ForumSections

[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 99

New comment

MetaDriver >> Если 4-угольник не квадрат - тогда единственен.

No, he's not. See Candid's post. Both diagonals of a quadrilateral are at equal angles to the sides of the rectangle they connect (they are perpendicular). But the diagonals are equal - so all the sides of the rectangles are equal. So they are also squares.

Next simple problem: given segments with lengths a, b, c. Construct a segment of length ab/c.

Mathemat >>:

Нет, не единственен. Смотри пост Candid'а. Обе диагонали четырехугольника расположены под одинаковыми углами к соответствующим сторонам (они перпендикулярны). Но они равны - поэтому все стороны прямоугольников равны. Значит, тоже квадраты.

Uh-huh. Convinced. :)

In fact, the most interesting thing is probably just now - figuring out all sorts of boundary conditions, degeneracies, etc.

Here is another curious question: can any (arbitrary) convex quadrilateral be "wrapped" in a square? Apparently not.

Of course not. If the diagonals are "highly unequal" (by a sqrt(2) factor or more), it won't work at all.

Mathemat >>:

Конечно, нет. Если диагонали сильно не равны (в sqrt(2) раз или больше), то вообще не получится.


Is there any more of this? :-)

Already there, see the first post on this page.

Mathemat >>:

Конечно, нет. Если диагонали сильно не равны (в sqrt(2) раз или больше), то вообще не получится.

Mm-hmm. It's also a very strong condition. You can weaken it considerably - and it still won't fit.

For example, if the diagonals are perpendicular but NOT equal (even a little) - it won't work.

Mathemat >>:Следующая простенькая задача: даны отрезки с длинами а, b, c. Построить отрезок длиной аb/c.

that's bullshit! a*b/c = Exp(log(a) + log(b) - log(c))

$-)

In principle, the marked points can not only be on the sides of the square, but also on its extensions. That's where it gets really wild.

2 MetaDriver: with a compass and a ruler. The ruler has no measure divisions.

Mathemat >>:

В принципе отмеченные точки могут быть не только на сторонах квадрата, но и на его продолжениях. Вот тут настоящий разгул получается.

Er... that wasn't the deal. Then the old solution won't work. Would you consider it a new problem?

2 MetaDriver: with a compass and a ruler. There are no graduations on the ruler.

That was a joke.

Anyway, it's not that easy. I haven't solved it yet.

1...9293949596979899100101102103104105106...628
New comment