[Guest]

A dog is in the center of a round swimming pool (radius 100 meters). A tiger, who can't swim, is on the edge of the swimming pool and tries to catch the dog. What is the maximum ratio of the tiger's running speed and the dog's swimming speed such that the dog can swim ashore without being caught by the tiger? What strategy (i.e. swimming route) should the dog use?

[Guest]

the dog will always be anle to swim ashore without being caught by the tiger because the tiger cant swim. When the dog is alreasy on shore he may get killed but the question says "swim ashore w/o being caught by the tiger," and the dog will always swim ashore safely

[Guest]

The answer is in the question...

']The dog should stay in the center of the pool, and wait until the Tiger (who can't swim,) jumps into the water, and then the dog may swim safely to shore, going in the opposite direction from which the Tiger jumped, who is now flailing helpless in the pool that he can't swim in.

Edited July 6, 2009 by racheletta

[Guest]

1:pi if the dog is in the center of the pool and starts swimming towards the farthest point from the tiger, the dog must travel the radius before the tiger travels half of the circumfrence.

[Guest]

This is not a trick question, so pls dont post answers as given by mensaman and racheletta.

NickyThump, nice try but pi is not the max ratio. There is another way for the dog to escape even when the speed of dog is greater (up to a certain limit) than pi times its speed.

Edited July 6, 2009 by DeeGee

[Guest]

I don't know about the mathematical ratio you are after, but as for the strategy, the dog should commence swimming in the opposit direction from the tiger. Then, as the tiger begins to run the circumference, the dog shouldturn and make a spiral toward the opposite shore, laboring to keep the center of the pool between him and the tiger at all times. If the tiger chagnes directions, so does the dog. Little by little he will grown closer to the shore, all the while keeping maximum distance between himself and the tiger.

[Quantum.Mechanic]

BTW, Tigers can swim.

[Guest]

Mmhh

BTW, Tigers not only do they swim, but they actually enjoy it... The dog has no chance!! However, I also calculated pi to be the ratio. (Think I've gotta try harder)

[DetroitMike]

The tiger can be 3.09 times faster than the dog and the dog can still get away. Assuming the tiger follows the circuference of the pool and waits for the dog's exit, the dog would swim 1 meter towards a side, then when the tiger is in line with that path, the dog swims 101m in the opposite direction. As soon as the dog makes it's move the tiger will have to run the 314m (the semicircle) to catch up to the dog's spot. Therefore, if the dog is to make it away the tiger can only travel 313m, leaving 1m for the dog. This ratio (313/101 = 3.09)is the maximum ratio of the tiger's speed to the dog's for a safe exit. Am I even close?

[Guest]

DetroitMike, the 3.09 ou mention is less than pi (3.14). If the dog were to just swin in the opposite direction, the tiger would need to have atleast pi times the speed to catch the dog at the shore.

For the others who mention that tigers can swim, assume that this tiger is not allowed to enter the pool.

[Guest]

ok heres wat i came up with...

the dog can reach the edge if it can travel at least 1/(pi+1) times as fast as the tiger. heres y i think that..

If the dog travels directly away from the tiger, and the tiger moves around the edge of the pool to get as close to the dog as possible, and dog will move in a spiral towards the outside until the dogs angular velocity=the tigers angular velocity, at which point the tiger will make the dog move in a circle, at which point the dog will have to make a break for the edge.

the difference in distance they have to travel depends on the radius of this circle.eg If the dog travels half as fast as the tiger, it will have the same angular velocity when the radius of the circle is half. Then the dog would have to swim to the edge opposite the tiger.

so the distance the dog has to travel is the difference in radius of the circles, and the distance the tiger makes is JUST under pi X big radius.

so i got an equation: k(R - (1/k)R) = piR

where k is the speed ratio, R is the radius (in this case 100, cancels so doesnt matter).

solving this gives k=1(pi+1) or pi+1... w.e..

Anyone get me? if im way off ignore me.

[Guest]

Dog goes to center of pool.

Dog must swim r distance in whatever direction is opposite the tiger. r = 100m

C = 100*2*3.14 = 628.3185m

The tiger must run half the distance of C, so it must run 314.159m.

By my calculations, the tiger must run more than pi times the speed of the dog to catch it.

TEST

Tiger runs (pi)m/s for 314.159m

Dog swims 1m/s for 100m

It takes 100 seconds for the tiger to circumnavigate the pool.

It takes the dog 100 seconds to cross the pool.

The dog must swim more than .31831 times the speed of the tiger.

If a maximum speed ratio of tiger to dog would be found, it would be slightly smaller than pi.

