DrPizza's Math and/or Science Challenge

[Mo0o]

Originally posted by: 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.

I think regardless of teh gearing, the friction force generated at the point of contact between teh wheels and the road is equal or less than the force being applied to the peddle. So unless you're just barely pulling at it, the bike should slide left regardless of whether the wheels turn or not (which would depend on the gearing)

 

[mugs]

It'll move slightly forward until the cranks are parallel to the ground, then it will be dragged backward because the rope is holding it back.

 

[destrekor]

Originally posted by: 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!

im going with C. 😀

 

[mugs]

Player 1 starts with 8
Player 2 is forced to take 7
Player 1 takes 6, giving him 14
Player 2 is forced to take 1
Player 1 takes 5, giving him 13 or 11
Player 2 must take either 2 or 4, but can't take both. Neither option makes player 2 the winner.
Player 1 takes whichever one player 2 didn't take and wins.

Player 2 can never win, nor can he avoid losing if player 1 uses this strategy. Every move player 2 makes is an attempt to prevent player 1 from reaching 15.

 

[Mo0o]

Problem 3:

Player 1) 9 5 8 7
Player 2) 6 1 2

I think thats the best way I have of ensuring player 1 wins. There's gotta be a more efficient route

 

[Atomic Playboy]

Originally posted by: mugs
Player 1 starts with 8
Player 2 is forced to take 7
Player 1 takes 6, giving him 14
Player 2 is forced to take 1
Player 1 takes 5, giving him 13 or 11
Player 2 must take either 2 or 4, but can't take both. Neither option makes player 2 the winner.
Player 1 takes whichever one player 2 didn't take and wins.

Player 2 can never win, nor can he avoid losing if player 1 uses this strategy. Every move player 2 makes is an attempt to prevent player 1 from reaching 15.

Even easier:
Player 1 starts with 8
Player 2 is forced to take 7
Player 1 takes 6, giving him 14
Player 2 is forced to take 1 or 9. Whichever he doesn't select, Player 1 can use to make 15.

 

[PowerEngineer]

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)

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.

Hhhhmmmm.... I'm betting I'll see the error in my ways as soon as I post this, but it seems to me that Cogman's answer might be right.

I'm looking at the 10-point circle with one point in the middle and the nine others equally spaced around its circumference. Assuming the bisecting line can't go through a point, then it seems to me that any line that that divides the nine circumference points into five and four points must also put the center point on the same side as the five circumference points. 5/9 of the circumference is greater than 180 degress, which means the center point is "inside the bowl" formed by the five circumference points and the line. This looks to work for all even numbers greater than or equal to six.

I'm probably just missing something obvious...

 

[mcmilljb]

Originally posted by: Atomic Playboy

Originally posted by: mugs
Player 1 starts with 8
Player 2 is forced to take 7
Player 1 takes 6, giving him 14
Player 2 is forced to take 1
Player 1 takes 5, giving him 13 or 11
Player 2 must take either 2 or 4, but can't take both. Neither option makes player 2 the winner.
Player 1 takes whichever one player 2 didn't take and wins.

Player 2 can never win, nor can he avoid losing if player 1 uses this strategy. Every move player 2 makes is an attempt to prevent player 1 from reaching 15.

Even easier:
Player 1 starts with 8
Player 2 is forced to take 7
Player 1 takes 6, giving him 14
Player 2 is forced to take 1 or 9. Whichever he doesn't select, Player 1 can use to make 15.

2 Ultimate 3 picks for player 1 win.

You pick 7, Opp picks 8.
You pick 6, Opp picks any number, doesn't matter.
You pick 2 or 9, whichever is left, game over! GG!

You pick 8, Opp picks 7.
You pick 6, Opp pick any number, doesn't matter.
You pick 1 or 9, which ever is left, game over! GG!

Player 2 can never win if Player 1 knows this trick.

