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Bullet Richocheting in a Room |
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| May25-12, 03:54 PM | #18 |
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Bullet Richocheting in a Room .Still, there is an infinite number of directions in which the gunner can shoot to hit me, so I don't see how it can solved with a finite number of bags. But I don't have to be right. |
| May25-12, 03:59 PM | #19 |
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| May25-12, 04:02 PM | #20 |
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| May25-12, 04:32 PM | #21 |
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I will assume that the location of the shooter and my location are off limits for punching bags.
Spoiler
In that case, it is easy to prove that countably many bags suffice. If the bullet is to strike me without ricochet, then it must be aimed at me and I can stop it with a single bag. If it is to strike me after a single ricochet, then there are no more than 4 directions that I need to protect against, one for each wall. I can protect myself with 4 bags. For two ricochets, there are 12 directions, first hitting one wall, then a different wall, then me. 12 bags suffice. For each subsequent number of ricochets, multiply the number of bags by 3. That is to say, a bullet can ricochet off of any wall except the wall it just ricocheted off of, and then hit me. So for any finite number of ricochets, I need a finite number of bags. And since there can only be a countable number of ricochets, there can only be a countable number of directions I need to protect against. That is to say, a countable union of finite sets is countable.
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| May25-12, 04:52 PM | #22 |
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| May25-12, 05:16 PM | #23 |
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| May25-12, 05:18 PM | #24 |
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I know. That's why I deleted it. Not soon enough though.
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| May25-12, 06:07 PM | #25 |
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I think it can be done with
Spoiler
only 7
bags. Just let me know if I'm right and if so I'll provide the proof. Otherwise it's back to the drawing board. Literally. |
| May25-12, 07:17 PM | #26 |
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| May25-12, 07:22 PM | #27 |
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I have an idea, I don't know how useful it may be. In the case where the two side lengths and the cosine of the angle of shooting are all commensurable (meaning the ratio of any two of them is a ratio of whole numbers, or equivalently there is some unit relative to which all three numbers are whole-number multiples), then the path of the bullet will be finite, in the sense that it will go back and hit the shooter eventually.
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| May25-12, 08:09 PM | #28 |
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I'm sorry, there was a flaw in my proof. However, I will prove the following.
Spoiler
If there is a finite number of bags that suffice for one placement of the shooter and yourself in the room, then the same number of bags suffices for all rooms and all placements. Pf. Suppose that a finite number of bags suffice for a particular layout. Draw the layout and call it the base cell. Now tesselate the plane with mirror images of the base cell. Draw a line from the shooter in the base cell to the copy of yourself in each of the cells. Each of these lines when folded and reflected back into the base cell will be the trajectory of a bullet that would have hit you were it not for a bag. Therefore, there will be a bag in its path between its endpoints. For any other configuration of room and people, leave all the copies of the shooter, yourself, and the bags as is, and erase the lines that represent the walls of the room. Now redraw a rectangle to represent the room in such a way that there is exactly one copy of the shooter and one copy of yourself within it and so that the the room has the correct dimensions and the poeple have the correct positions. Then tesselate the plane with the walls of this new room. You may need to fold and reflect bags so that one copy of each bag ends up in the new room. Since no change was made to the trajectories, they still are interupted on their flight from shooter to yourself by a bag and when folded and reflected across the walls of the new rooms they still represent the trajectories of bullets. I thought that I had found a particulary nice symmetric layout of room and people that required 7 bags. However, I was able to find a trajectory that missed all 7 of these bags so I have no proof of how many it takes. However, I think the idea of tesselating the plane makes it easy to look for symmetrical layouts that are easy to work with. |
| May26-12, 06:36 PM | #30 |
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Only one punching bag should suffice iif you can also choose your own location.You would place the bag between you and the shooter so he cannot aim at you directly.
My hint is that it is a case of odd/even numbers, and whether we are working with points or the same discrete size of bullet/punching bag doesn't play into the solution at all. PS. of course the size if the bullet/bag mutilpyied by a whole number must equal the length of a wall which would also be true for the set of points. |
| May26-12, 11:57 PM | #31 |
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The solution is to be provided in 2 dimensions.
Spoiler
Put the punching bag in a diagonal position in a 'corner' of the 2 dimensional room with you behind the bag.
mathal As the puzzle is presented, an infinite number of bags is required even in a 2 dimensional room.-my spoiler is wrong. |
| May27-12, 05:28 PM | #32 |
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Here's a related problem. How many people will post wrong answers without even reading the OP? Hopefully it's finite.
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| May27-12, 07:14 PM | #33 |
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| May27-12, 09:27 PM | #34 |
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