[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 252
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Mathemat писал(а) >>
A reconnaissance aircraft flies in a circle of radius 10 km, centred at point A, at a speed of 1000 km/hour. At some point, a missile with the same speed as the aircraft is launched from point A and controlled so that it is always on the straight line connecting the aircraft with point A. In what time will it catch up with the aircraft? 69
This problem also lacks initial data. If we consider the plane and the rocket as mathematical points, then the rocket will never catch up with the plane, though it will approach it at ever closer distances (until it runs out of gasoline). (This is the case when Zeno rules:). If the moment of 'meeting' is considered to be approaching at some finite distance between their centres (L), then the 'meeting' will happen in a finite amount of time (T). I.e. T=f(L). What function exactly, I will not say, only it is clear that at L tending to 0, T tends to infinity.
There is enough data. Olympiad problem, the solution is given.
P.S. A rocket does not have to have the same tangential speed as an aeroplane.
Давайте задачку оформим по уму ?
What's there to do?
The first one moves until it hits a wall, then stops.
The second one also goes to the wall, and then bypasses the wall.
The trajectory is optimal.
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About the plane. After a few simple calculations:
dS/dt = v*sqrt(1 - S^2/R^2) - all we have to do is to take the integral and substitute the numbers.
P.S. Ракета не обязана иметь ту же тангенциальную скорость, что у самолета.
So she can't (without violating the conditions of the problem).
TheXpert, if they are deaf-blind and don't communicate, they won't be able to know which one of them is standing and which one is moving after meeting the wall. The task is pointless.
With the rocket you have a very similar solution. The missile flies on an arc of a circle with half of its diameter and meets the plane when it has flown a quarter of its circumference (the missile flies half of its circumference). The answer is Pi/200 hours, i.e. about a minute.
TheXpert, если они слепоглухонемые и не общаются, они не смогут узнать, которому из них стоять, а которому двигаться после встречи со стеной. Задача бессмысленна.
If they know the problem, then the letters can be assigned :). In any case, in the one-dimensional plane there is a better chance of finding it.
With the rocket you have a very similar solution. The missile flies on an arc of a circle with half its diameter and meets the plane when it has flown a quarter of its circumference (the missile half its circumference). The answer is Pi/200 hours, i.e. about a minute.
Yeah right, but head-on... Man, I knew there was an ambush and an easier way. And how do you prove that a trajectory is a semicircle?
Four beetles, A, B, C and D, sit in the corners of a square with a side of 10 cm (Figure 51). Beetles A and C are male and B and D are female. They start crawling simultaneously: A to B, B to C, C to D, and D to A. If all the beetles crawl at the same speed, they will describe four identical logarithmic spirals that intersect in the centre of the square. What distance will each beetle crawl before they meet?
Four balls can be placed so that each one touches three others. Five coins can be arranged in such a way that each coin will touch the other four
Can six cigarettes be arranged so that each one touches the other five? Cigarettes can't be bitten :)
Seven?
A rectangular triangle is inscribed in a quarter circle as shown in Fig. 48. Using only the data given in the drawing, can you calculate the length of hypotenuse AC? You have one minute to think about it!
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Two maths buddies meet:
- How are you, how are you doing?
- Everything is good, two preschoolers' sons are growing up.
- How old are they?
- The product of their ages equals the number of pigeons near this bench.
- This information is not enough for me.
- The older one looks like his mother.
- Now I know the answer to your question.
How old are the sons? (The answer is logical and unambiguous)
Is it possible to arrange the six cigarettes so that each one is in contact with the other five?
The hypotenuse is 12. But thought for more than a minute.
Score. Really could be easier. How about seven?