How many punching bags do you need to survive a bullet ricocheting in a room?

[BobG]

Spoiler

Five.

You don't want the shooter to shoot directly at you. Other than directly at you, there are only four directions the shooter can shoot and hit you.

Rebounds from the wall come off the wall at the same angle they approached the wall. The sum of the angle on the side wall plus the end wall has to equal 90 degrees. (In other words, if the bullet approached and left the side wall at 30 degrees, it will approach and leave the end wall at 60 degrees.

That means a shot will not cover the entire area of the rectangle no longer how many times it bounces. You could select an angle that will cover a huge percentage of the rectangle, but, it can never cover the entire rectangle and there are still only four possible directions that you can be hit from.

Essentially, you could consider the rebounds as the mirror image of an infinitely expandable rectangle and the path of the bullet could eventually be expanded into a large parrellelogram that constantly repeats.

And, because the departure from the wall has to equal the approach to the wall, there's only four possible indirect directions the shooter can shoot that will actually hit you. And four instead of two because you also have to account for your mirror image on the other side of the shooter (in other words, the bullet can actually travel two different directions along the same line).

 

[Jimmy Snyder]

But a bullet might ricochet many times without ever striking 2 adjacent walls in succession.

 

[BobG]

Wouldn't matter. All distances would be expanded by the same factor until the size of the rectangle and the distance between the two persons were large enough for the parallelogram to fit. You could alternatively calculate each of the rebounds for cross bounces, but the paths of alternate bounces have to be parallel to each other and we're only worried about direction - not which bounce hits him.

 

[Jimmy Snyder]

Wouldn't matter. All distances would be expanded by the same factor until the size of the rectangle and the distance between the two persons were large enough for the parallelogram to fit. You could alternatively calculate each of the rebounds for cross bounces, but the paths of alternate bounces have to be parallel to each other and we're only worried about direction - not which bounce hits him.

You are not allowed to expand distances. The shape of the room and the placement of shooter and yourself are fixed before the bags are placed. By ricocheting off of oppostite walls, the bullet may approach you from infinitely many different angles.

 

[BobG]

You are not allowed to expand distances. The shape of the room and the placement of shooter and yourself are fixed before the bags are placed. By ricocheting off of oppostite walls, the bullet may approach you from infinitely many different angles.

Expanding the dimensions is just to find an easy way to calculate the direction the shooter has to fire in order to hit you. Once shot, the bullet only travels in four directions over and over and over. All the parallel paths can be treated as a single path.

But six or eight might be the right answer (five definitely is wrong since I made an assumption that might not be true). You have two opposite directions along each side of the parallelogram, plus directly at you, which is the sum of the vectors forming the two sides, plus directly away from you provided the line formed by you and the shooter isn't perpendicular to the wall (the bullet actually has to hit you before it hits the shooter).

I think the difference of the two vectors might hit you, as well, with there being two different differences since subtraction isn't commutative (in other words, you can travel two different directions along the difference, as well). But the problem with the difference idea is that the shooter and you have to be at opposite corners of whatever parallelogram you form. Intuitively, I think the difference is guaranteed to miss you.

Keep in mind the possibilities are restricted by the fact that the angle of approach to a wall has to equal the angle of departure. The bullet can't approach the wall at a 30 degree angle and then bounce off at a 50 degree angle, so really aren't an infinite number of parallelograms that intersect your point. You have to toss out all of the parallelograms that can't possibly occur, and that turns out be almost all of them.

So, for example, if I had a 10x8 rectangle, and the shooter were located at coordinates 1,2 and I were located at coordinates 8,5; then any combination of vectors that added up to 7+3i or -7-3i (or some scalar multiple that gives me the same directions) would hit me. That's an infinite number of vector combinations.

For example, the bullet travels 2+5i, bounces off at 3-9i, and then bounces off the other wall at 2+7i. The bullet hits me.

The lengths of any parellel vectors can just be added together as if they were one vector, since they're traveling the same direction. Additionally, for each vector I add together I also have a vector along the other side that has to be added to the other side of the parallelogram. I don't care that adding them together gives me a fictional vector that travels beyond the rectangle. That still leaves me with an infinite number of vectors that can hit me.

For example, the bullet travels 2+5i, bounces off at 1-18i, then bounces off at 4+10i. It's the same result as if you just added the two parallel vectors together (6+15i), creating a two vector solution of (6+15i) + (1-18i).

