## Bullet Richocheting in a Room

 Quote by lugita15 Remember, your location and his location are fixed, and you can put the punching bags wherever you want. If you want to prevent him from shooting directly at you, just put a bag between you and him.
OK, I misread the problem. Positions are known beforehand and fixed, somehow I missed that part .

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.

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 Quote by Borek OK, I misread the problem. Positions are known beforehand and fixed, somehow I missed that part . 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.
The thing is, there may be a finite set of points through which all of the infinitely many ricocheting paths go.

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 Quote by collinsmark The riddle statement specifically said, "This being a math problem, you should treat the room as two dimensional, and yourself, the bullet and the punching bags as points." What that implies is that the gun shooter and bullet (before the bullet is shot, if it's not already inside a bag) can be treated as a single point. The bullet stops the moment the bullet location is exactly equal to to the location of a bag or you. Placing a punching bag at the location of the shooter satisfies the mathematical requirement for stopping the bullet. Think of it as placing the shooter, gun, and bullet inside a single punching bag, if that helps. [Edit: or alternately, think of it as sticking a single punching bag inside the bullet (which happens to be at the exact location as the shooter and gun). In either case, the bullet and bag intercept, from a mathematical perspective. The requirement is satisfied. The bullet is stopped.]
The bullet stops when it HITS a bag, not when it starts from or leaves a bag.
 Blog Entries: 1 Recognitions: Gold Member 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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 Quote by Jimmy Snyder 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.
That's clear enough, but the hard part would presumably be to prove that finitely many bags suffice.

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 Quote by Jimmy Snyder Moreover: If neither I nor my assailant are touching a wall and are not both in the same place, then it is equally easy to prove that countably many bags are necessary. Since we are not in the same place, either our x or y (or both) coordinates differ. Say our x coordinates differ. Then with a single ricochet off of a wall that is parallel to the x direction I can be hit. With two ricochets, I can be hit with a bullet that hits first one, and then the other wall parallel to the x direction, then me. The bullet will make a zig-zag path. And so on with a bullet that alternates ricocheting off of the two walls parallel to the x direction and making a zig-zag path to me. Then for each n, there is a path with n zigs and zags that approaches me with the last zag at a different angle closer and closer to 90 degrees. Each of these paths requires a punching bag and so no less than countably many will suffice.
The thing is, the paths can intersect. So just because you have distinct paths doesn't mean you need distinct punching bags for each path. It may be that there is a finite set X of points such that for any path P, among the infinitely many ricocheting paths, there exists a point in X such that P passes through that point.
 Blog Entries: 1 Recognitions: Gold Member I know. That's why I deleted it. Not soon enough though.
 Blog Entries: 1 Recognitions: Gold Member 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.

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 Quote by Jimmy Snyder 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.
The right answer is not known yet. But it would still be nice to hear what you got.
 Blog Entries: 1 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.
 Blog Entries: 1 Recognitions: Gold Member 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.

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 Quote by TheNewGuys I'm Gunna say 100 punching bag should secure your safety. Just jam them all in there.
The punching bags are all points.
 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.
 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.
 Blog Entries: 1 Recognitions: Gold Member 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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 Quote by mathal As the puzzle is presented, an infinite number of bags is required even in a 2 dimensional room.
What is your justification for this? Just because there are infinitely many paths does not mean that there are infinitely many bags, does it? You don't need one punching bag for each path, because you can place a bag at the intersection point of multiple paths. Isn't it possible that all the infinitely many paths intersect in finitely many points? That is to say, can't there be a finite set of points such that any one of the infinitely many paths passes through at least one of the points in the finite set?

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 Quote by 256bits 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.
Are you saying that you want to be able to choose you're location given the shooter's shooting angle, or do you want to choose your location and the punching bag location BEFORE the shooter decides what angle to shoot at? If the former, then the problem becomes trivial, because you just put the bag in whatever angle the shooter is shooting at. If the latter, how will one punching bag suffice? What if the shooter shoots at you indirectly?
 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.
If we had discrete bullets and punching bags, then the problem could have an entirely different solution, because as TheNewGuys said we can just jam all the bags in there.