[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 81
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Выбираем в квадрате точку
рисуем две прямые пересекающиеся в этой точке соблюдая правило 2/3
Вопрос - можно ли провести третью прямую через эту точку соблюдая 2/3
навскидку - нет
hee
you can have infinitely many.
Yeah, well, the ninth will always be at that point.
How to prove it beautifully I don't know.
Draw two centre lines in the square (lines connecting the centres of opposite sides of the square). Recall how to calculate the area of a trapezoid through the length of the centre line.
Проведи две средние линии в квадрате (линии, соединяющие середины противоположных сторон квадрата). Вспомни, как вычисляется площадь трапеции через длину средней линии.
Yeah, I get that through the square, I'm just lazy.By the way, the restriction to divide exactly into two trapezoids is not necessary. It just complicates the reasoning a bit, but the answer remains the same. But for now the problem is solved for trapezoids.
P.S. The area of a trapezoid S = 1/2 * h * m, where h is the height, m is the length of the midline. It is the same for a triangle, since a triangle is a special case of a trapezium.
Есть квадрат. Мы пересекаем его 9 прямыми, каждая из которых делит его по площади в соотношении 3:2. Доказать, что хотя бы три из них пересекаются в одной точке.
The impression is that it is easier to refute. Let us define the construction algorithm this way: draw a vertical line dividing the area in the ratio 3:2, let its "lower" and "upper" coordinates be x0 = 0.4*a, here a is the side of the square. Now let us draw another "solved" line through point x0-dx on the basis, it is easy to see that at the top it gets to point x0+dx and intersects with the first one exactly at half-height. Obviously there can be an infinite number of such lines and they will all intersect at one point, exactly (0.4*a, 0.5*a). But since we are doing a refutation, we can take only two lines from this set. Symmetrically, we can get three more such sets, i.e. 6 more lines and 3 more intersection points: (0.6*a, 0.5*a), (0.6*a, 0.5*a), (0.5*a, 0.4*a), (0.5*a, 0.6*a).
Now we are at the culmination, we have 8 lines pairwise intersecting in four points. And we need at least one more "solvable" line, but not falling in any of these points. To do this, we recall that the trapezoid-trapezoid partitioning is not the only variant, there are also 4 triangle-pentagon variants. Let's do this: draw the diagonal of the square and start to move away from it in parallel until the ratio of areas is equal to the sought one. The area of the smaller triangle (isosceles and right-angled) will be (k*a)*(k*a)/2 = 0.4*a*a . We find k and frankly rubbing our hands we see that it is equal to the square root of 0.8. The reason of our gladness is clear, the equation of the straight line passing through points (k*a, 0) and (0, k*a) looks like y = sqrt(0.8)*a - x and because of this remarkable line this ninth line cannot pass through the four special points found earlier
P.S. Eh, so unfair, which means only for trapezoids :). At least now we can see that this restriction is mandatory. And for two trapezoids - yes, there are only four sets, for each of them any line passes through its "central" point and hence any ninth line will fall into the intersection of at least two previously found ones.
You got something wrong, k = 2/sqrt(5) - and generally less than 1, by the way :)
And the case of a triangle with a pentagon is no different from two trapezoids.
You have solved the problem, you just messed up a bit with the richmetic.
P.S. I was wrong too: the case of triangle and pentagon is different. It seems to get 4 points there too, only different. Like (1/sqrt(5), 1/sqrt(5)), (1 - 1/sqrt(5), 1/sqrt(5)), (1/sqrt(5), 1 - 1/sqrt(5)), (1- 1/sqrt(5), 1 - 1/sqrt(5)). Or is it not?
P.P.S. Yes, I fucked up with this case. But never mind.
Что-то ты напутал, k = 2/sqrt(5) - и вообще меньше 1, кстати :)
А случай треугольника с пятиугольником ну никак не отличается от двух трапеций.
Задачу ты решил, просто напортачил немного с рихметикой.
Not out of eight, not out of 0.8. Not with arithmetic, but with grammar :)
P.S. And how did you get your ugly one? k = 2/sqrt(5) :)
P.P.S. I'll correct the solution, so that people won't get nervous for nothing, they will read it earlier
Just like you have the root of 0.8. It's the same thing.
Так же, как у тебя корень из 0.8. Это ж то же самое.
:)
P.S. OK, let's get out of this thread before it's too late.
P.S. Я тоже ошибся: случай с треугольником и пятиугольником другой. Там, похоже, тоже 4 точки получаются, только другие. Или нет?
No, that trick doesn't seem to work there, you get asymmetrical triangles for the increments