Jump to content
Fantasy Football - Footballguys Forums

Math Puzzles (from FiveThirtyEight - new puzzle every Friday)


Recommended Posts

Easiest explanation:

...

Proof by Induction.

...

These are both great posts. This was one of those days that I had only a few short and scattered moments to think about the puzzle and yet somehow I managed to vastly overcomplicate it.

Agree - both of those are really good.

I "proved" it empirically by writing a simulation, but wasn't sure exactly what the math was going to be.

Link to comment
Share on other sites

New puzzle, and solution to last week's.

There is a duck paddling in a perfectly circular pond, and a Nova Scotia duck tolling retriever prowling on the pond’s banks. The retriever very much wants to catch the duck. The dog can’t swim, and this particular duck can’t take flight from water. But the duck does want to fly away, which means it very much wants to make it safely to land without meeting the dog there. Assume the dog and the duck are both very smart and capable of acting perfectly in their best interests — they’re both expert strategists and tacticians. Say the duck starts in the center of the pond, and the dog somewhere right on the pond’s edge. To ensure that the duck could never escape safely, how many times faster would the dog have to be able to run relative to how fast the duck can swim? (Assume the dog is a good boy, yes he is, and thus very patient. The duck can’t simply wait him out. Also assume both animals can change direction without losing speed, and that the duck can take flight safely immediately after reaching shore if the dog isn’t there.)

Link to comment
Share on other sites

New puzzle, and solution to last week's.

There is a duck paddling in a perfectly circular pond, and a Nova Scotia duck tolling retriever prowling on the pond’s banks. The retriever very much wants to catch the duck. The dog can’t swim, and this particular duck can’t take flight from water. But the duck does want to fly away, which means it very much wants to make it safely to land without meeting the dog there. Assume the dog and the duck are both very smart and capable of acting perfectly in their best interests — they’re both expert strategists and tacticians. Say the duck starts in the center of the pond, and the dog somewhere right on the pond’s edge. To ensure that the duck could never escape safely, how many times faster would the dog have to be able to run relative to how fast the duck can swim? (Assume the dog is a good boy, yes he is, and thus very patient. The duck can’t simply wait him out. Also assume both animals can change direction without losing speed, and that the duck can take flight safely immediately after reaching shore if the dog isn’t there.)

Seems like the minimum is Pi times faster

Link to comment
Share on other sites

The duck can go from any edge to the other edge in a straight line, which would mean the dog would have to run half of the 2 pi r circumference of the circle in the same time the duck swam the diameter.

If the duck starts in the center, then he can always swim from the center to the exact opposite of the side the duck is on. So he would have to run half the circumference in the time it took for the duck to go the radius.

But that's not the solution.

Because if the duck sees the dog run left, he can swim towards the edge but also further in the same direction in a swirly pattern.

This is hard.

Link to comment
Share on other sites

Seems like the minimum is Pi times faster

That's the minimum if the duck employs a simple strategy of "swim in the opposite direction of where the dog is at the beginning, and try to beat him to the other side." But I'm sure the duck can do better than that, which means the dog must be faster than that.

Link to comment
Share on other sites

Because if the duck sees the dog run left, he can swim towards the edge but also further in the same direction in a swirly pattern.

This is hard.

Yeah this seems like it might require some calculus or something. I was thinking the same thing as you, the duck should have some strategy where, at every moment, it swims in the opposite direction of wherever the dog is at that exact moment. Which would probably result in some kind of spiral to the shore.

Link to comment
Share on other sites

Seems like the minimum is Pi times faster

That's the minimum if the duck employs a simple strategy of "swim in the opposite direction of where the dog is at the beginning, and try to beat him to the other side." But I'm sure the duck can do better than that, which means the dog must be faster than that.

I am sure there is something else to it - but the dog is always furthest away when he has to run half the circumference. And the Duck is closest to the opposite shore of the dog when it is in the middle - right?

