DrPizza's Math and/or Science Challenge

[DrPizza]

Let's say that on a piece of paper, I mark 400 distinct, separate points. Is it possible to place the points in a manner that would prevent someone from being able to divide them exactly in half with a straight line (half are on one side of the line, half are on the other side of the line, no points lie on the line.)
Prove it.

As this is math, we're talking about points and lines, not giant 1/2 inch across marks on a paper.

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2nd question added (I'm going to fill in for someone at the pizza shop tonight for something different to do on my summer vacation, so won't be around for a while to post answers/tell people they're right or wrong.)

Physics question. I tie a rope to the pedal of a bike as in this picture:
bicycle.jpg
Someone lightly touches the bike seat, *only* to keep the bike upright and balanced, not to aid it in its motion. If I pull on the rope backwards (direction of the arrow), in which direction(s) will the bike move? Forward? Backwards? Forward, then backwards? Depends on how well the chain is lubricated? Explain.
The bike is on a level surface, etc. There are no trick things about this question.

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3rd question, and then I'm gone for the night:
2 players take turns picking one of the digits from 1 to 9. Once a digit has been selected, it can not be selected again. (write each digit on top of one of 9 coins, and select coins if you want to do this physically)

The first person to be able to add some or all of the digits he's selected to get 15, wins.

Is there a strategy such that the first person can always win? Can the 2nd person ever win? Is there a strategy for the 2nd person so that he never loses?

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Number 3 has been answered. (It was easy) - Here's a modification to #3 to change it a little:
#4. They take turns. If 15 is the sum of any *three* of their cards (9 & 6 doesn't win), is there a way for player 1 to force the win?

#5 Two ropes have been tied to the roof of the Superdome and dangle down to the ground. The ropes are tied 1 foot apart. The length of each rope is 250 feet. A man armed with only his rope climbing skills & a sharp knife decides to climb up the ropes and steal as much of the rope as possible. He knows that if he falls more than 30 feet, he'll break his leg and be caught. What is the maximum amount of rope that he can steal? (Assume the obvious that would make this a puzzle - no ladders, no climbing across the ceiling, etc.)

 

[Crusty]

Put all the points in the same spot

 

[mugs]

My answer is no unless you can put two points on the same spot, but I can't prove it.

 

[Analog]

Prizes?

 

[DrPizza]

Originally posted by: Crusty
Put all the points in the same spot

There, I made it clearer in the OP.

 

[darkxshade]

does the line have to be straight?

is it an actual math/geometry question or is it more like a riddle where there's some loophole like putting dots on the other side of the paper on the same spot(which should qualify as distinct spots)?

 

[RedArmy]

Yeah, I was just about to ask if the line had to be straight or if it can curve.

 

[actuarial]

I started thinking about this, and I decided maybe not enough info? If the 400 points don't need to be distinct, then yes as I can mark 400 points at spot (0,0). If distinct, and curvy lines allowed, then no as there is always a space between points and a curvy line allows too much freedom.

Quick Answer (I don't know if this is rigorous enough for you). I am going to simplify it by talking about only straight lines, and assume that if I arrange the dots in a grid you can only split rows/columns or move along a diagonal.

In which case, yes. I would make a grid which is 19 points tall, 19 points wide, with the last column extending upwards to hold the remaining 39 points.
You cannot draw a line along a column, as one half would necessarily be only rows of 19, and 200 is not divisible by 19 evenly. This same logic applies along a row.
For diagonals, for a line with negative slope, we can look at the bottom left half. There is no way to make 200, as including the longest diagonal row (19 long) as your last row you would get 190, while including an additional diagonal (18 long) you would get 208. If talking about a positive slope, we can look at the lower right half. Including the longest diagonal of the main cube would again give 190, while including another diagonal (19 long due to the extended final row) you would get 209.

I hope I haven't left any glaring holes, and if I simplified the problem too much let me know. I'm not smart enough to do the super abstract anymore!!!

 

[Cogman]

Yes it is possible. Make a circle with 399 of the points (Such that every point is evenly spaced) and place a point in the very center of that circle. The points can now not be divided evenly without the line crossing one of the points.

I suppose there are probably other ways to do this, but this works just as well as any other way.

 

[DrPizza]

There, clarified that it's a *straight* line. I found a way to draw a straight line in the suggested solutions by actuarial & Cogman, so no correct solution has been given yet.

 

[actuarial]

Originally posted by: DrPizza
There, clarified that it's a *straight* line. I found a way to draw a straight line in the suggested solutions by actuarial & Cogman, so no correct solution has been given yet.

Can you tell me how? If you're asking for MY proof I should be allowed to ask for your refutation!

