Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 55
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At thirty degrees : 10/sqrt(3)+2.5+5+3.1415*5*7/6 = 31.59891936
So it fits.
Yes, something is wrong, too different from 32.
The problem is deliberately made to gradually come to a value a little less than 32. And here it comes out as much as 400 meters less.
At 30 degrees, I have this:
BH = 5*(1-sin(x)).
2. Angle ABH is also equal to x, so
AB = BH/cos(x) = 5*(1-sin(x)) / cos(x) = 5 * s1, where
s1 = (1-sin(x)) / cos(x).
3. OA = sqrt(AB^2 + OB^2) = 5*sqrt( s1^2 + 1 ).
4. The length of the arc on the circle is equal to
S_circ = 5 * ( pi + x ), hence the full path
S = S_circ + (OA + AB + KL) =
= 5 * ( pi + x ) + 5 * (s1 + sqrt(s1^2 + 1 ) + 1 ).
Computer calculations show that the minimum of this function is precisely observed at x = pi/6 (i.e. 30 degrees) and is equal to
S = 5 * ( 7*pi/6 + sqrt(3) + 1 ) ~ 31.986211.
All the way is OAVKL.
Yes, something is wrong, too different from 32.
The task is deliberately designed to gradually come to a value just under 32. And here it's 400 meters less.
At 30 degrees, that's what I have:
The function of x is highly non-linear and complicated.My correction = 31.9856707 = 15 / sqrt(3)+5 + 3.1415*5*7/6 == (5+2.5) / (sqrt(3) / 2) + 3.1415*5*7/6 + 5
This is at 30 degrees
Sorry, of course.
Your "tasks" ("1+1=2", "A and B sat on a pipe...") - is, as I understand it, some desire to show the participants that they are doing bullshit here and are unable to solve even the simplest problems, which you are cracking like nuts.
It seems to me that your efforts are not very productive. And there's nothing for you to do here, judging by the level of your tasks...
P.S. By the way, your interview with Irishka turned out pretty good.
TheXpert: When will somebody draw mine about 30m? Or draw and pass?
Erm... that seems to be the best way to go. The impression is that the brick has to be thrown every time from the same height as the ball, and at the same time as it, i.e. resonantly pumping energy into the whole system.
Okay, I'm convinced. I've had to unscramble 2048 realities. 1023 of them have had mega-brains drinking beer for a long time.
The remaining 1,025 are still fighting. And only one of those 1025s has an honest coin.
Uh... that seems to be the best way. It seems that a brick has to be thrown every time from the same height as the ball and simultaneously with it, i.e. resonantly pumping the whole system with energy.
No. It's enough to bump into each other, optimum when the brick reaches the ground and the ball only bounces off it.
But one has to throw it several times. How many times is a matter of calculating.
No. It's enough to bump into each other, optimally when the brick reaches the ground and the ball only bounces off it.
But you'll have to throw it more than once. How many times is a matter of opinion.
But here we have to calculate specifically, i.e. take into account the frequency and phase of oscillation of the plane. Or did I overdo it again?
Yeah. It turns out that if the mass of a ball compared to a brick tends to zero, then six bricks thrown from 1 meter is enough.
Only bricks (starting from the second one) have to be shot through with laser gun immediately after impact, so that the ball on its way back could pass through the hole.
Andrei, does your solution use lasers?
(4)
80 megabrains stood in the form of a rectangle 10×8. In each longitudinal row the tallest one was found, and the lowest one was a megamogon with a dog. Then they found the lowest one in each transverse row, and the tallest among them was a megamorg wearing a hat. The question is, who is taller: the megamoggle with the dog or the one with the hat?
(3)
There are two armies of megamogs: pointed and dull-tipped. Each army has 2*N people. Each megabrain has a gun that can kill at most one enemy when fired. Megabrains follow the rules of combat: first shoot the sharp-tipped ones, then shoot the blunt-tipped ones and then shoot the sharp-tipped ones again. After these three volleys the battle ends. Question: what is the maximum number of mega-brains that could have died in this battle? Justify that this number is the maximum.
(4)
At one megashop, there is a megaritual held on the last day of the megachurch: the megastudents go out into the hall and stand around their mega-cupboards, where they keep their clothes. On the first whistle, each student opens his or her mega-cupboard; on the second whistle, the mega-students close the even numbered mega-cupboards (i.e. mega-cupboard numbers 2, 4, 6, etc.). On the third whistle, the megastudents change the door position of every third mega-cupboard (i.e. close it if it was open and vice versa). This happens with mega-cupboards 3, 6, 9, etc. On the fourth whistle, the state of the door of every fourth mega-cupboard changes, etc. There are a total of N mega-students in the mega-school. At the Nth whistle, the megastudent who stands next to the number N megastudent (and only that megastudent) changes the door position of his megastudent. How many mega-cupboards are open after that?
And a reminder of a few problems which were posted here yesterday. Not all of them have been solved.Ага. У меня получилось, что если масса шарика по сравнению с кирпичом стремится к нулю, то достаточно шести кирпичей сброшенных с 1 метра.
The logic is as follows:
After the first impact, as we know the ball bounces at 1/2 the speed of the brick (mass is neglected).
On further impacts it is further accelerated by the velocity of the brick.
i.e. sequence: 1/2 of the speed of the brick, 3/2, 5/2, 7/2, 9/2, 11/2, 13/2, etc.
To get up to 30m, you need to accelerate to sqtr(30)*(the velocity of the brick at the bottom of its trajectory)
This is approximately 11/2