DrPizza's Math and/or Science Challenge

[darkxshade]

What about something like this on the larger scale



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3 parallel lines 133 dots across with one final dot in between 2 of those lines in the first spot?

 

[DrPizza]

Hmmm, yes it can be divided Terzo. Try again!

 

[ElFenix]

Originally posted by: 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.

if you can put 4 points on a page so that a line can't be drawn with half on one side and half on another, then you can put 400 points on a page so that....

 

[AmberClad]

As far as I can tell, it can only be done with an odd number of points assuming all points are distinct .

Originally posted by: 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.

Oh, how I detested that proofs & theorems crap in high school.

Thankfully, all of the college level math courses I took were for engineers (i.e. applied math), so I don't recall ever having to do another proof in college.

 

[chuckywang]

I'm almost positive the answer is no. I believe the proof has to do a linear transformation (think about rotating the piece of paper) that puts all the points at a distinct vertical axis. Then you just draw a vertical line that has 200 points on one side and 200 points on the other.

That's not a formal proof of course, just a thought. What do you think Dr Pizza? On the right track?

 

[Chronoshock]

I believe it is impossible

Consider an arbitrary collection of points
Draw a line across the points and collapse them from 2 dimensions into one (by moving the points perpendicular to the line)
A cut may be made on this 1 dimensional projection that splits them into two even sets
The only way a cut cannot be made is if all points occupy an odd number of projected points in this 1-dimensional space
Points coalesce into a single projected point if they fall along the same line
For a finite set of points there is a finite set of lines
There is an infinite set of dividing lines along which to make a 1-dimensional projection
By the pigeon hole principle it is impossible.

 

[KK]

put one dot on the back of the paper.

 

[sdifox]

depends on size of paper

2nd question depends on how far you pull.

Assuming the person is still there to steady the bike, as long as the distance pulled back is not greater than the arc the pedal will make to horizontal (9 o'clock) position, it goes forward. After that depends on if the force applied is greater than the friction force of the real wheel (well, just the patch touching ground)

 

[TheoPetro]

nvm

 

[JDMnAR1]

Originally posted by: ElFenix
if you can put 4 points on a page so that a line can't be drawn with half on one side and half on another, then you can put 400 points on a page so that....

This sounds vaguely like "If you can dodge a wrench, you can dodge a ball".

On question 2, wouldn't it depend on the force being applied to the rope and how it relates to the rotational force necessary to turn the pedals?

 

[Terzo]

Back with another guess...once again, haven't really reasoned it out but it seems like it might work.
An isosceles triangle that has 39 points for it's base, with each layer above it having 2 less points (39, 37, 35, etc).

 

[Mo0o]

For problem 2 i say it moves to the left.

The bike has a force A (the rope) moving it to the left and a force B moving it to the right (friction force from the tires). The friction force B on the tires should be mu*A (assuming total transfer), and since the coefficient of friction is <1, the net force should be to the left.

So i think the wheels should turn a bit, but overall the bike skids left

I think...

 

[yhelothar]

The bike is in a really high gear, so it would take a significantly greater torque to turn the pedals. If the torque required to turn the pedals is greater than the friction between the tires and the surface of the road, then the pedals won't turn at all, and the tires will just skid backwards.

 

[her209]

Assuming the rope is taut at all times, then the bike will move forward until the pedal is horizontal to the ground (on the same line as the rope) then backwards.

 

[KMc]

The bike moves to the right, here's why...

A multi-speed bike has a freewhweeling rear wheel, duh, we all know this. Ok, so if you just push the bike forward, it will roll independent of the pedals...blah blah blah.

But, if you push a bike backwards, the cranks turn backwards too - seriously, go home and try it. So that means if you pull to the left on this rope, the bike can't go to the left because that would require the pedal to move to the right (crank turning backwards). That means the forces resolve to clockwise torque on the crank and the bike would move toward the right.

 

[Sumguy]

Meh, I've never had to do proofs, but I'll give it a shot. Lazy style.

1) Draw an even number of points on a piece of paper in a way that they are easily separated into even parts. I went up to ten before seeing the obvious pattern (just to be sure it continued).
2) Draw ten points randomly and separate them into even parts. Count the total number of possible separating lines.
2a) I noticed some sort of relationship between the points that could be explained mathematically...but I'm too lazy and I already have the really simple version, so fuck that.
3) Came up with this

f(fuckit)=x/2

Where "fuckit" is the number of possible lines that can separate the number of points in half (provided that you aren't a douche and line a few up into a straight line), and x is the number of points.

4) Provided the point-pointer is an asshole and lines a few points up into a line and then throws some random dots around for the lulz, the equation becomes

f(youdick)=(asshole)+(x/2)

Where "youdick" is the number of possible lines, "asshole" is the total number of lines created by points (not separable lines), and x is the total number of points not on those lines.

Note: This only applies if there is an even number, 4 or more, on a line. If its an odd number than refer to the "fuckit" equation.

5) Noticing that theres a "super-general" equation you can write for all possible lines given all possible configurations for a given number of points, I also noticed I'm far too uninterested in writing that since I've already came to my own conclusion. By extrapolation or whatever, you can't draw the line. Of course I can be wrong.