TEST

314.159m /3.14x(m/s) = 100.0507s/x (using a cut decimal as a smaller number)

100m /x(m/s) = 100s/x

So the greatest ratio (t to D) for dog survival would be (alittlelessthanpi) to 1.

This is now a trick question, as pi is commonly accepted as having infinite decimals.

Also, unless there is sombody nearby conducting this experiment (with a gun in hand) or the dog can run faster than the tiger, or the tiger is restricted in it's running distance, or the tiger is friendly...the dog will die.

[Guest]

Lets call the dog D and the tiger T.

How about: When the dog is in the centre, it begins swimming away from the tiger. As the tiger moves, the dog moves so that it moves along the line DT away form the tiger. This way if the tiger took an optimum path the dog would move in a curve (I’m guessing a spiral). Now I have a feeling that in order to find the length of this curve I would need some maths ability that I don’t yet have.

Can anyone help me with this idea?

Edited July 6, 2009 by psychic_mind

[Guest]

Lets call the dog D and the tiger T.

How about: When the dog is in the centre, it begins swimming away from the tiger. As the tiger moves, the dog moves so that it moves along the line DT away form the tiger. This way if the tiger took an optimum path the dog would move in a curve (Im guessing a spiral). Now I have a feeling that in order to find the length of this curve I would need some maths ability that I dont yet have.

Can anyone help me with this idea?

A curve in direction, keeping the tiger on the opposite side constantly?

Based on a mental image, I don't think this can be done step by step with measuring (as the goal is to find a speed). I can't graph this on paper, but in a graphing program, have a segment between two points, with the T point on the circle, and the D point moving along the segment away from the T point. Measure the path traced by D and the movement of T, to calculate out a speed.

From swimming experience, if the dog is too slow, the tiger will move to the side it is close to, blocking it.

Trace out the path and distances and then calulate out the speed ratio.

I will give this a try. Hope the methos helps.

EDIT: Ugh... no graphing program on my laptop. It will take me a while to get one.

Edited July 6, 2009 by Out4Blood4

[Guest]

I am now thinking that if the dog does that method I described above then it does not matter how fast the tiger runs as it ensures that the dog is always getting further away. I could be wrong, if the tiger can run so fast that it forces the dog to swim in circles but I’m not sure if this can happen.

[Guest]

Lets call the dog D and the tiger T.

How about: When the dog is in the centre, it begins swimming away from the tiger. As the tiger moves, the dog moves so that it moves along the line DT away form the tiger. This way if the tiger took an optimum path the dog would move in a curve (I’m guessing a spiral). Now I have a feeling that in order to find the length of this curve I would need some maths ability that I don’t yet have.

Can anyone help me with this idea?

(IF I REMEMBER MY FORMULAE CORRECTLY!!!)

I the tiger can travel 1+pi times as fast as the dog for the dog to still make it.

As the dog gets further from the origin, it's angular velocity slows for some constant velocity v.

This means that the dog can only stay opposite the tiger until some radius r' at which its angular velocity becomes equal to that of the tiger...At that point, i think the dog needs to make a run for it.

The dog keeps its angular velocity the same as the tiger's until some radius r'... as the dog gets further from the origin, Then, from r' to r, has the same amount of time that the tiger needs to travel pi units. And in fact, the simplistic variation of this idea is the same as some of the first answers, in that if the dog swims directly to the opposite side of the pool as the tiger, it needs to be able to swim 1/pi times as fast as the tiger.

I'm hoping that i remember correctly that angular velocity = velocity / radius.

So we know that the dog needs to travel r-r' units in the same time as the tiger travels pi*r units

sooo

(r-r')/V_d = r*pi/V_t

Now, if the angular velocities are the same:

V_t/r = V_d/r'

so

r' = r*V_d / V_t

(r-(r*V_d / V_t))/V_d = r*pi/V_t

V_tr - V_dr = pi*V_dr

V_t = V_d(pi + 1)

I hope i remembered all that right...

[Guest]

The tiger can be 3.09 times faster than the dog and the dog can still get away. Assuming the tiger follows the circuference of the pool and waits for the dog's exit, the dog would swim 1 meter towards a side, then when the tiger is in line with that path, the dog swims 101m in the opposite direction. As soon as the dog makes it's move the tiger will have to run the 314m (the semicircle) to catch up to the dog's spot. Therefore, if the dog is to make it away the tiger can only travel 313m, leaving 1m for the dog. This ratio (313/101 = 3.09)is the maximum ratio of the tiger's speed to the dog's for a safe exit. Am I even close?