Edited by Sinn Fein
Link to comment
Share on other sites

If the dog starts at 12 o clock, and the duck starts in the center, he will swim towards six of clock.

If the dog runs towards one, the duck will change directions slightly and swim towards seven. No matter how fast the dog runs, the duck will always be able to course correct when it's at the center.

That's because it's much faster for the duck to change directions the closer he is to the center. So we can probably ignore that. Because no matter where the duck is, the last part of the swim is where the dog has maximum information, and there's no way for the duck to hide it's final landing spot.

Link to comment
Share on other sites

And the Duck is closest to the opposite shore of the dog when it is in the middle - right?

Seems like that would depend on their relative speeds. Imagine the lake has a radius of 10m. If the duck can use a strategy to swim in the opposite direction of where the dog is at any moment, then he can start swimming away from center, and continuously veer towards the opposite side of where the dog is, in which case he's still opposite from the dog but now he's less than 10m away from the shore.

Link to comment
Share on other sites

Scaling: suppose that the pond radius is 1, that the duck speed is 1 and the dog speed is d (that’s the only scale definition we need because d is always relative to the duck speed).

Intuition is that, at start, if d is bigger than pi than the dog can travel to any point opposite once the duck goes into that direction.

But the duck can make sure that he’s further from the dog by swimming to a distance 1/d from center and then swim in a smaller circle (center of this small circle same as the center of the pond) from there. Given that there is no time limit, the duck will eventually find himself exactly opposite the dog (since he can swim around that small circle as fast as the dog can adjust on the edge of the pond).

Once the duck is opposite the dog, his distance to the shore is: 1 - 1/d and the dog has to race a distance of pi at speed d. Solving for d, my guess is that it has to be greater than pi+1.

Edited by JayMan
  • Like 1
Link to comment
Share on other sites

Let’s say the pond has radius R, and that the dog runs X times faster than the duck.

The duck can swim R/X towards the shore in a straight line. (Doesn’t matter which direction, or where the dog goes at the moment). You can now imagine that he’s sitting on the edge of an “inner” circle with radius R/X. This circle is exactly X times smaller than the pond, and since the duck is X times slower than the dog, they can move around their respective circles at the same exact speed.

At any point inside that inner circle, the duck can travel faster around the center than the dog can. Thus the duck can effectively position himself opposite the dog. e.g. imagine the duck is slightly closer to the center. He’s now on a circle that’s slightly smaller, so he can travel on his circle faster than the dog can travel on his. Therefore, he can (eventually) position himself directly opposite of the dog.

So basically the duck can position himself on a circle of radius R/X, opposite from the dog. From this position, the duck must travel a distance of R - (R/X) to get to shore in a straight line, and the dog must travel a distance of R*pi.

If the dog’s distance is less than X times farther than the duck’s distance, then the dog will catch the duck. In other words, if X > (R*pi)/(R - (R/X)) then the dog will catch the duck. Cancel out the R’s and rearrange and we get that if X > pi + 1, the dog will catch the duck.

  • Like 2
Link to comment
Share on other sites

Why does the dog have to move around the pond, when the duck is swimming in his smaller circle? If the dog is Pi faster, he does not really need to leave his side of the pond until the duck commits to cross the center, right? The Dog simply heads to the shore closest to the duck's circle.

Link to comment
Share on other sites

Why does the dog have to move around the pond, when the duck is swimming in his smaller circle? If the dog is Pi faster, he does not really need to leave his side of the pond until the duck commits to cross the center, right? The Dog simply heads to the shore closest to the duck's circle.

I don't think I understand your question completely. Here's a slightly different explanation that might make sense?

Picture when the duck is really close to center, he can swim in a circle around the center faster than the dog can run around the pond.

Now also picture that when the duck is really close to shore, the dog can run around the pond faster than the duck can swim in a circle around the center.

There's some point in between where that change occurs - a distance from center at which the duck can swim around center at the same exact speed as the dog can run around the pond.

