Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 72
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Why is the heating convex? I have it almost straight - but I could be wrong. Well the heat transfer, but it's very small, and does contribute to the concavity, but not the convexity.
TheXpert: У треугольника не может быть несколько центров описанных окружностей.
Can we clarify which of the points in the figure is the centre of the described point (I don't think any of them are)?
Can we clarify which of the points in the figure is the centre of the described point (I don't think any of them are)?
Where all three circles intersect.
It's not the centre of the blue, even with the naked eye you can see...
Right :)
Can you explain which of the points in the picture is the centre of the circumcircle (I think none of them are)?
I first tried to build a proof on the inversion - it did get a triangle and a circumcircle.
The inversion circle is a circle with the centre at the intersection of all three reds. The radius is equal to the radius of each red.
That's just the way out of it:
TheXpert:A triangle cannot have more than one centre of circumcircle.
TheXpert:The statement of the problem is still unclear.
TheXpert: The heating is uniform and the heat transfer is increasing.
Yeah, that's right.
Nope, not half. And please be more specific about the centre of mass. The centre of which lines intersect?
(5) There are three boxes in front of you, one of which contains a candy, and the presenter knows which one. The presenter always tells the truth, but can only answer the question about the falsity or truth of some judgement "yes" or "no". How can you find out where the candy lies by asking the presenter just one question? The presenter's answer is constructed according to the laws of mathematical logic.
(4) Brainiac is shaped as a regular triangle. The inner boundary divides it into two states of equal area. Describe the shape and location of the boundary if it is known to be continuous and has the shortest possible length.
(4) A circle painted in 2 colours, red and blue, is given. 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.
In the beginning I tried to build a proof on the inversion - it did produce a triangle and a circumcircle.
The inversion circle is a circle centred on the intersection of all three reds. The radius is equal to the radius of each red.
Let A, B, C be the centres of red circles. The elementary proof is that DK is the median perpendicular in the triangle ABC (this follows from the fact that AODB is a rhombus). Similarly, EL and FM are median perpendiculars to the other two sides. According to the known property, all these three perpendiculars must intersect in the centre of the circumcircle, which is unique. But this does not yet prove that the point of intersection is exactly point O, i.e. that there is no small curvilinear triangle in its place instead of just a point.
My proof drawing is a little more complicated. But it is also drawn in Paint. I don't show the text yet.
I didn't use inversion in the proof, although I really wanted to.