Edit: Whoops, original "youdick" equation was wrong.

 

[ivan2]

well for number one, we can think of a scenario that you can NOT divide the dots evenly. the only case when this will happen is that no matter how you orient or move your line, 2 or more of the dots will always line up on the same line. since you only have a finite number of dots, but you have an infinite ways to orient or move your line, you will always be able to find a place that you can evenly divide them in half.

for the second one concur with astroidea. most likely the force pulling back the bike will overcome the friction of the wheel. if the guy's got a chicken hand that could not overcome the friction, the bike will not move.

 

[Mo0o]

Originally posted by: KMc
The bike moves to the right, here's why...

A multi-speed bike has a freewhweeling rear wheel, duh, we all know this. Ok, so if you just push the bike forward, it will roll independent of the pedals...blah blah blah.

But, if you push a bike backwards, the cranks turn backwards too - seriously, go home and try it. So that means if you pull to the left on this rope, the bike can't go to the left because that would require the pedal to move to the right (crank turning backwards). That means the forces resolve to clockwise torque on the crank and the bike would move toward the right.

Why can't the wheels turn clock wise but the bike skid left?

 

[Tremulant]

I think the bike would skid backwards and then fall over towards the left.

edit: changed right to left, just realized the bike is being pulled from below it's center of gravity.

 

[DrPizza]

Originally posted by: Chronoshock
I believe it is impossible

Consider an arbitrary collection of points
Draw a line across the points and collapse them from 2 dimensions into one (by moving the points perpendicular to the line)
A cut may be made on this 1 dimensional projection that splits them into two even sets
The only way a cut cannot be made is if all points occupy an odd number of projected points in this 1-dimensional space
Points coalesce into a single projected point if they fall along the same line
For a finite set of points there is a finite set of lines
There is an infinite set of dividing lines along which to make a 1-dimensional projection
By the pigeon hole principle it is impossible.

I suppose this and chuckywang's answer are close enough. Similar but simpler explanation that doesn't involve projections: for every pair of points, draw a line through the points (extend it beyond the points in both directions.) It's possible that some lines may even pass through more than 2 points. There are, at most, 400*399 / 2 distinct lines which connect pairs of points (less if 3 or more points are collinear.)

Starting off the paper, pick a point which does not lie on any of those lines. Using this point as an anchor for a line, sweep through the 400 points, one point at a time, until you've passed through 200 of them. 200 will be on one side, 200 on the other side. At the point you pass the 200th, you can't pass 2 points at the same time - to do so, you would have had to have been on one of the original lines, which you're not.

 

[destrekor]

Question 2:

Without a means of seeing exactly how rearward force on a bike pedal would work in real life, my guess is the bike does one of two things, but first, it depends on the gear assembly. Can the bike be wheeled backwards with the pedals remaining stationary?
If so, as I said, the bike will do one of two things: either move forward slightly as the pedal [with the rope attached] moves the chain in a normal forward fashion, but the rope being pulled would prevent it from being able to continue in a full pedal motion, and would begin to pull the back backwards, or fall over depending on the way the chain/gear assembly works.

Or, the bike would simply begin to move backwards immediately. The rearward pulling force on the pedal would likely be greater than a single-pedal attempt at chain rotation, and without the added benefit of downward force on the pedals to ease the process of chain/gear rotation, and thus would simply pull the back back.

With no means of doing anything of that sort, and with a lack of super physics knowledge, I cannot begin to even guess at calculations in order to provide a solid choice between my two answers. However, one of them is right.

Answer 1:

If this piece of paper is divided into four quadrants, you could place, say, 299 of the points in the bottom left quadrant, in a circular formation (might not matter), 20 in a tight square in the furthest bottom-left corner, 49 points in a long rectangular formation in the top half of the upper left quadrant, half of the remaining (16) in a tight scatter group in the bottom-right corner, and the other 16 scattered all over the upper-right quadrant.

 

[KMc]

Originally posted by: Mo0o

Originally posted by: KMc
The bike moves to the right, here's why...

A multi-speed bike has a freewhweeling rear wheel, duh, we all know this. Ok, so if you just push the bike forward, it will roll independent of the pedals...blah blah blah.

But, if you push a bike backwards, the cranks turn backwards too - seriously, go home and try it. So that means if you pull to the left on this rope, the bike can't go to the left because that would require the pedal to move to the right (crank turning backwards). That means the forces resolve to clockwise torque on the crank and the bike would move toward the right.

Why can't the wheels turn clock wise but the bike skid left?

Well, I guess it all comes down to what gear the bike is in, how much the bike weighs and what the surface is made of. On this bike the chain is on the large chainring, but also on a fairly large cog on the rear, so the gearing may not be really high.

 

[vshah]

Originally posted by: her209
Assuming the rope is taut at all times, then the bike will move forward until the pedal is horizontal to the ground (on the same line as the rope) then backwards.

i agree

 

[Number1]

The bike will be dragged backward with the back wheel spinning forward until the pedal is parallel to the ground. then the wheel stops spinning forward.

 

[DrPizza]

For the bike question, the following answers have been submitted so far:
to the left
to the right
it depends
to the right, then to the left.

One of those answers is correct!