No matter what the radius of the pond is, or what their relative speeds are, this point exists (and it's precisely R/X away from center). The duck's strategy is to get to that point, and be positioned on the opposite side of center from the dog. From there he just swims straight to shore and hopes to get there before the dog can. The mathy bits above show that if X is greater than (pi + 1), the dog will catch the duck.

So to (maybe) answer your question, the dog really doesn't have to move until the duck reaches that inner circle. The duck's going to get there anyway, and is going to position himself opposite from the dog. The only thing that matters is from that point forward, if the dog is more than (pi+1) times faster than the duck, he'll catch him; if he isn't, he won't.

Link to comment
Share on other sites

Why does the dog have to move around the pond, when the duck is swimming in his smaller circle? If the dog is Pi faster, he does not really need to leave his side of the pond until the duck commits to cross the center, right? The Dog simply heads to the shore closest to the duck's circle.

I don't think I understand your question completely. Here's a slightly different explanation that might make sense?

Picture when the duck is really close to center, he can swim in a circle around the center faster than the dog can run around the pond.

Now also picture that when the duck is really close to shore, the dog can run around the pond faster than the duck can swim in a circle around the center.

There's some point in between where that change occurs - a distance from center at which the duck can swim around center at the same exact speed as the dog can run around the pond.

No matter what the radius of the pond is, or what their relative speeds are, this point exists (and it's precisely R/X away from center). The duck's strategy is to get to that point, and be positioned on the opposite side of center from the dog. From there he just swims straight to shore and hopes to get there before the dog can. The mathy bits above show that if X is greater than (pi + 1), the dog will catch the duck.

So to (maybe) answer your question, the dog really doesn't have to move until the duck reaches that inner circle. The duck's going to get there anyway, and is going to position himself opposite from the dog. The only thing that matters is from that point forward, if the dog is more than (pi+1) times faster than the duck, he'll catch him; if he isn't, he won't.

Nice analysis. And use of the term "mathy bits".

Link to comment
Share on other sites

There's some point in between where that change occurs - a distance from center at which the duck can swim around center at the same exact speed as the dog can run around the pond.

You're onto something here. The duck doesn't need to go the full radius of the pond, only a fraction of it. He can position himself some distance from the center and still have the dog on the far bank. Lets for a second say that point is halfway between the center and the outside - and the pond has a diameter of 20m (therefor a radius of 10m, and this duck can get 5m from shore with the dog as far from him as possible).

The dog needs to go 31.415m (half the circumference) in the same time (or less) than the duck swims that last 5m. 31.415 / 5 = 6.283 times as fast (which is 2 pi).

Now if the duck can only get 2m from center - he'd need to swim the last 8m straight. The dog would have to go the same 31.415m in the same or less time than the duck swims 8m. 31.415 / 8 = 3.97x as fast.

It all will depend on how far from center the duck can go and still have the dog as far from him as possible.

Link to comment
Share on other sites

There's some point in between where that change occurs - a distance from center at which the duck can swim around center at the same exact speed as the dog can run around the pond.

You're onto something here. The duck doesn't need to go the full radius of the pond, only a fraction of it. He can position himself some distance from the center and still have the dog on the far bank. Lets for a second say that point is halfway between the center and the outside - and the pond has a diameter of 20m (therefor a radius of 10m, and this duck can get 5m from shore with the dog as far from him as possible).

The dog needs to go 31.415m (half the circumference) in the same time (or less) than the duck swims that last 5m. 31.415 / 5 = 6.283 times as fast (which is 2 pi).

Now if the duck can only get 2m from center - he'd need to swim the last 8m straight. The dog would have to go the same 31.415m in the same or less time than the duck swims 8m. 31.415 / 8 = 3.97x as fast.

It all will depend on how far from center the duck can go and still have the dog as far from him as possible.

