Puzzle time - don't waste braindead liberals

[LuFins Dad]

Are the brain dead liberals the same size as Larry? They’re likely vegan, and tiny little things...

[Klaus]

Indeed. So tiny that they have no extension.

[jon-nyc]

Does Larry have to worry about shooting himself?

[jon-nyc]

I’m guessing it’s a countably infinite answer that is related to the primes but I still need to work on it.

[Klaus]

No, it's a very finite number.

[jon-nyc]

Ok. I think I can see that now... more tomorrow

[Klaus]

This puzzle can be solved using high school math and geometry knowledge. Actually, basically no math/geometry is involved apart from knowing how a mirror reflects a laser (incoming angle = outgoing angle).

I can add a few more hints if it is too hard (well, it is hard).

[jon-nyc]

Don’t add any hints

[bachophile]

Link to video

[jon-nyc]

I’ve almost got this just need to figure out one last piece when I can get to a desk

[Klaus]

@jon-nyc said in Puzzle time - don't waste braindead liberals:

I’ve almost got this just need to figure out one last piece when I can get to a desk

Reminds me of a famous quote from Fermat from 1637.

I have discovered a truly marvelous proof of this, which this margin is too narrow to contain

The proof of his famous conjecture was written down in 1995 after thousands of mathematicians spend their whole life working on the maths necessary to make that proof.

[jon-nyc]

Finally I think I have it.

Lets call A0 Ax's actual position and A1 through A4 his reflected position off of each of the four mirrors. Same idea with L0-4.

Draw out those 10 points, and draw the 25 line segments that start from Ax and all 4 of his reflections and go to Larry and all four of his reflections. Now, all but one of these segments will extend outside the arena, but the ones that do are the equivalent of a 'bank shot' encompassing one or more reflections in the mirrored surface.

You can translate the midpoint of each of these segments to it's equivalent point within the arena, bu recognizing that it is the mirror image of the point across the nearest wall of the arena.

This will get you 25 points within the arena. You soon notice that 9 of these fall out of duplicates. So you are left with 16. They seem to form a neat grid.

I've not yet shown here that those 16 points are sufficient, though I convinced myself that as long as you choose the midpoints, you will also block equivalent shots between more distant reflections, which correspond to more zig-zaggy bank shots between Larry and Ax.

For example, here I have shown visually how Ax shooting Larry's reflection in the bottom mirror is equivalent to him shooting actual Larry with the bank shot:

But more than that, you see that the midpoint will also block the case where Ax shoots Larry using the same two walls, but does three banks instead of 1. Note this corresponds to Ax's reflection A1 shooting Larry's reflection L2:

That point will stop all of the shots using just those two walls using an odd number of reflections. A distinct point will stop all that use an even number of reflections.

[jon-nyc]

That wasn’t correct. I need to fix it and repost.

[Klaus]

You are close. Many right ideas in your draft.

[jon-nyc]

I know it’s the second example I drew out which I didn’t like.

[jon-nyc]

I am confident my answer is right, just trying to show it’s exhaustive

[Klaus]

I think it's not exhaustive. You get only 11 distinct BDLs that way, not 16.

[jon-nyc]

I got 16. 25 possible segments, 9 fell out as redundant.

[Klaus]

9 are obviously redundant. But there are 4 more that are redundant. Do you want me to illustrate that?

[jon-nyc]

Not yet