 

[mugs]

Originally posted by: Atomic Playboy

Originally posted by: mugs
Player 1 starts with 8
Player 2 is forced to take 7
Player 1 takes 6, giving him 14
Player 2 is forced to take 1
Player 1 takes 5, giving him 13 or 11
Player 2 must take either 2 or 4, but can't take both. Neither option makes player 2 the winner.
Player 1 takes whichever one player 2 didn't take and wins.

Player 2 can never win, nor can he avoid losing if player 1 uses this strategy. Every move player 2 makes is an attempt to prevent player 1 from reaching 15.

Even easier:
Player 1 starts with 8
Player 2 is forced to take 7
Player 1 takes 6, giving him 14
Player 2 is forced to take 1 or 9. Whichever he doesn't select, Player 1 can use to make 15.

Yeah that's faster. There are several ways to do it, but they'll all involve forcing player 2 to block you until you can create a situation where you have two ways to win and he can't block both. Like tic-tac-toe.

 

[mcmilljb]

Originally posted by: mugs

Originally posted by: Atomic Playboy

Originally posted by: mugs
Player 1 starts with 8
Player 2 is forced to take 7
Player 1 takes 6, giving him 14
Player 2 is forced to take 1
Player 1 takes 5, giving him 13 or 11
Player 2 must take either 2 or 4, but can't take both. Neither option makes player 2 the winner.
Player 1 takes whichever one player 2 didn't take and wins.

Player 2 can never win, nor can he avoid losing if player 1 uses this strategy. Every move player 2 makes is an attempt to prevent player 1 from reaching 15.

Even easier:
Player 1 starts with 8
Player 2 is forced to take 7
Player 1 takes 6, giving him 14
Player 2 is forced to take 1 or 9. Whichever he doesn't select, Player 1 can use to make 15.

Yeah that's faster. There are several ways to do it, but they'll all involve forcing player 2 to block you until you can create a situation where you have two ways to win and he can't block both. Like tic-tac-toe.

But you can force a non win for player 1 in tic-tac-toe.

 

[Atomic Playboy]

Originally posted by: mcmilljb

Originally posted by: Atomic Playboy
Player 1 starts with 8
Player 2 is forced to take 7
Player 1 takes 6, giving him 14
Player 2 is forced to take 1 or 9. Whichever he doesn't select, Player 1 can use to make 15.

2 Ultimate 3 picks for player 1 win.

You pick 7, Opp picks 8.
You pick 6, Opp picks any number, doesn't matter.
You pick 2 or 9, whichever is left, game over! GG!

You pick 8, Opp picks 7.
You pick 6, Opp pick any number, doesn't matter.
You pick 1 or 9, which ever is left, game over! GG!

Player 2 can never win if Player 1 knows this trick.

This can also be done with the 9 + 6 combination.

Player 1 takes 6.
Player 2 takes 9.
Player 1 takes 7.
Player 2 has to take either 2 or 8. Whichever is left makes Player 1 the winner.

As long as player 1 starts with 6, 7 or 8, they will always win.

 

[mugs]

Originally posted by: mcmilljb

Originally posted by: mugs

Originally posted by: Atomic Playboy

Originally posted by: mugs
Player 1 starts with 8
Player 2 is forced to take 7
Player 1 takes 6, giving him 14
Player 2 is forced to take 1
Player 1 takes 5, giving him 13 or 11
Player 2 must take either 2 or 4, but can't take both. Neither option makes player 2 the winner.
Player 1 takes whichever one player 2 didn't take and wins.

Player 2 can never win, nor can he avoid losing if player 1 uses this strategy. Every move player 2 makes is an attempt to prevent player 1 from reaching 15.

Even easier:
Player 1 starts with 8
Player 2 is forced to take 7
Player 1 takes 6, giving him 14
Player 2 is forced to take 1 or 9. Whichever he doesn't select, Player 1 can use to make 15.

Yeah that's faster. There are several ways to do it, but they'll all involve forcing player 2 to block you until you can create a situation where you have two ways to win and he can't block both. Like tic-tac-toe.

But you can force a non win for player 1 in tic-tac-toe.

Yeah, but the strategy is the same.