Of course, if the dog is 2pi times faster - and the duck is 5m from the center in my example - his circle has a circumference of 31.415m. The dog, 6.28x times faster can run the full 62.83m circumference much faster than the duck can swim his 31.415m circle - meaning the duck couldn't get that far from the center in the first place (with the dog on the opposite shore).

Edited by matttyl
Link to comment
Share on other sites

"The duck's strategy is to get to that point, and be positioned on the opposite side of center from the dog. From there he just swims straight to shore and hopes to get there before the dog can"

I'm not convinced this is the optimal strategy for the duck. If the duck is at the west end of the inner circle when the dog is at the east end, and the dog takes the north route, why would the duck head straight west and not veer to the south?

I think the duck's optimal strategy is to head toward the point on the pond directly opposite the dog at all times, which would be some kind of spiral.

Link to comment
Share on other sites

"The duck's strategy is to get to that point, and be positioned on the opposite side of center from the dog. From there he just swims straight to shore and hopes to get there before the dog can"

I'm not convinced this is the optimal strategy for the duck. If the duck is at the west end of the inner circle when the dog is at the east end, and the dog takes the north route, why would the duck head straight west and not veer to the south?

I think the duck's optimal strategy is to head toward the point on the pond directly opposite the dog at all times, which would be some kind of spiral.

because that's a longer route, giving the obviously faster dog more time to catch him.

  • Like 1
Link to comment
Share on other sites

"The duck's strategy is to get to that point, and be positioned on the opposite side of center from the dog. From there he just swims straight to shore and hopes to get there before the dog can"

I'm not convinced this is the optimal strategy for the duck. If the duck is at the west end of the inner circle when the dog is at the east end, and the dog takes the north route, why would the duck head straight west and not veer to the south?

I think the duck's optimal strategy is to head toward the point on the pond directly opposite the dog at all times, which would be some kind of spiral.

Well, I guess the point is that if the dog's speed exceeds the given multiple of the duck's speed, it doesn't really matter what the duck does. The dog will catch him. You're right that the duck can prolong his inevitable capture by spiraling towards the shore instead of going straight for it but either way he'll be caught because the dog can go around the center of the lake faster than the duck can (beyond the border of the inner circle).

Link to comment
Share on other sites

Ok, here goes nothing....

Duck starts in the center of a 10m diameter pond. He needs to swim 5m (radius, r) faster than the dog can run the 15.708m (half circumference). Therefor dog must be at least 15.708 / 5 = pi times faster to have any shot from the start. The final answer will be > pi.

The duck, in this situation, can go out from a center far enough that the circumference of the circle he's swimming is the diameter of the full circle. Meaning he can go 1.5915m out from the center of the pond. He's now swimming a circle of 2 x 1.5915m x pi = 10m circumference. He can swim *half that* (my mistake from initially) circle in exactly the same amount of time than the dog can run a half circumference of the pond. The duck is going around in circles, and the dog is always at the far side of the pond from him.

The duck, though, is now only 5m - 1.5915m = 3.4085m away from the shore. He can swim that distance faster than the dog can run the 15.708m. At this point the dog would need to be 15.708 / 3.4085 = 4.6085x faster. Of course, if that were the case, the duck wouldn't have been able to swim that far from the center and keep the dog on the far side - because he couldn't swim 5m half-circles as fast as the dog could run 15.708m pond half circumferences.

The ultimate answer is between pi and 4.6085.

Edited by matttyl
Link to comment
Share on other sites

12 hours ago, Ignoratio Elenchi said:

I thought we had this solved but a friend of mine who's smarter than I am told me my answer isn't right.  I assume he's correct but I'm not 100% sold on his explanation. 

 

 

I’ve been thinking about this also – trying to see if it was possible for the duck to escape even if the dog moved at more than pi+1 times the duck speed – that we got to by the smaller concentric circles to make sure the dog was opposite the duck.

