Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 74

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There will be a sector of a circle. How to prove it I don't know, but part of a circle definitely has a shorter length than a segment.
 
Mathemat:

(4) Given a circle, coloured in 2 colours - red and blue. Prove that no matter how exactly it is coloured, it is always possible to inscribe an isosceles triangle into it such that its vertices are the same colour.

Suppose this is not the case. Find on the circle the points 1 and 2 of the same colour, let it be red. Let us draw a line perpendicular to the chord 1-2 through its midpoint. It passes through the centre of the circle and intersects it at points 3 and 4. Since triangles 1-2-3 and 1-2-4 are isosceles, points 3 and 4 are blue. Draw the diameter 5-6 which is perpendicular to the diameter 3-4. Triangles 3-4-5 and 3-4-6 are isosceles, hence points 5 and 6 are red. We draw chords parallel to 3-4 through points 1 and 2, and obtain points 7 and 8 at the intersection with the circle. Triangles 1-5-8 and 2-6-7 are isosceles, hence points 7 and 8 are blue. However, now in the isosceles triangle 4-7-8 all vertices are blue, which cannot be. Come to the contradiction, the problem is solved.


 
ilunga:
IMHO it's not going to be straight =) and it can be proven without being tedious at all
i'll try to make a proof... Just in case, I'll prepare a plate of ashes)))
 
alsu:
I'm going to try and make a case for it... I'll prepare a plate of ashes just in case)))
Compare at once with the arc. I solved this problem once.
 
TheXpert:
Compare at once with the arc. I solved this problem once.
Compare, the arc is longer)))) can you draw a schematic picture, because I don't understand the thought process
 
alsu:
Suppose that's not the case. Find points 1 and 2 of the same colour on the circle, albeit red. Let us draw a line perpendicular to chord 1-2 through its centre. It passes through the centre of the circle and intersects it at points 3 and 4. Since triangles 1-2-3 and 1-2-4 are isosceles, points 3 and 4 are blue. Draw the diameter 5-6 which is perpendicular to the diameter 3-4. Triangles 3-4-5 and 3-4-6 are isosceles, hence points 5 and 6 are red. We draw chords parallel to 3-4 through points 1 and 2, and obtain points 7 and 8 at the intersection with the circle. Triangles 1-5-8 and 2-6-7 are isosceles, hence points 7 and 8 are blue. However, now in the isosceles triangle 4-7-8 all vertices are blue, which cannot be. We come to a contradiction, the problem is solved.

It's beautiful. But it's complicated. It's funnier on the menu. Decorate any one-colored arc with three dots, two at the edges and a third in the middle. Connect them with straight lines. You get an isosceles triangle.)

// Don't tell me all the arcs are infinitesimal, I'll split them all in half anyway. ;-)

 
alsu:
I compared it, the arc is longer)))) can you draw a schematic picture, because I don't follow the thought process
ah, no need, I imagined it)))) maybe so, maybe so
 
MetaDriver:

It's beautiful. But it's complicated. The menu is funnier. Let's decorate any one-color arc with three dots, two on the edges and a third in the middle. Connect them with straight lines. We get an isosceles triangle.)

// Don't tell me all the arcs are infinitesimal, I'll split them all in half anyway. ;-)

I'll colour it this way: I'll mark the starting point and go clockwise with arcs of 1 radian, marking red-blue-red-blue-... Owing to irrationality of pi there will be irrational number of segments in a circle, hence the whole circle will be painted in infinite time, and for any two points of one colour there will be a point of another one which is between them. In other words, this method of colouring does not allow "any one-coloured arc" because there are none. (Somehow this construction is similar to "cantor dust", imho)
 
alsu:
I'll paint this way: Mark the starting point and go clockwise with arcs of 1 radian, marking red-blue-red-blue-... Because of irrationality of pi there will be irrational number of segments in the circle, hence the whole circle will be coloured in infinite time, and for any two points of one colour there will be a point of another one which is between them. In other words, this method of colouring does not allow "any one-coloured arc" because there are none. (Somehow this construction is similar to "cantor dust", imho)
Your coloring will be hole-like (you can prove it) - it will have unpainted points, which contradicts the conditions of the problem.
 
alsu:
I'll paint this way: I'll mark the starting point and go clockwise by arcs of 1 radian, marking in turn red-blue-red-blue-... Because of irrationality of pi there will be irrational number of segments in the circle, hence the whole circle will be painted in infinite time, and for any two points of one colour there will be a point of another one which is between them. In other words, this method of colouring does not allow "any one-coloured arc" because there are none. (Somehow this construction is similar to "cantor dust", imho)

Refutation:

Let's draw two arcs of length Pi/3 of radian from any point on the circle "coloured" by this "method" and at the same time let's construct an isosceles triangle on these points (its two sides lengths will be equal to R). :)

Obviously, only one corner of it is in the shaded point (the reverse contradicted the statement about irrationality of Pi). So, as it turns out, there are at least twice as many holes on this circle as there are shaded points. :))

// What's in inverted commas read with a snide tone.

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