 

[CoinOperatedBoy]

1. Mathematically, it is impossible. A line of non-zero length can contain infinitely many points and 2D space can be infinitely subdivided so there will always be a straight line that can bisect an even-numbered set of distinct points (although not always symmetrically).

However, if we are actually talking about a piece of paper and points made from a writing implement like a pen, it's possible to imagine that two points could be drawn close enough together that the human eye cannot detect any space between them through which a line could pass without touching one or both points. If you say the points are no longer distinct if we cannot detect the separation, or if the writing tool is considered infinitely fine (since any "point" drawn by a traditional pencil or pen could be magnified and revealed to actually be a shape that takes up a non-zero amount of the paper's finite 2D space), then this stipulation is irrelevant.


2. First, this is under the assumption that the force due to the rope is constant. The force from the rope on the pedal will have a two-fold effect: the portion of the force tangent to the path of the crank arm will pull the pedal clockwise (which will transfer energy to the crank + chain + wheel system and cause a forward force on the bike), while the portion of the force parallel to the ground will pull the bike backwards. Since the only force propelling the bike forward is a result of the tangential force on the crank arm from the rope*, the forward force will never exceed the backward force, only come close to equalling it when the pedal is at the bottom-most position on the crank's path. If the resultant total backward force on the bike also exceeds the opposite force of friction, the bike will slide backwards. Otherwise, it will only remain stationary, but will never move to the right.

I have no idea if this is correct.

* And, I suppose, a small force from gravity pulling the pedal down, but that force will negligible.

 

[DrPizza]

#1: answered completely.
#2: I'll leave it a little longer

#3: Answered. This was meant to be easy enough so that even mugs could answer it. 😛

#4 added (which is a modified #3)

and #5 added

 

[mrjminer]

#1:

My guess is that you make multiple crosses. Like everyone else, I'm not going to draw out 400 dots, but I think that using multiple connected crosses (they wouldn't have to be equivalent in the number of dots they use) you might be able to create something that is not equally divided by a straight line. Then again, I'm not drawing out hundreds of dots to test this 😀

#2:
I'm not a physics person or a bike person, so it might depend on how much force the person is putting on the seat to hold the bike up--or the type of brakes--or the distribution of force based on uses of triangles--etc...

My guess is that the bike goes upward. My reasoning is that the force would unequally distributed to the rear tire, as the pedal is not able to spin to distribute the equivalent force to the fron tire. I think it's similar to when you brake harshly.

 

[ivan2]

with the ropes he can take them all. just go up there, cut one down, tie them together. basically you can make the rope just hang on the ceiling instead of tied to it and come down on 2 ropes.

 

[Demon-Xanth]

#5:
He can steal all but about two feet needed to tie the knots.

Step 1: climb up rope #1 as far as possible
Step 2: cut rope #2 as high as possible
Step 3: cut rope #1 as high as possible that lets you tie a bowline
Step 4: tie a bowline in rope #1
Step 5: tie the cut ropes together with a square knot
Step 6: feed the rope through the loop in the bowline
Step 7: climb down the tied ropes
Step 8: pull the ropes through the loop, leaving only the small bowline at the top
Step 9: RUN!

 

[darkxshade]

What kind of mechanism is used to hold the ropes on the roof?

 

[DrPizza]

Originally posted by: darkxshade
What kind of mechanism is used to hold the ropes on the roof?

Tied. For the sake of argument, there is nothing to loop the ropes over, loop the ropes through, etc. The only thing you get to interact with are the ropes & the knife.

 

[ivan2]

Originally posted by: DrPizza

Originally posted by: darkxshade
What kind of mechanism is used to hold the ropes on the roof?

Tied. For the sake of argument, there is nothing to loop the ropes over, loop the ropes through, etc. The only thing you get to interact with are the ropes & the knife.

well u can cut one rope close to the top and then tie a loop with the remainer of that rope still tied to the ceiling, it changes almost nothing.

 

[darkxshade]

Originally posted by: ivan2

Originally posted by: DrPizza

Originally posted by: darkxshade
What kind of mechanism is used to hold the ropes on the roof?

