Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 202
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You can't do that. A ruler can only connect 2 points - draw a line through them. A compasses can draw a circle through 2 points. These are different tools.
A ruler may only be able to connect 2 points, but in the right hands it can easily be transformed into a compass.)
I hope the ruler from the problem has a right angle, otherwise my whole construction falls apart )
a ruler may only be able to connect 2 points, but in skilful hands it easily turns into a compass)
I hope the ruler has a right angle from the problem, otherwise my whole construction is falling apart )
the link to the problem says only hardcore straight lines...
I'm having a hard time withthe highlight- I don't understand why?
Yeah - there's a solution to that point:
Mathemat: Эта 5 делит большое основание пополам,
MigVRN:
I'm stuck withthe highlighted one- I don't understand why?This is one of the properties of a trapezoid. Смотри вики http://ru.wikipedia.org/wiki/%D2%F0%E0%EF%E5%F6%E8%FF#.D0.9E.D0.B1.D1.89.D0.B8.D0.B5_.D1.81.D0.B2.D0.BE.D0.B9.D1.81.D1.82.D0.B2.D0.B0, свойство 6.
I see that you have already found it.
P.S. By the way, I don't like the first proof at all: the halving of one of the bases applies as something that is already given. But the halving must be proved for both bases simultaneously: it can turn out that the line passing through points O and Q divides the bases not in half, but in equal proportions.
I haven't got into the second one yet. But it seems to be the same shit, just in a different sauce.
In short, both proofs prove the following: if the intersection points of side extension and diagonal intersection of a trapezium, and the midpoint of one of the bases lie on one line, then the midpoint of the second base also lies on the same line. But this is not identical to the statement of the theorem.
P.S. Can you point me to the resource where this "proof" is posted?
P.P.S. I was wrong. At least the first proof is correct.
A brutal problem (for those wishing to learn how to correctly generalise solutions):
There are12 candles in a magical candlestick, arranged in a circle. Some of them are lit. The magic is that if one candle is lit or extinguished, the two adjacent candles will also change their state: the unlit ones will light up and the burning ones will go out. A position is considered "divine" if one can get a completely burning set from it. Otherwise, it is "diabolical".
1) Specify an arithmetic way to distinguish between divine and devilish positions.
2) If B is the set of all divine positions and D is the set of all diabolical positions, then which is greater: B or D? // substantiate. positions translated one into another by rotation are considered the same.
HELP: In the trailer there's a button engine on Excel, which will simplify your search for a solution // there's a solution implemented, but it's coded, so you can't peek :)
--
Note. I've already written here, that for the number of candles divisible by 3 the solution doesn't always exist. But when I tried to find the solvability condition for a multiple of 3, my brain went crazy. To my surprise the solution turned out to be not easy at all (at least for me) and I had to throw away several quite plausible hypotheses before I managed to find the right solution.
A brutal task (for those wishing to learn how to summarise solutions correctly):
What a pervert. OK, I'll think about it.
If I find and justify the solution, maybe I should post it on the same resource as a sequel to the original 13-candle problem.
What a pervert. OK, I'll think about it.
:)
I'll explain my motivations, why I, in fact, stuck to the girl's problem: recently I am strongly interested in the topic of solvability/insolvability. This is after I discovered that clarifying restrictions and degrees of freedom of any system greatly increases my ability to "industrially exploit" it... ;)
If I find and justify the solution - maybe I should post it on the same resource as a sequel to the original problem about 13 candles?
No problem.
I also added there: ... // justify. positions translated one into another by rotation are considered the same.
P.S.: As it turns out, the condition "positions which are transferred one to another by rotation are counted the same" is a complete nightmare. But let it stay... // as if to make life easier... :) :)
But here I will also add a simpler question:
Let's consider positions that are "magically" translated into each other as belonging to the same "magic class".
3) How many magic classes are there in total ? 3a) What is the ratio of their sizes ?
Mathemat:
OK - figured out how - I'll post the solution with pictures later...
No way - it was a false path :) No solution yet...
I also added there: ... // justify. positions translated one into another by rotation are considered the same.
P.S.: As it turned out, the condition "positions translated one into another by rotation are counted the same" is a complete nightmare. But let it stay... // as if to make life easier... :) :)
But here I will also add a simpler question:
Let's consider positions that are "magically" translated into each other as belonging to the same "magic class".
3) How many magic classes are there in total ?
Well... you haven't said everything.
There's also 'reflected in the mirror'. You seem to categorise them as different classes, I would categorise them as one. Anyway, it's a matter of taste. You may have to recall geometry with its equivalence transformations.
And if we generalize, then not only by modulo 3, but by any prime. But that would be too much... The main question is still the first one.