square.