However, only vectors where the approach angle and the departure angle will be equal will actually occur. There's an infinite number of vector combinations that don't meet that requirement, and once I've subtracted out the infinite number of vectors that don't meet the requirement, only six remain that meet both requirements.

In other words, if the bullet travels a 2+5i, it isn't going to bounce off at 3-9i or 1-18i. It's going to bounce off at 2-5i (or some multiple, such as 4-10i). And once it finally reaches the side wall, it's going to bounce off at -2+5i or -2-5i (depending on which direction the bullet was traveling when it hit the side wall). So only those vectors that can meet that requirement, plus add up to 7+3i are valid solutions. (2+5i is not a valid solution for the dimensions and locations of the shooter and me).

 

[256bits]

If off an angle of reflection of 45 degrees the bullet hits you, then the bullet will also hit you with an angle of reflection of 22.5, 11.25, ... All the angles can never be covered.

 

[BobG]

If off an angle of reflection of 45 degrees the bullet hits you, then the bullet will also hit you with an angle of reflection of 22.5, 11.25, ... All the angles can never be covered.

The numbers are off unless both the shooter and the target are against one of the walls, but the principle is correct.

After playing with it a little bit, I think this is better handled as a wave problem(except with a sawtooth wave instead of a sine wave), which is sort of what you did with your solution. But, just as with a sine wave, you could get an infinite number of angles approaching the target by varying the wavelengths and the amplitudes (and the wavelengths that aren't harmonics of the shooter-target wavelength just miss and miss and miss).

But, yes, there will be an infinite number of harmonic wavelengths.

(Too bad. That seemed so cool to realize that every solution was just a parallelogram that folded over whenever it hit a wall, which it is, I guess, just with an infinite number of parallelograms.)

 

[mathal]

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 anyone of the infinitely many paths passes through at least one of the points in the finite set?

As the puzzle is two dimensional and all the 'pieces' are 'points', in the corner, 2 points, in the 'outside' points of a square would preclude any point of the same 'size' reaching you (exclude the diagonal as contact would necessarily be made with one or both of the other bags first.). As long as a regressive analysis of this solution to infinitely small 'points' is allowed this should suffice as a solution.
mathal

 

[lugita15]

As the puzzle is two dimensional and all the 'pieces' are 'points', in the corner, 2 points, in the 'outside' points of a square would preclude any point of the same 'size' reaching you (exclude the diagonal as contact would necessarily be made with one or both of the other bags first.).

I have no idea what you're saying. What is the "size" of a point?

 

[AlephZero]

One way to get a handle on this is to imagine the walls are perfect mirrors, and ask yourself how many images of the shooter you would see.

You know the image in a mirror is the same distance behind the mirror as the object is in front. So plot the 4 images of the shooter in the 4 mirrors, then the images of the images in the 4 mirrors, then the images of the images of the images, etc.

You can then trace the path of the bullet backwards from you an image of the shooter, and reflect it off the walls till it reaches the "real" shooter.

It's fairly obvious that a finite number of bags will only work in special cases, where a infinite number of multiple reflections all "line up" behind each other when viewed from your position.

 

[mathal]

Originally Posted by mathal
As the puzzle is two dimensional and all the 'pieces' are 'points', in the corner, 2 points, in the 'outside' points of a square would preclude any point of the same 'size' reaching you (exclude the diagonal as contact would necessarily be made with one or both of the other bags first.).
As long as a regressive analysis of this solution to infinitely small 'points' is allowed this should suffice as a solution.

I have no idea what you're saying. What is the "size" of a point?

a point can be considered to be a circle of radius 0 at a particular coordinate position. The limit approaching 0 is 1 divided by infinity. If the argument I presented for finite circles in a corner of the room is accepted then divide the radius of the point 'circles' by 2 an infinite number of times.
The 'puzzle' is more philosophical than anything else. When is a point, not a point-when it has names and -properties a point can't posess- such as absorbing a bullet.

mathal

 

[legend_b0bby]

pi, because it is the amount needed to cover the circumference of a point. or 4 because a point cannot be cut into 0.14159265358979...

 

[Whovian]

pi, because it is the amount needed to cover the circumference of a point. or 4 because a point cannot be cut into 0.14159265358979...

The "circumference" of a point (considering a point as a circle with 0 radius) is 0. Also, again, the bags are points. How would 3 punching bags cover 3 units of distance if they're pointlike?

 

[lugita15]

Steve Landsburg, who originally posted the puzzle, now posted his solution:
http://www.thebigquestions.com/2012/06/12/tuesday-solution/