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#1 Re: Puzzles and Games » Spaceship and the planet » 2010-08-27 05:59:21
I also got the problem not defined well. I' assumed the ship coming form infinity heading towards the planet. In trying to find the optimal algorithm for finding the inhabitant, I assumed it stopped at the distance d and moved the way I described in post #4.
#2 Re: Puzzles and Games » Spaceship and the planet » 2010-08-27 00:14:26
Assuming that the inhabitant doesn't want to be seen, I've found a way the spaceship can move so to always spot the inhabitant. This gives the lower limit for v/u in this case. I would like to hear your opinion on it. Here it goes...
Set up a spherical coordinate system with the origin at the center of the planet. Let the spaceship be at the distance
from the center of the planet, with angular coordinates and . If the spaceship is point A and the origin is point O, draw a tangent from the spaceship to the planet and call it point B on the planet - there are two of these, any one will do. Now, denote the angle AOB by . Since BOA is a right angle, we have . Now, let the ship circle around the z-axis with the velocity . The time it takes for one complete revolution is . At the same time the ship sweep the strip of the planet of width . So, in order to see the inhabinant, it has to go through all the strips taking care that the inhabitant cannot escape, i.e. the time neccesary for him to cross the strip should be greater or equal to the time of revolution. So . Using the relation above, one has . Since the relation must hold for all , we take the worst case scenario . Then the only free parameter is and we can choose it so to minimize the left hand side of the inequality. Numerically, one gets .Does this look ok?
#3 Puzzles and Games » Spaceship and the planet » 2010-08-19 00:18:14
- Heirot
- Replies: 6
A spherical planet has only one inhabitant that can move freely on the surface of the planet with speed u. A spaceship approaches the planet. Let the maximal speed of the spaceship be 10u. Show that the spaceship can always see the inhabinant regardless of the way he moves.
#4 Re: Puzzles and Games » Square screw & hexagonal key » 2010-08-14 03:17:03
Bob, you said it all! If a wrench can do it, so can a hex key. Now, is your three points solution unique?
#5 Re: Puzzles and Games » Square screw & hexagonal key » 2010-08-13 01:40:08
Hi, Bob,
I agree with you on the concentric case. What's your conclusion on the non concentric case? I think, due to symmetry, the non concentric case must also fall within the range of the inequality you gave.
#6 Puzzles and Games » Square screw & hexagonal key » 2010-08-08 22:57:26
- Heirot
- Replies: 11
Suppose that you have a screw with square head of side
that you want to undo. Unfortunately, the only tool you have around is a key with a hexagonal opening of side . What relations must and satisfy for you to succed in undoing the screw?#7 Re: Help Me ! » Tricky integral of a rational function » 2010-08-08 21:49:05
I agree with you that the indefinite integral would be more challenging to calculate. Because of the limits of integration one can use methods of contour integration or something of the sort that would greatly simplify the calculation. I was hoping that the special choice of the coefficients in the denominator makes that choice possible. E.g. expand the lower limit to minus infinity and then use the semicircle contour. It would pick up singularities in the upper half-plane (which are zeros of the denominator with positive imaginary part), summing their residues to give the integral. Now, since all of the poles are simple ones, I was hoping to express the sum of their residues in the terms of the coefficients of the denominator via Viete formulas. Do you think that could work?
#8 Re: Help Me ! » Tricky integral of a rational function » 2010-08-08 19:11:15
I doubt that Mathematica runs any algorithm because it takes as much time to do this integral as it takes to calculate integral of x. Perhaps this integral if from the tables. How could I check that?
Btw, I know nobody ever would spend a lot of time thinking about this problem, but it's like a puzzle for me, especially when the solution is so elegant.
#9 Help Me ! » Tricky integral of a rational function » 2010-08-08 04:29:36
- Heirot
- Replies: 86
Hi all!
I have a very hard time verifying this integral:
None of the standard methods work because of the irreducibility of both the numerator and the denominator (the denominator can be factored out as a product of two polynomials of sixth degree, but that's not very helpful). On the other hand, Mathematica seems to have no problem evaluating the integral. So I must be missing something. It seems that the coefficients in the denominator are carefully chosen because changing any one of them gives Mathematica a hard time and it doesn't evaluate the integral at all.
Any suggestions will be greatly appreciated 
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