 

[Cogman]

Originally posted by: DrPizza
There, clarified that it's a *straight* line. I found a way to draw a straight line in the suggested solutions by actuarial & Cogman, so no correct solution has been given yet.

? How so, with the point in the center, it will always be off by one as the line can't pass directly through the center. (Thus the center point will shield one of the other points from an even divide (It works with 10 points).

 

[Mo0o]

edit: oops nm

 

[Auggie]

cogman, even then, 399 points in a circle is diisible by three, meaning that the circle would not be bilaterally symetrical. then you could just nudge the line going through the center point a tiny bit so that it didn't pass through the very center point, but was still on one side of the other 200 points.

i firmly believe the answer is that there is no way to arrange 400 points so that a line cannot separate them into even groups of 200.

 

[Kirby]

Originally posted by: DrPizza
Let's say that on a piece of paper, I mark 400 distinct, separate points. Is it possible to place the points in a manner that would prevent someone from being able to divide them exactly in half with a straight line (half are on one side of the line, half are on the other side of the line, no points lie on the line.)
Prove it.

As this is math, we're talking about points and lines, not giant 1/2 inch across marks on a paper.

Yes, take away the pencil from the person who draws the line.

 

[her209]

Can I prove whether it can or cannot be done with 4 points instead of 400?

 

[Albatross]

put 399 on one side and one point on teh other side

 

[fuzzybabybunny]

Hookers and Blow?

 

[mugs]

Originally posted by: Cogman

Originally posted by: DrPizza
There, clarified that it's a *straight* line. I found a way to draw a straight line in the suggested solutions by actuarial & Cogman, so no correct solution has been given yet.

? How so, with the point in the center, it will always be off by one as the line can't pass directly through the center. (Thus the center point will shield one of the other points from an even divide (It works with 10 points).

Draw it (with 10 points)

Or just draw the 9 points around the perimeter, pick any two of the "openings" that would results in a 5/4 division, and realize how many different lines you could draw that would pass through those two openings. They don't all go through the center. You could place your center dot in a position that would force it to be on the "5" side with those two openings, but then when you pick two other openings it would be on the "4" side.

 

[Cogman]

Originally posted by: mugs

Originally posted by: Cogman

Originally posted by: DrPizza
There, clarified that it's a *straight* line. I found a way to draw a straight line in the suggested solutions by actuarial & Cogman, so no correct solution has been given yet.

? How so, with the point in the center, it will always be off by one as the line can't pass directly through the center. (Thus the center point will shield one of the other points from an even divide (It works with 10 points).

Draw it (with 10 points)

ah, crap. Ok, then I'll just revise it.

398 points on the perimeter of the circle and put one point in the center, and the other to the right of that point, and close enough that there is no distance between the two points (thus no passing between the two points.

 

[Chronoshock]

Originally posted by: Cogman
398 points on the perimeter of the circle and put one point in the center, and the other to the right of that point, and close enough that there is no distance between the two points (thus no passing between the two points.

As noted in the OP, treat all points in the mathematical or physics sense (occupying no space)

 

[Crusty]

Originally posted by: Cogman

Originally posted by: mugs

Originally posted by: Cogman

Originally posted by: DrPizza
There, clarified that it's a *straight* line. I found a way to draw a straight line in the suggested solutions by actuarial & Cogman, so no correct solution has been given yet.

? How so, with the point in the center, it will always be off by one as the line can't pass directly through the center. (Thus the center point will shield one of the other points from an even divide (It works with 10 points).

Draw it (with 10 points)

ah, crap. Ok, then I'll just revise it.

398 points on the perimeter of the circle and put one point in the center, and the other to the right of that point, and close enough that there is no distance between the two points (thus no passing between the two points.

If there is no distance between the two points aren't they in the same spot then?

 

[AmberClad]

Are we talking about a 2-D plane here? Do all the points have to be on the same side of the sheet of paper? Can you fold up or roll up the sheet of paper?

 

[Terzo]

I'm not about to draw out 400 points by hand, so I'm not 100% sure about this.
Assuming you have a plane on the x-y axis.
Take 200 points, starting at the orgin, and place them along the x-axis (y=0).
Take one point, and put it below that line (x=0, y=-1)
Take the remaining 199 points and place them on the y axis (y=1,2,3..., x-0)
I haven't thought about it too much, but i'm pretty sure this can't be evenly divided by a straight line.

*edit*
I just realized i didn't consider diagonal lines, but trying to imagine that situation is hurting my head too much. Is there a site which has graph paper you can write on?

 

[DrPizza]

Originally posted by: her209
Can I prove whether it can or cannot be done with 4 points instead of 400?

Depending on the proof, it can probably be easily visualized with 4 points & generalized to 400 points. 6 points may be more advantageous for it to really sink in though.