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[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 210

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There's an old geometry problem here:

How do you divide an angle into three equal parts using a compass and a ruler?

xeon писал(а) >>

There's an old geometry problem here:

How do you divide an angle into three equal parts using a compass and a ruler?

http://rutube.ru/tracks/884542.html?v=990340ca393c92a01c1a1bd4f9b900be&autoStart=true&bmstart=0

Trisection of an angle
(from Lat. tri-, in compound words, three and sectio, cutting, dissecting), a problem of dividing an angle into three equal parts. At first, they solved it by means of the simplest geometrical tools - compass and ruler (without division, considered as a tool for drawing straight lines). This was possible, however, only in some cases (e.g. for angles of 90° and 90°/2n, where n is a natural number). The strict proof of impossibility of exact T. y. in general case by means of a compass and a ruler (i.e. insolubility in quadratic radicals of the cubic equation to which T. y. is reduced) was given only in the 19th century.
Richie >>:

http://rutube.ru/tracks/884542.html?v=990340ca393c92a01c1a1bd4f9b900be&autoStart=true&bmstart=0

Wapchette articulated attachments are often called for in construction tasks.

Why the hell don't they come in standard preparatory sets? Probably to make life easier for students. :)

That's what the protractor and other moulds are for.

Maybe some Germans make articulated ones, how do you know?

Mathemat >>:

Может, шарнирные какие-нибудь немцы и делают, откуда ты знаешь?

I don't think so. We would have known about it. You could Google it at your leisure.

Probably a good tradition from the time of Archimedes. :)

I typed in "ready-made" into Yandex-in-pictures. For a long time I stared blankly at the sets, familiar from my childhood, with no sign of ideological progress...

Finally found one cool picture in a row. (See above) I felt less regret for the time spent... :)

About the trisection ...

1. Find A'

2. Find the centre of BA' and CA'.

3. Draw straight lines from A to the midpoints of BA' and CA'.


Seems to be equal ;)

MaStak >>:

Насчёт трисекции ...

1. Найти A'

2. Найти середину BA' и CA'

3. Провести прямые из A в середины отрезков BA' и CA'.

.....

Вроде равны ;)

Good one! ;)

Now make the original angle close to the deployed angle (160 degrees) and repeat the feat, please... :)

// Hinges rule, after all.

The unfolded one can be broken into sharp ones )

Just make sure there are more circles.

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