With the same parameters (duck swimming along that smaller circle to make sure the dog is opposite him diagonally on the shore), I’m guessing that there is a way for the duck to ensure the dog has to run further than the exact opposite pi distance – by going tangential (to his smaller circle) in the opposite direction the dog is running at the exact moment they are diagonally opposite (Tough to explain without an image). This means that the duck has to swim a longer distance than if he heads straight to the shore – but it also means that the dog has to run a lot more distance than the half-circumference.

Not sure it’s that easy to prove – but if others want to chime in on this.

  • Like 1
Link to comment
Share on other sites

Not 538, but this is a pretty interesting set of questions asked of British schoolkids.  They get pretty hard toward the end.  I think I got 8/10 - missed 8 and 10 (should have gotten 10, but the probabilities in 8 always get me).

https://www.theguardian.com/science/2016/feb/15/can-you-solve-it-are-you-smarter-than-a-british-13-year-old

Link to comment
Share on other sites

3 hours ago, JayMan said:

I’ve been thinking about this also – trying to see if it was possible for the duck to escape even if the dog moved at more than pi+1 times the duck speed – that we got to by the smaller concentric circles to make sure the dog was opposite the duck.

 

With the same parameters (duck swimming along that smaller circle to make sure the dog is opposite him diagonally on the shore), I’m guessing that there is a way for the duck to ensure the dog has to run further than the exact opposite pi distance – by going tangential (to his smaller circle) in the opposite direction the dog is running at the exact moment they are diagonally opposite (Tough to explain without an image). This means that the duck has to swim a longer distance than if he heads straight to the shore – but it also means that the dog has to run a lot more distance than the half-circumference.

 

Not sure it’s that easy to prove – but if others want to chime in on this.

 

This is the gist of his answer.  It seems like he knows what he's talking about but it doesn't make complete sense to me.  Basically, the answer is that rather than our solution having the duck swim straight for shore once reaching the "inner circle", it should instead swim in a straight line 90 degrees from this line (tangent to the "inner circle").  There's some handwaving and somehow this means that the dog has to take the long way around the lake, meaning the dog would have to be something like ~4.6x the duck's speed to catch it.  

It doesn't really make sense to me because I don't see why, if the duck breaks left, the dog doesn't also go left (perhaps after waiting some infinitesimal amount of time to see which way the duck goes), rather than going 3/4 of the way around the circle to the right.  

Edit: Eh, suddenly I think I get it.  Perhaps the duck doesn't immediately start going left (or right) but instead continues straight ahead out of the circle.  The dog can't just wait him out, because the duck would just keep swimming towards the shore while the dog remained frozen on the opposite side, waiting for the duck to turn.  So the dog has to commit to a certain direction, and once he does the duck breaks the other way. 

What is the crux of this solution, is that if the straight line between duck and dog ever passes through the center again, then they're in the same position as before, except now the duck is closer to shore (i.e. the duck is now on a larger "inner circle").  So once the dog commits and the line between duck and dog is on one side of center or the other, the dog can't reverse course.  

If I'm right then I think the whole "tangent to the circle" bit is what threw me because it seemed to imply that the duck immediately chose a direction.  But instead if the duck goes out a little from the circle, the dog is forced to pick sides first and then the duck can react.  The distance is arbitrarily small so by definition the path probably is still technically tangent to the circle but I think that's the bit of reasoning I couldn't make sense of, until just now.  

Edited by Ignoratio Elenchi
Link to comment
Share on other sites

1 hour ago, Sand said:

Not 538, but this is a pretty interesting set of questions asked of British schoolkids.  They get pretty hard toward the end.  I think I got 8/10 - missed 8 and 10 (should have gotten 10, but the probabilities in 8 always get me).

https://www.theguardian.com/science/2016/feb/15/can-you-solve-it-are-you-smarter-than-a-british-13-year-old

7/10 :bag:  But I did them all quickly in my head because I was impatient.  If I'd actually used the 24 minutes allotted or whatever I definitely would've gotten at least 9, probably 10 of them.  I'm the opposite of you I think, I'd definitely have gotten 8 but might've still gotten 10 wrong just by being careless.  

