MrRubix's Riddle Thread

[One Winged Angel]

Re: MrRubix's Riddle Thread

Back to the drawing board :/

[s0ulst0n3]

Re: MrRubix's Riddle Thread

so Shash Soulstone and laggy at night around the campfire working their wood
while laggy pays to enjoy others wood because he cant get his own wood
ok

MrRubix_MK5 i hope you had a camcorder this will be hotter then paris

Hawt.

[Fractal_Monkey]

Re: MrRubix's Riddle Thread

Reach, I think it's 51.

If there are six points all connected to each other by lines, then there will be 15 lines in the diagram. The maximum number of intersections of those 15 lines would be if each line intersected the other 14, which gives sum(1:14) = 105.

However, at each of the six given points, there are 5 intersecting lines. So we have to subtract the maximum possible number of intersections of 5 lines (10) at each point, and replace it with the one forced mass intersection. This gives a total maximum of 105 - (6x9) = 51 intersections.

Don't ask for a diagram, my solution doesn't give an arrangement of points that satisfies these conditions.

[Reach]

Re: MrRubix's Riddle Thread

Reach, I think it's 51.

If there are six points all connected to each other by lines, then there will be 15 lines in the diagram. The maximum number of intersections of those 15 lines would be if each line intersected the other 14, which gives sum(1:14) = 105.

However, at each of the six given points, there are 5 intersecting lines. So we have to subtract the maximum possible number of intersections of 5 lines (10) at each point, and replace it with the one forced mass intersection. This gives a total maximum of 105 - (6x9) = 51 intersections.

Don't ask for a diagram, my solution doesn't give an arrangement of points that satisfies these conditions.

I don't need the diagram. This is correct and the best way to solve the problem. Usually I would give the problem with more dots so people can't draw diagrams, but allowing them to do so is a good trick to keep people from solving it, since most people jump right to diagrams and can only end up getting 49 or 50 intersections...not the optimal 51.

Exceptionally good thinking.

Since Rubix hasn't posted a new puzzle, I'll post one:

Five identical cubes, each having a volume of 1 cubic inch, are fused together to form a solid object. Adjoining cubes are fused together in such a way that their contiguous faces cover one another entirely without any overlap. No cube is connected to the rest of the object at only a finite number of points or lines. How many different shapes could these 5-cubic-inch objects have?

[MrRubix_MK5]

Re: MrRubix's Riddle Thread

I had 51 actually, miscounted one of my points. I used a diagram. Not very elegant, but it worked.

[MahouMinachan]

Re: MrRubix's Riddle Thread

it's weird, because I used to play this game Block Out, and it used blocks exactly what Reach is talking about. I can only remember 17 of them though...

[MrRubix_MK5]

Re: MrRubix's Riddle Thread

Are we assuming that a shape is all that matters? (ie. an arrangement isomorphic to another is considered the same arrangement? Do the cube positions themselves matter? As in, if we had two cubes, would we say there's only 1 position?)

[Reach]

Re: MrRubix's Riddle Thread

Are we assuming that a shape is all that matters? (ie. an arrangement isomorphic to another is considered the same arrangement? Do the cube positions themselves matter? As in, if we had two cubes, would we say there's only 1 position?)

If there were two cubes, there would be only one distinct shape. The rule is essentially this: If a piece can be rotated through three dimensional space to fit exactly another piece, they are identical shapes and this counts as only one shape.

Otherwise the answer would become ambiguous, since there are a lot of configurations that appear different but rotate to become the same.

[MrRubix_MK5]

Re: MrRubix's Riddle Thread

New riddle up

[MrRubix_MK5]

Re: MrRubix's Riddle Thread

Reach: I'm getting 29 shapes here

Will explain logic if correct

[Reach]

Re: MrRubix's Riddle Thread

Three men begin atop a mountain slope during a snowy night. Each man sits on top of their own flatbed cart on wheels (think Bugs Bunny cartoon, here). One of the men, Man #3, however, is stationed within a mountain tunnel, where no snow can pass through. Each man/cart is identical in every way, and their distances down to the bottom of the slope is the same. Each cart, at time 0, has no snow. They all start to go down the slope. Man #1 decides to sleep and does not do anything about the snow that falls onto his cart. Man #2 is constantly sweeping snow off his cart perpendicular to the direction of motion. Man #3, in the tunnel, obviously does nothing as he has no snow. Who makes it down to the bottom of the slope first, second, and third, and why?

Well, this requires a few assumptions on the physics end. I'll assume we can neglect friction in the tunnel. I'll assume assume we can neglect air resistance in the tunnel. Thus, Man 3 will beat Man 2, as Man 3 should be approximately the same weight, should experience less friction and MUCH less air resistance.

Now we have to consider whether man 1 will beat man 2. This could be complicated, depending on how we look at the physics, but I will assume that man 1 will beat man 2 for the following reason: If friction is treated as coulombic, man 1 should start going faster due to his gain in mass. However, this depends on how fast you're going. =/

Finally, man 3 should beat man 1 because his terminal velocity will be significantly higher (Man 3 does not face wind resistance under my assumption).


Thus Man 3 > Man 1 > Man 2.

This could be wrong, depending on whether or not my assumptions are correct.



Rubix, 29 is correct. Dispatched that one quickly D:

[MrRubix_MK5]

Re: MrRubix's Riddle Thread

I had made a few edits already to the post, Reach, but no, friction won't matter here.

Again I can only say one is correct if the answer and logic is correct, so I will say no, that answer's not quite right

[Reach]

Re: MrRubix's Riddle Thread

I had made a few edits already to the post, Reach, but no, friction won't matter here.

Again I can only say one is correct if the answer and logic is correct, so I will say no, that answer's not quite right

Well assuming no air resistance changes things, since now the man in the tunnel loses his primary advantage for realistically creaming the other two.


Man 1 should win in this case. Realistically, it would depend on how long the hill was, but given his mass continues to increase and he faces no air resistance he should cruise past the other two. I could probably work this out mathematically but I won't waste time.

Now who wins between man 2 and 3 is a bit trickier in this scenario. I would say pushing mass off perpendicular to the direction of motion like that is going to slow you down. Man 3 should beat man 2.

So man 1 > man 3 > man 2


If not, then I'll let someone else tackle this, as I have trouble neglecting variables and then making sense of the problem.

[MrRubix_MK5]

Re: MrRubix's Riddle Thread

Hint: Since this is a problem about mass and speed, all other variables negligible, think about a conservation law that includes these two variables.

[Reach]

Re: MrRubix's Riddle Thread

Hint: Since this is a problem about mass and speed, all other variables negligible, think about a conservation law that includes these two variables.

I considered momentum immediately. I don't like this problem. You can probably tell based on my puzzle style that I prefer very concretely defined variables XD