Puzzle for Halflife2.net

[ericms]

How about a puzzle Halflife2.net? Don't worry guys, it doesn't require anything that you don't already know.

Consider any five points in the interior of an unit square (one with side length 1), show that at least one of the distances between two of these points is less than sqrt(2)/2.

 

[Acepilotf14]

 

[Cavalry]

How about a puzzle Halflife2.net? Don't worry guys, it doesn't require anything that you don't already know.

Consider any five points in the interior of an unit square (one with side length 1), show that at least one of the distances between two of these points is less than sqrt(2)/2.

All I got was squares squirt a distance of 2 units.
I have no idea what the question was

 

[The Monkey]

 

[theotherguy]

Um, isn't that easy? Just put five points in a row from (0,0.000001) to (0,0.000005). All of the points will be less than sqrt(2)/2 apart...

EDIT: I suppose you might have meant that I must prove that in ALL cases AT LEAST one of the points must be less than sqrt(2)/2 apart, which is a bit more difficult, but still simple.

In a unit square, the maximum allowable distance between points is sqrt(2). If you place four of the five points at the corners, you will have some distances of 1, and some distances of sqrt(2). The problem comes with placing the fifth point. To try and break the rule, we will put the point the farthest possible from all other points, that is, the center of the square. At this point, it will be exactly sqrt(2)/2 away from all the other points, therefore if we allow points ON the square rather than just IN the square, this is the one and only arrangement which will break your rule. But since you specified IN the square, then the points cannot be at the corners. They would have to be at points (0,0.99999999999), (1,0.999999), etc. At this point, the distance between the center point and the ones very near to the corners will be very close to sqrt(2)/2, but it will still be less by a minute amount.

Since this is the "breaking point" or "ideal scenario", all other arrangements will yield differences less than those in the ideal scenario. No matter how far apart you place the points, there will always be at least one which has a distance between another point less than sqrt(2)/2

 

[Reginald]

We're not gonna just do your homework for you. Unless you threaten to flush our head down the toilet or give us a wedgie. Then we may consider.

 

[ericms]

Um, isn't that easy? Just put five points in a row from (0,0.000001) to (0,0.000005). All of the points will be less than sqrt(2)/2 apart...

This question asks NO MATTER WHERE you place five points in an unit square there will always be at least one distance between two of them that is less than sqrt(2)/2. So that is only one possible case.

All I got was squares squirt a distance of 2 units.
I have no idea what the question was

I'm asking you to show that no matter how you place 5 points inside a square of side length 1 that there always exists a distance between at least two of them that is less than sqrt(2)/2.

 

[theotherguy]

This question asks NO MATTER WHERE you place five points in an unit square there will always be at least one distance between two of them that is less than sqrt(2)/2. So that is only one possible case.

okay, edited my previous post to make the proof for you.

 

[Acepilotf14]

__________
|n......p..|
|...q.....|
|_g___l___|

Do I Win?

 

[ericms]

Um, isn't that easy? Just put five points in a row from (0,0.000001) to (0,0.000005). All of the points will be less than sqrt(2)/2 apart...

EDIT: I suppose you might have meant that I must prove that in ALL cases AT LEAST one of the points must be less than sqrt(2)/2 apart, which is a bit more difficult, but still simple.

In a unit square, the maximum allowable distance between points is sqrt(2). If you place four of the five points at the corners, you will have some distances of 1, and some distances of sqrt(2). The problem comes with placing the fifth point. To try and break the rule, we will put the point the farthest possible from all other points, that is, the center of the square. At this point, it will be exactly sqrt(2)/2 away from all the other points, therefore if we allow points ON the square rather than just IN the square, this is the one and only arrangement which will break your rule. But since you specified IN the square, then the points cannot be at the corners. They would have to be at points (0,0.99999999999), (1,0.999999), etc. At this point, the distance between the center point and the ones very near to the corners will be very close to sqrt(2)/2, but it will still be less by a minute amount.

Since this is the "breaking point" or "ideal scenario", all other arrangements will yield differences less than those in the ideal scenario. No matter how far apart you place the points, there will always be at least one which has a distance between another point less than sqrt(2)/2

Yeah, this is enough. You can also break the unit square down into four congruent squares and argue that since at least two of the points are within one of these squares the maximum distance between them (if contained on the square's vertices) is sqrt(2)/2.

I'm in a puzzle sort of mood.

 

[Acepilotf14]

My turn to make a puzzle..
Uhm..
How many times can you subtract five from twenty-five?

 

[Dan]

How about this one: Prove that hexagonal close packing is the most dense possible arrangement for an array of spheres.

 

[ericms]

Because the boat was inside a submarine. Sorry, I'll not joke around.

Acepilotf14 deleted the puzzle so now my post makes no sense.

 

[theotherguy]

My turn to make a puzzle..
Uhm..
Four men were in a boat on the lake. The boat turns over, and all four men sink to the bottom of the lake, yet not a single man got wet! Why?