Link to comment
Share on other sites

On 2/12/2016 at 8:55 AM, BroadwayG said:

No way I'm going to be able to do the math, but my final answer will be the duck's optimal target will be the opposite point of a chord containing the dog and duck location.

Not sure if this is still optimal or if it's better for the duck to target the shore directly opposite the dog.  Still feel good about the fibonacci/golden spiral path for the duck.

Link to comment
Share on other sites

I think now the optimal path is basically an L shape.  Say the duck is at center and the dog is at the south shore.  The duck should swim north until it reaches the edge of the inner circle with radius R/X (where X is the ratio of dog speed to duck speed, and R is the radius of the pond).  Technically the dog could chase and force the duck to do more of a spiral path within this circle but it's pointless, the dog knows the duck can position itself opposite of center anyway so why waste the effort?  Once the duck reaches the edge of this inner circle, the dog has to pick a direction to chase.  Say he goes counterclockwise.  Then the duck immediately makes a 90 turn left and heads straight for shore.  The dog has to ~2/3 of the way around the circle to where the duck will land.  This works out to the dog needing to be a little more than 4.6x the duck's speed to catch it.  

Link to comment
Share on other sites

28 minutes ago, Ignoratio Elenchi said:

I think now the optimal path is basically an L shape.  Say the duck is at center and the dog is at the south shore.  The duck should swim north until it reaches the edge of the inner circle with radius R/X (where X is the ratio of dog speed to duck speed, and R is the radius of the pond).  Technically the dog could chase and force the duck to do more of a spiral path within this circle but it's pointless, the dog knows the duck can position itself opposite of center anyway so why waste the effort?  Once the duck reaches the edge of this inner circle, the dog has to pick a direction to chase.  Say he goes counterclockwise.  Then the duck immediately makes a 90 turn left and heads straight for shore.  The dog has to ~2/3 of the way around the circle to where the duck will land.  This works out to the dog needing to be a little more than 4.6x the duck's speed to catch it.  

So the dog will turn around.  The dog will always do what's in his own best interest.  The dog will always go the shortest distance at the time to be at the point on the shore the least distance from where the duck is in the pond.

Link to comment
Share on other sites

2 minutes ago, matttyl said:

So the dog will turn around.  The dog will always do what's in his own best interest.  The dog will always go the shortest distance at the time to be at the point on the shore the least distance from where the duck is in the pond.

That's one of the pieces I had to put together in my head earlier.  I explained it a bit in a previous post, but that's counterproductive for the dog.  Imagine a straight line between the duck and dog.  If that line crosses the center of the pond, then at that exact moment they're in the same relative position as they were before the dog started running, except now the duck is even closer to the shore than he was before.  So he just reacts from that point as he did earlier.  

Once the dog chooses a direction to run, he "drags" the imaginary line between himself and the duck to one side of the center.  Once he's done this, it's counterproductive to let that line cross back over the center of the pond again so he has basically no choice but to keep running in the same direction.  At that point he's either fast enough to catch it or he isn't.  (Which kind of makes the whole puzzle moot - if the duck and dog are these perfect strategists who know each other's speeds and can react perfectly and instantaneously, then they both know from the outset whether the dog's going to win or not.  So the whole charade of optimal paths and whatnot seems ultimately pointless.)

 

 

Link to comment
Share on other sites

1 minute ago, Ignoratio Elenchi said:

That's one of the pieces I had to put together in my head earlier.  I explained it a bit in a previous post, but that's counterproductive for the dog.  Imagine a straight line between the duck and dog.  If that line crosses the center of the pond, then at that exact moment they're in the same relative position as they were before the dog started running, except now the duck is even closer to the shore than he was before.  So he just reacts from that point as he did earlier.  

