Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 200
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Another one, quite practical.
The terror of the Megamogg village by the damned occupiers continues. This time, having caught Megamogg, the occupiers gave him an ordinary full bottle of water and a carbon ruler, demanding that he count the volume of the bottle, otherwise death. Megamraz examined the bottle carefully: it was not shaped, flat, flat-bottomed, with no label. He performed a few actions and gave an answer. How had he managed it?
Weight - 3.
FAQ:
- What an angle piece is, I hope it's clear to most people. It is a ruler in the form of a right triangle with divisions on the cathetuses,
- the walls of the bottle are very thin, so you can ignore the volume,
- the bottle comes with an airtight cap (such as a cork),
- at first, the bottle is filled to the brim with water. The water can be poured out, but the poured water cannot be used again,
- the neck of the bottle may have an arbitrary, very nasty shape - for example, this (this is my drawing of the whole bottle in my own solution of the problem):
A trapezium (arbitrary) is given. How can one ruler (without divisions) divide the bottom base of a trapezoid into 3 equal parts?
MigVRN, please explain.
I understand it this way:
- Draw diagonals 1 and 2 of the large trapezoid,
- Then construct the extensions of sides 3 and 4, and from their intersection point construct 5 through the intersection point of the diagonals. This 5 divides the large base in half,
- draw 6 and 7,
- and now what? Prove, for example, that point 8 divides half the base of the original trapezium in the ratio 2:1.
MigVRN, clarify, please.
I understand it this way:
- Draw diagonals 1 and 2 of the large trapezoid,
- Then construct the extensions of sides 3 and 4, and from their intersection point construct 5 through the intersection point of the diagonals. This 5 divides the large base in half,
- and then we draw 6 and 7,
- Now what? Prove, for example, that point 8 divides half of the base of the original trapezium in the ratio 2:1.
It seems to me a very ancient problem, literally, it's better to wield a string.
I'm up to the star of David ))
MigVRN, please explain.
I understand it this way:
- Draw diagonals 1 and 2 of the large trapezoid,
- Then construct the extensions of sides 3 and 4, and from their intersection point construct 5 through the intersection point of the diagonals. This 5 divides the large base in half,
- draw 6 and 7,
- and now what? Prove, for example, that point 8 divides half the base of the original trapezium in the ratio 2:1.
You can do it with a string, and you can do it with a ruler, too.
It comes down to the construction of an equilateral triangle.
I think it's a very old problem, literally, it's better to use twine.
I'm up to my eyeballs in David's star ))
You can divide any segment into any number of equal parts with a whip (compass) and a straight stick :)
Any segment can be divided into any number of equal parts with the help of twine and a straight stick :)
please divide by 3 )
The twine and the compass are gone, my solution is by reducing it to a right triangle.
I'll have to think about the proof, but it's 100% divisible - I measured it in AutoCad :) The whole tricky part is having a straight line parallel to the bottom base.
Yes, that's the main fuss, I understand.
Did you google this drawing by any chance?
A drawing without a justification is not a solution.
Yes, that's the main difficulty, I understand.
You didn't google this drawing, did you?
No - I drew it myself - by connecting all the possible points in sequence and cutting off the excess :)
A drawing without justification is not a solution.