You could argue that a pocket of air was trapped beneath the boat, but MythBusters proved this impossible. The boat would stay afloat while the men sank.

You could also argue that they were wearing diving suits.

Or that the lake is not filled with water, but air.

 

[Raziaar]

 

[theotherguy]

How about this one: Prove that hexagonal close packing is the most dense possible arrangement for an array of spheres.

Do you mean in 2D or 3D?

In any case, the maximum number of circles (or spheres) that you can tessellate around another one is six. This translates to all dimensions. The pattern will tessellate outwards into infinity, but the structure will be the same, simply because you can only fit a finite number of spheres around another sphere.

 

[Dan]

Here's another one from biophysics: Suppose that a protein molecular chain is 100 units long. Each link in the chain can have two possible configurations like a stereo isomer. If the molecule can sample 20 different configurations every second, what is the statistical average amount of time that it will take for the molecule to achieve a particular form necessary for it to perform its function? Then, if each link is randomly oriented, what is the average end-to-end length and the standard deviation of that length of the protein?

 

[Dan]

Do you mean in 2D or 3D?

In any case, the maximum number of circles (or spheres) that you can tessellate around another one is six. This translates to all dimensions. The pattern will tessellate outwards into infinity, but the structure will be the same, simply because you can only fit a finite number of spheres around another sphere.

That's just a statement, not a proof though. And 2d doesn't tessellate into 3d the same way. That would be true if we were talking about cylinders instead of spheres. The actual proof is very complicated and I only put the question up as a joke.

Here's the wiki on Kepler's conjecture: http://en.wikipedia.org/wiki/Kepler_conjecture

 

[theotherguy]

Here's another one from biophysics: Suppose that a protein molecular chain is 100 units long. Each link in the chain can have two possible configurations like a stereo isomer. If the molecule can sample 20 different configurations every second, what is the statistical average amount of time that it will take for the molecule to achieve a particular form necessary for it to perform its function? Then, if each link is randomly oriented, what is the average end-to-end length and the standard deviation of that length of the protein?

Ugh, statistics.

100 units x 2 configurations=200 possible configurations.

there is exactly one form which will perform as necessary.

To find out the expected amount of time, we have to make a probability model.

P(correct isomer)=1/200. Assuming that the arrangements are independent, we have to find the expected value of this arrangement using a probability model. If we're trying one isomer every 20 seconds, and then not trying it again, the probability of getting the correct isomer increases over time.

so the probability model looks like this: P(20 seconds)-1/200, P(40 seconds)-199/200*1/199, P(60 seconds)-199/200*198/199*1/198, etc. This is an extremely tedious calculation, but if you multiply all of these numbers together, you will get the expected number of seconds. I really do not feel like doing this, so I assume there is a simpler way.

I also have no idea how to do the length part. Have you specified the length of each isomer?

 

[ericms]

Oh ok, joke. Lulz. I've been going back through previous years of a certain mathematics exam and here was one that I never ended up solving until someone suggested conic sections. It's a really cool idea. Needs calculus though.

A dart thrown at random, hits a square target. Assuming that any two parts of the square of equal area are equally likely to be hit, find the probability that the point hit is nearer to the center than to any edge.

Also, you can fix the target in the coordinate plane and assume it has side length two and is centered at the origin.

 

[Dan]

I also have no idea how to do the length part. Have you specified the length of each isomer?

You can leave the length in unit monomers.

 

[theotherguy]

You can leave the length in unit monomers.

k, haven't taken biochemistry, but I have taken chemistry and I thought that isomers were the same length save for a rearrangement of the positions of its component parts. And in that case the chain would be exactly the same length with every arrangement. I guess what I'm trying to say is, I have no idea how to do this puzzle :cheese:

 

[Crushenator 500]

This thread fails

I prefer riddles and logic puzzles to maths puzzles personally

 

[Gormanmod]

How does I make sence?

 

[Acepilotf14]

The first riddle I put up didn't really make sense. The answer was 'Because they were married and not single', so I don't get it. The one I edited in is better.

 

[Dan]

A dart thrown at random, hits a square target. Assuming that any two parts of the square of equal area are equally likely to be hit, find the probability that the point hit is nearer to the center than to any edge.

Also, you can fix the target in the coordinate plane and assume it has side length two centered at the origin.

Well, you can break the square down into eighths because of symmetry. So now you are just analyzing a right angle triangle. We'll define y to be the position along one side, and x to be the position down the other side of the triangle. The dividing line, between closer to centre and closer to edge will be such that x^2 = (1-x)^2+y^2. This works out to a quadratic which can be integrated to get an area.

I got 22% chance of hitting nearer to the centre than the edge.

 

[ericms]

The first riddle I put up didn't really make sense. The answer was 'Because they were married and not single', so I don't get it. The one I edited in is better.