Once the dog chooses a direction to run, he "drags" the imaginary line between himself and the duck to one side of the center.  Once he's done this, it's counterproductive to let that line cross back over the center of the pond again so he has basically no choice but to keep running in the same direction.  At that point he's either fast enough to catch it or he isn't.  (Which kind of makes the whole puzzle moot - if the duck and dog are these perfect strategists who know each other's speeds and can react perfectly and instantaneously, then they both know from the outset whether the dog's going to win or not.  So the whole charade of optimal paths and whatnot seems ultimately pointless.)

 

 

So the dog is Werner Herzog? Or the duck?

Link to comment
Share on other sites

Maybe I'm off, but the way I read the question was "what's the minimum speed of the dog relative to the duck so that the dog would catch him?"  So the sake of argument lets say that's 3.5x.  But that's not necessarily the actual speed of the dog relative to the duck - he may only be 3x the speed of the duck - so the duck flies away.

Link to comment
Share on other sites

1 minute ago, matttyl said:

Maybe I'm off, but the way I read the question was "what's the minimum speed of the dog relative to the duck so that the dog would catch him?"  So the sake of argument lets say that's 3.5x.  But that's not necessarily the actual speed of the dog relative to the duck - he may only be 3x the speed of the duck - so the duck flies away.

Right - I'm not sure if you think that disagrees with my answer but that is the question.  And the answer is ~4.6x.  If the dog speed is slower than that (relative to the duck) then the duck has a strategy to escape.  If the dog is faster than that, it can catch the duck.

Link to comment
Share on other sites

On 2/12/2016 at 3:27 PM, matttyl said:

Ok, here goes nothing....

 

Duck starts in the center of a 10m diameter pond.  He needs to swim 5m (radius, r) faster than the dog can run the 15.708m (half circumference).  Therefor dog must be at least 15.708 / 5 = pi times faster to have any shot from the start.  The final answer will be > pi.

 

The duck, in this situation, can go out from a center far enough that the circumference of the circle he's swimming is the diameter of the full circle.  Meaning he can go 1.5915m out from the center of the pond.  He's now swimming a circle of 2 x 1.5915m x pi = 10m circumference.  He can swim *half that* (my mistake from initially) circle in exactly the same amount of time than the dog can run a half circumference of the pond.  The duck is going around in circles, and the dog is always at the far side of the pond from him. 

 

The duck, though, is now only 5m - 1.5915m = 3.4085m away from the shore.  He can swim that distance faster than the dog can run the 15.708m.  At this point the dog would need to be 15.708 / 3.4085 = 4.6085x faster.  Of course, if that were the case, the duck wouldn't have been able to swim that far from the center and keep the dog on the far side - because he couldn't swim 5m half-circles as fast as the dog could run 15.708m pond half circumferences. 

 

The ultimate answer is between pi and 4.6085.

So I'm going to try pi + 1 really quick to see how it plays out - same size pond and all.

Duck can go 1.2073m away from center of pond and still have dog on opposite shore.  Duck is swimming his circle (2 pi r = circumference of 7.58m) in x period of time, and dog is running his 31.4159m circles in 4.1415 (pi +1) times that speed.  [7.58m x 4.141592 = 31.41m]

Duck is now 5m - 1.2073m (3.7927) away from shore, with dog on opposite shore.  Duck can go 3.7927m in EXACTLY THE SAME AMOUNT OF TIME that dog can run 15.708m (half circumference of pond) - because 15.70796 / 4.141592 = 3.7927m.

Problem solved.  Pi plus 1.  Final answer Regis.

Link to comment
Share on other sites

2 minutes ago, matttyl said:

Problem solved.  Pi plus 1.  Final answer Regis.

That's what I thought before:

On February 12, 2016 at 10:10 AM, Ignoratio Elenchi said:

Cancel out the R’s and rearrange and we get that if X > pi + 1, the dog will catch the duck.

Now I'm pretty sure it's not, and that ~4.6x is the right answer (I haven't worked it out exactly but my new answer is a little over 4.6x).  

