[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 95
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the cat doesn't fall down.
>>Sveta's coming over and she'll show you :)
Даю подсказку моего решения:
If I'm not mistaken in the conversion, you have only 7 independent equations for 8 unknowns. So you can build as many rectangles as you want from them, but how are they better than as many rhombuses as you want?
You have to add the condition of equality of sides, and this will lead either to trigonometric functions or to the second order. So the usual analytical solution would be obtained.
Or is there still room for a feat here?
P.S. Yeah, I see, it's already second order.
P.P.S. Yeah, and trigonometry at the same time. Seems to me that one thing is preferable, but maybe that's just the condition of the future focus? We'll have to wait.
If I'm not mistaken in the conversions, you only have 7 independent equations for 8 unknowns.
Already added :)
Уже добавил :)
I don't think it will change anything, the problem is that a and d will always remain as paired sums. That is, no angle in the form a1 = f(b1,b2,...,c1,c2,...) can be obtained from this set, it will always be a1+d3 = f(b1,b2,...,c1,c2,...). That means an infinite number of solutions using only conditions for angles. You can only decouple them by involving equations derived from the conditions for the sides, but there is a trap prepared in the form of trigonometry and/or second order.
Trigonometry and second order are constructed according to the theory of construction with a compass and ruler. What Richie has written is obvious. But there's a much simpler solution, judging by the comments of those in the know. OK, no more hints needed.
Уже добавил :)
it does not solve the problem.
Quote: next problem (boring maths again, Richie). You marked a point on each side of the square and erased the square itself. Reconstruct it.
At least if you take the problem literally.
In my opinion there is no single solution, you can build a lot of squares, with different lengths of sides, if the size of the side is given then there is a chance :-)
Trigonometry and second order are constructed according to the theory of construction with a compass and ruler. What Richie has written is obvious. But there's a much simpler solution, judging by the comments of those in the know. OK, no more hints needed.
There is a simpler solution, maybe even a compass. I remember we solved such a problem at school some time ago, but it was very long ago and I don't remember. But I do remember that it was not a system of equations :)
Есть более простое решение, может быть даже циркулем.
How is it that you don't know the solution after all?
P.S. At school we learned a curious theorem, like this: any construction by a finite number of steps with a compass and a ruler is feasible with a ruler alone - provided that one circle of arbitrary radius with a marked centre is drawn.
And here's more: According to the Mohr-Mascheroni theorem, any figure that can be drawn with a compass and a ruler can be constructed with a single compass. A line is considered constructed if two points are given on it.
You don't know the solution after all?
I gave the solution above: https://www.mql5.com/ru/forum/123519/page94,
but I don't remember and don't know the simple solution, and there it is.
На мой взгляд здесь нет одного решения, можно построить множество квадратов, при этом с различной длинной сторон, если б был дан размер стороны тогда шанс есть :-)
No, in general the conditions for the corners are rectangles, the conditions for the sides are rhombuses, and only their intersection is a square. This is solved graphically, the question is whether the solution is exact or approximate. Here is what I described before will be exact only if you specify a way to build the exact trajectory of the vertices of rhombuses. Without it, the vertices of the rhombus can be brought as close as you want to the geometric place of the vertices of the rectangles, that is, to the circles, but it will be an approximate solution.