You didn't, really? The boat flipped over for a good reason.

 

[Dan]

k, haven't taken biochemistry, but I have taken chemistry and I thought that isomers were the same length save for a rearrangement of the positions of its component parts. And in that case the chain would be exactly the same length with every arrangement. I guess what I'm trying to say is, I have no idea how to do this puzzle :cheese:

The length measured along the chain would be the same, but the end to end length depends on how it is folded. There would be a very small chance that the isomer is stretched in a perfectly straight line, making the end to end length the sum of it's component monomers.

 

[Pesmerga]

I don't get it.

 

[theotherguy]

The length measured along the chain would be the same, but the end to end length depends on how it is folded. There would be a very small chance that the isomer is stretched in a perfectly straight line, making the end to end length the sum of it's component monomers.

So this must be another joke, because I can't imagine any way to compute all the possible protein folds without massive supercomputers and a very long time.

 

[Dan]

Ok here is a good one that is a combination of logic and computer knowledge and cleverness:

An evil genius kidnaps the princess of Moldova. He sends a letter to the police chief saying that he will execute the princess unless the chief can guess a specific number between 1 and 2000 which the evil genius is thinking of. He explains further that the chief can post in the newspaper an ad containing up to 15 yes or no questions about the number he is thinking of, and he will send a reply with his answers. But the catch is that he may or may not lie about one of the answers. What 15 questions can the police chief ask to ascertain the exact number the evil genius is thinking of and get the princess returned safely?

 

[Dan]

So this must be another joke, because I can't imagine any way to compute all the possible protein folds without massive supercomputers and a very long time.

It's actually quite simple (the phrased question, not actual proteins). It's just about probabilities and diffusion. Try to model it as a random walk.

 

[Raziaar]

He needs only one... "Do you realize that we care nothing for this stupid princess because monarchism is ancient and worthless?"

Perp gives himself up, lets princess go because it doesn't matter if she lives or dies.

 

[ericms]

Well, you can break the square down into eighths because of symmetry. So now you are just analyzing a right angle triangle. We'll define y to be the position along one side, and x to be the position down the other side of the triangle. The dividing line, between closer to centre and closer to edge will be such that x^2 = (1-x)^2+y^2. This works out to a quadratic which can be integrated to get an area.

I got 22% chance of hitting nearer to the centre than the edge.

Yes, it's about that. Though I'm not understanding the reason for splitting it into eights?

The idea behind it is that the center is the focus of four parabolas each having a directrix on their respective side.

 

[Dan]

He needs only one... "Do you realize that we care nothing for this stupid princess because monarchism is ancient and worthless?"

Perp gives himself up, lets princess go because it doesn't matter if she lives or dies.

Didn't you ever watch the Princess Diaries?

 

[Dan]

Yes, it's about that. Though I'm not understanding the reason for splitting it into eights?

The idea behind it is that the center is the focus of four parabolas each having a directrix on their respective side.

An eighth is the smallest unit you can symmetrically divide a square into. It just simplifies things, because if you solve a symmetrical unit which is repeated 8 times in a square, then you've solved the whole square.

 

[theotherguy]

Ok here is a good one that is a combination of logic and computer knowledge and cleverness:

An evil genius kidnaps the princess of Moldova. He sends a letter to the police chief saying that he will execute the princess unless the chief can guess a specific number between 1 and 2000 which the evil genius is thinking of. He explains further that the chief can post in the newspaper an ad containing up to 15 yes or no questions about the number he is thinking of, and he will send a reply with his answers. But the catch is that he may or may not lie about one of the answers. What 15 questions can the police chief ask to ascertain the exact number the evil genius is thinking of and get the princess returned safely?

Well, if he weren't able to lie about it, the police chief could essentially do a binomial search. He could ask "is the number greater than 1000?" and continue on, dividing it until two until the number he gets is the correct one.

However, this won't work, because he would have to ask him the questions after receiving an answer, and there is a chance that the man could be lying. I'm stumped.

 

[ríomhaire]

The furthest distance two points on a square can be apart is at two opposite corners. On a unit square this distance happens to be root two over two (by Pythagoras' theorem) so if you put four points at the four corners then the last point must be closer to one of these points than root two over two.

 

[Dan]

Well, if he weren't able to lie about it, the police chief could essentially do a binomial search. He could ask "is the number greater than 1000?" and continue on, dividing it until two until the number he gets is the correct one.

However, this won't work, because he would have to ask him the questions after receiving an answer, and there is a chance that the man could be lying. I'm stumped.

Think about it some more. If you are still stumped after a while I can give some hints.

 

[ericms]

An eighth is the smallest unit you can symmetrically divide a square into. It just simplifies things, because if you solve a symmetrical unit which is repeated 8 times in a square, then you've solved the whole square.

Oh ok, sorry that was a dumb question of me to ask. You just split one of the conics with an axis and integrated from 0 to the intersection of it and y=x,-x?