 

  • Like 1
Link to comment
Share on other sites

Also, for what it's worth - the difference in the radius of the pond (5m) and the radius of the smaller circle the duck is swimming (1.2073m) is 3.7927m - as mentioned above.  Twice that amount is 7.58, which is the circumference of the smaller circle the duck is swimming.  Not sure if that's the "check" for this problem, but it made me say "huh, interesting."

Link to comment
Share on other sites

15 minutes ago, Ignoratio Elenchi said:

That's what I thought before:

 

I guess I missed your post on Friday, but I think that's spot on.  If the dog is any faster than that, he'll always catch the duck.  If the dog is any slower than that - he'd be able to make a larger inner circle from the start and thus have a smaller distance to travel to shore and escape. 

Link to comment
Share on other sites

8 minutes ago, matttyl said:

I guess I missed your post on Friday, but I think that's spot on.  If the dog is any faster than that, he'll always catch the duck.  If the dog is any slower than that - he'd be able to make a larger inner circle from the start and thus have a smaller distance to travel to shore and escape. 

If the dog is, say 4.5x faster than the duck, the duck can escape, using the method I described above.  The duck swims to the edge of the inner circle, waits for the dog to start running and then swims 90 degrees in the opposite direction.  The dog can either keep going (in which case he has to go 2/3 of the way around the pond), or backtrack (in which case the duck-dog line will cross the center of the pond again, and we're right back where we started except with the duck even closer to shore than it was before).  Either way the duck gets away.  

4.6x is the speed at which this strategy doesn't work for the duck, and the dog will always catch it.  

Edited to add: "2/3" and "4.6x" above are rough approximations.  I'll work out the exact figures later.  

Edited by Ignoratio Elenchi
Link to comment
Share on other sites

Just now, Ignoratio Elenchi said:

If the dog is, say 4.5x faster than the duck, the duck can escape, using the method I described above.  The duck swims to the edge of the inner circle, waits for the dog to start running and then swims 90 degrees in the opposite direction.  The dog can either keep going (in which case he has to go 2/3 of the way around the pond), or backtrack (in which case the duck-dog line will cross the center of the pond again, and we're right back where we started except with the duck even closer to shore than it was before).  Either way the duck gets away.  

4.6x is the speed at which this strategy doesn't work for the duck, and the dog will always catch it.  

The inner circle is smaller if the dog is that much faster, though.  R of this circle will be around 1m. 

Also, what's the distance to shore for the duck after making the 90 degree turn?  It's not that much less than r of the pond (5m in my example).  If dog is 4.5x faster, he can do ~20m in the time it takes duck to swim to shore, and dog can switch directions and "backtrack" immediately when duck makes that 90 degree turn.  He'd only have about ~10m to go.

Link to comment
Share on other sites

14 minutes ago, matttyl said:

The inner circle is smaller if the dog is that much faster, though.  R of this circle will be around 1m. 

Also, what's the distance to shore for the duck after making the 90 degree turn?  It's not that much less than r of the pond (5m in my example).  If dog is 4.5x faster, he can do ~20m in the time it takes duck to swim to shore, and dog can switch directions and "backtrack" immediately when duck makes that 90 degree turn.  He'd only have about ~10m to go.

Radius of inner circle is 5/4.5 = 1.11m.  Perpendicular distance to shore from the edge of this circle is 4.875m.  The distance around the perimeter to this point (the long way around) is 22.44m, but the dog can only travel 4.5*4.875 = 21.94m in that amount of time.  

The dog cannot backtrack.  Once he does, he'll get to a point where the center of the pond is directly between him and the duck again, which is the position we were just in before the dog started running.  So we're starting over, except now the duck is even closer to the shore (i.e. the inner circle is even larger) than it was before.  Clearly this is a losing strategy for the dog as well.  

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Restore formatting

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

  • Recently Browsing   0 members

    No registered users viewing this page.

×
  • Create New...