Remember the bullet problem from two weeks ago? If not, I’ll give you a few moments to refresh your memory.
Okay. Are we ready?
The explanation I liked best was Jeffrey’s. Let me try to illustrate it. (Warning: There’s no way, I think, to make this instantly clear. It will take a little work to understand it. Only you can assess the opportunity cost of your own time!)
The gray rectangle is the room you’re in.
The blue dot is you. The red dot is the shooter.
With mirrors on all the walls, you’ll perceive yourself as standing in an infinite grid. The graph shows 16 of the grid rectangles, but the pattern continues forever in every direction.
You’re standing at some point (a,b). The shooter is at some point (p,q). You’ll see copies of that shooter at every point with coordinates (2m ± p, 2n ± q) where m and n range over all integers. Sixteen of those infinitely many points are shown in the picture.
Any shot that hits you will appear to come in a direct line from one of these shooters. For example, it might appear to come from the point (4+p, 6-q). The midpoint of that bullet’s trajectory is (2 + a + p⁄2, 3 + b – q⁄2). More generally, the midpoint of any trajectory is at (m + a ± p⁄2, n + b ± q⁄2) for some integers m and n.
Now if m is even, then points with horizontal coordinate m + a ± p⁄2 look exactly like points with horizontal coordinate a ± p⁄2. If m is odd, then points with horizontal coordinate m + a ± p⁄2 look exactly like points with horizontal coordinate 1 + a ± p⁄2. Likewise with vertical coordinates.
To sum up, the midpoint of any trajectory is of the form (m + a ± p⁄2, n + b ± q⁄2), and the location of this point in the original room depends on four things: the parity of m, the parity of n, and the choices for the two ± signs. There are exactly 16 ways to make those four choices, hence exactly 16 possible midpoints for the bullet’s trajectory. Those 16 midpoints are where you want to put your punching bags.
In other words: Draw straight lines from the 16 black dots in the picture (including the black dot that the red shooter is standing on) to your blue self. Mark the midpoints of these 16 lines, reflect them over into the gray rectangle as needed, and that’s where you place your punching bags.
I think it’s also reasonably clear that (for a shooter in general position) there’s no more economical way to block all 16 of these basic trajectories, so that 16 really does solve the problem.
But why are you spending time on recreational puzzles like this one? Yesterday’s puzzle actually matters. Get to work!
Note: The first version of this post contained an error, which Ken B pointed out in comments. This is a corrected version, which I hope I’ve now got right!
Follow up note: The second version contained a more minor error, which AMarchant pointed out. That’s fixed now too.