I find it rather silly that my answer was so rudely discarded, when all these other folks are making mathematical equations, even when it's clear that, Yes, perhaps the dog gets out of the pool before getting caught, but once it's on dry land, it's done for! No WAY can that dog outrun the tiger - I don't care what the ratio to pi is or not! Although, the riddle itself is now moot, as the others pointed out what I thought was "suspension of disbelief" for the sake of the game - if the Tiger can swim, that little fact right there negates the entire riddle....ho-hum...Try to give us one with some real wit next time? lol

[Guest]

the dog must swim directly away from the tiger's position toward the opposite shore where the tiger is.

then at the hafway(R/2) the dog must change his course perpendicular to his original course away from the tiger's current position.

it will give the tiger more distance to travel(1.17pi)than the dog.

[Guest]

That is what i got, it seems right to me, i didnt explain it very well tho

[Guest]

sry, i got the same as psychic mind.

Edited July 7, 2009 by dnthrtme

[Guest]

tpaxatb definitely got it. the trick was finding the distance from the center of the pool that the dog is swimming in a circle at the same rate as the tiger is on the circumference.

[Guest]

Dog goes to center of pool.

Dog must swim r distance in whatever direction is opposite the tiger. r = 100m

C = 100*2*3.14 = 628.3185m

The tiger must run half the distance of C, so it must run 314.159m.

By my calculations, the tiger must run more than pi times the speed of the dog to catch it.

TEST

Tiger runs (pi)m/s for 314.159m

Dog swims 1m/s for 100m

It takes 100 seconds for the tiger to circumnavigate the pool.

It takes the dog 100 seconds to cross the pool.

The dog must swim more than .31831 times the speed of the tiger.

If a maximum speed ratio of tiger to dog would be found, it would be slightly smaller than pi.

TEST

314.159m /3.14x(m/s) = 100.0507s/x (using a cut decimal as a smaller number)

100m /x(m/s) = 100s/x

So the greatest ratio (t to D) for dog survival would be (alittlelessthanpi) to 1.

This is now a trick question, as pi is commonly accepted as having infinite decimals.

Also, unless there is sombody nearby conducting this experiment (with a gun in hand) or the dog can run faster than the tiger, or the tiger is restricted in it's running distance, or the tiger is friendly...the dog will die.

Well, I now I understand the mathematics of this riddle (thank you!) I am also glad that you came to the same conclusion about the livelihood of the dog once these formulas are tested; it brings me back to my initial answer.. Since the terms of this riddle state "tigers cannot swim," then in order for the dog to stay safely away from the tiger, it must simply remain in the pool until the tiger jumps towards it. The dog would now swim to the nearest point out of the pool, while the tiger drowns. The dog has now outwitted the tiger.

[Guest]

dnthrtme and tpaxatb have the ratio very close to the answer. But not quite the answer!

The answers they have is pi+1 = 4.14, and the actual answer is 4.6

The maximum speed ratio is 1 / cos t, where t satisfies:

tan(t) - t = pi = 3.1416

Edited July 7, 2009 by DeeGee

[Guest]

dnthrtme and tpaxatb have the ratio very close to the answer. But not quite the answer!

The answers they have is pi+1 = 4.14, and the actual answer is 4.6

The maximum speed ratio is 1 / cos t, where t satisfies:

tan(t) - t = pi = 3.1416

I was thinking about it last night, and I realized that there is no reason for the dog to go in a straight line to the shore once it reaches r'. Once it reaches that point, it can spiral out from there in such a way that the angular velocity of the dog from r' to r decreases at a constant rate from the tiger's angular velocity to zero...With the condition that it cannot matter if the tiger changed directions as long as the dog always swims in the exact same direction..since the tiger, if it changes directions, has to re-cover the ground he had already traversed...

Basically, it sounds like from r' to r, we are finding what the v component is such that the deceleration rate from the angular velocity of t to zero is constant. (although, this now moves r' closer to the center of the pool, since we go slower).

So i'm thinking, the dog needs to go to r', keeping the center point between the tiger and himself at all times, spiral out (here's where we need to find the correct angle...) is in the same direction as the tiger. The tiger will always start to overtake the dog after r' (you will not be able to keep the center point between the dog and the tiger any more), but if the tiger changes directions, the dog should change directions, since the tiger will have to retrace the steps it took and the dog is that much closer to the edge.

I dunno. I really don't remember how to do this, it's been 15 years since i've even touched this stuff in school...

[Guest]

The name of this "riddle" is incorrect - as this is obviously a mathematical problem, it has nothing to do with wit.

[Guest]

The name of this "riddle" is incorrect - as this is obviously a mathematical problem, it has nothing to do with wit.

Considering it is in the section entitled "Logic/Math Puzzles"....