Tied. For the sake of argument, there is nothing to loop the ropes over, loop the ropes through, etc. The only thing you get to interact with are the ropes & the knife.

well u can cut one rope close to the top and then tie a loop with the remainer of that rope still tied to the ceiling, it changes almost nothing.

This

Pretty much cut as much as you need to create a tight loop. With the longer rope, tie a knot on one end, loop the other end through so that the knot gets caught in the loop once the entire rope is through. Move to that new rope, cut the other rope entirely, tie that to the knot but before the loop. So it's rope b tied to knot of rope a before loop of remaining rope a... since your weight is on long rope a, the knot holds you in place so you don't fall. Propel down, grab rope B and pull, since the knot is on the other side of the loop at rope B, you get both ropes through the loop. So you can have roughly all of the rope depending on how much rope you need to make a loop.

The only reason I ask about the mechanism, is that if there was already a loop in place, I would simply say 500 ft.



edit: here's a simple picture diagram

------------------o-o-o-O--------------------

first o: knot of rope A
second o: knot created from typing rope B to A
third o: loop created from small portion of rope A
O: You as you hang up there.

 

[darkxshade]

For #4:

Player A cannot force a win, Player B can always force a tie.

There are only 10 combinations that can be used to create 15 from 1-9:

951
942
861
852
843
762
753
681
672
654

Naturally, the best way to win is to pick numbers that can be used in the most combinations... that would yield the even numbers which can each be used 4 times:

2, 4, 6, 8


*corrected* There's 2 ways to win, either A achieves 15 with odd numbers only or 2 evens and an odd. So actually picking even isn't the best way to win but it's the best way to trick the opponent.


Person B can counter by picking an even number as well, preferably 2 or 4, whichever is available.

At this point, it doesn't matter anymore... of the 10 combinations, 5 uses two even numbers and the remaining 5 must use 3 odd numbers which is no longer an option as an even had already been selected so player A would have to start from scratch which player B can counter easily.

If A picks another even number, B shouldn't select the remaining even because you can't make 15 with 3 even numbers anyway. Instead B should block A by selecting the one odd number that yields 15 and he gets the upper hand because no matter what, this odd number will force A to select the remaining even number to create 15.

For ex.
A selects 4
B selects 2
A selects 6(10 total) or 8 (12 total)
B selects 5(to block) or 3 (to block)
B now has either (2+5) = 7 or (2+3) = 5
A is forced to select the remaining even number whether it be 6 or 8 to block B. B now has the upper hand.


If for the 2nd pick, A decides to go with an odd number instead, the game is also over as he would have to select the one that forces B to select the remaining even number because 15 cannot be achieved with even odd odd.

For the sake of argument, if A wanted to start the game by selecting an odd number, B can counter by selecting an odd. The point is to see how player A wants to proceed with getting 15... all odds or 2 evens and one odd. In any case, B can force a tie.

 

[CoinOperatedBoy]

Originally posted by: darkxshade
For #4:

Player A cannot force a win, Player B can always force a tie.

There are only 10 combinations that can be used to create 15 from 1-9:

951
942
861
852
843
762
753
681
672
654

Two of your combinations were already stated in a different order. There are 8.

 

[darkxshade]

Originally posted by: CoinOperatedBoy

Originally posted by: darkxshade
For #4:

Player A cannot force a win, Player B can always force a tie.

There are only 10 combinations that can be used to create 15 from 1-9:

951
942
861
852
843
762
753
681
672
654

Two of your combinations were already stated in a different order. There are 8.

Heh, you're right, I wasn't really paying attention to that... I was more trying to explain the whole thing with fewer words... I guess this will suffice:

The only 2 ways to win is 3 odd numbers or 2 evens and an odd.

So the strategy person B should use is to follow suit with A on the first pick, it's the 2nd pick that we determine A's strategy. So if A selects even even, then continually block with odds. If A selects even odd, then block with even and force A to block you. If A selects odd odd, continually block by selecting odds.

 

[silverpig]

Seems they've all been solved.

And the bike goes backwards.