Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 152
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I'll try it again to see if it doesn't suck.
It's easy to see the vertices as being pointed if you draw corresponding semicircles corresponding to rectangular ones. I'll show you the drawing.
P.S. No more doubts. See figure below. If the value of an angle whose value is doubtful extends beyond the semicircle, it is acute. Great, Avals!
The main doubts were about the angles KAL and OAK (and similar ones which are symmetric to them on the right-hand side). See the picture below.
lazarev-d-m: if to pick on the condition of the problem, a right angle is a right angle, not an acute angle, hence, by drawing the diagonals in the square we solve the problem, if not to pick on, then Avals, presented the solution
No, it's not a quibble. A rectangular triangle is always rectangular and not pointed. But the last figure shows that all corners can be made acute in Avals construction.
No, it's not a nag. A rectangular triangle is always rectangular and not pointed.
This is essentially 'two diagonals, but with some epsilon'. You can make the AB segment as close to the centre of the square as you like (but you have to make it smaller too). And then the figure won't be as clear.
P.S. The T-shirt problem just became 5 (a couple of days ago it was exactly 4).
Mathemat:
P.S. The T-shirt problem has just started to weigh 5 (it was definitely 4 a couple of days ago).
Well, it's quite complicated, despite the simplicity of the answer.
Well, yeah, it's kind of complicated. But I haven't got it in yet (haven't looked at it):
Two: the probability is obviously p(2) = 1/2.
N people:
We apply the full probability formula:
P(B) = Sum( P(B | A_i) * P(A_i) ).
Here {A_i} is the complete group of pairwise incompatible events.
a) The newcomer wears the jersey of the First. Everyone else will wear theirs. The probability is 1/N.
b) If the Rookie wears the jersey of the Last, it is an adverse event. Probability is 1/N.
c) The Rookie wears the jersey of neither the First nor the Last. The total probability is 1/N*Sum( p(n), n = 2...N-1).
Hence p(N) = 1/N + 1/N*p(N-1) + 1/N*p(N-2) + ... + 1/N*p(2) = 1/N*(1+p(N-1)+p(N-2)+... +p(2)) =
= 1/N*(1+p(N-1)) + 1/N*(p(N-2)+...+p(2)) =
= 1/N*(1+p(N-1)) + (N-1)/N * (1/(N-1)*(1+p(N-2)+...+p(2)) - 1/(N-1)) =
= 1/N*(1+p(N-1)) + (N-1)/N * (p(N-1) - 1/(N-1)) =
= 1/N + 1/N*p(N-1)) + (N-1)/N * p(N-1) - (N-1)/N * 1/(N-1)) =
= p(N-1) = const = 1/2.
Well, yeah, it's kind of complicated. But I haven't had it counted yet (haven't looked at it):
Well you are a giant. I, while trying to write the induction, 5 times completely confused and eventually gave up. Although I knew that it is quite possible and already knew the solution (by hand I counted the probabilities at N=2, 3, 4 and 7 (for the final check)).
;)
I am puzzling over a problem like this.
There is a chart, let it be a candlestick chart for simplicity.
How do I draw a line that crosses as many candles as possible?
The easiest thing that comes to mind is to draw a horizontal line, go through all the values and count the number of crossings, then bend it and repeat.
Stupid, slow, don't like it.
What are your options?
I am puzzling over a problem like this.
There is a chart, let it be a candlestick chart for simplicity.
How do I draw a line that crosses as many candles as possible?
About this very criterion - I am afraid it is not very simple. And sometimes this straight line will be not very similar to a trend line.
But to draw a linear regression line (not a curve, but a straight line) - it is possible.
About this very criterion - I'm afraid it's not that simple. And sometimes this straight line will not be too similar to a trend line.
But to draw a linear regression line (not a curve, but a straight line) - it is possible.
With linear regression, everything is plain and simple. There is no question.
The similarity with the trend line is also unnecessary because there are parts of the chart where there will be more than one such line and possibly with different directions.
My association with such a line is as an analogue of density. Or even the direction of density in a selected area.
All in all it is an interesting task. ;)