[Guest]

Soldiers of a regiment are ready to march.

They form a square of 20 meters x 20 meters.

The commander's dog, stands in the middle of the front line.

The regiment starts to march at constant speed, and the dog starts to walk the perimeter of the regiment clockwise at constant speed at the same time.

The dog is trained in such a way that, when the regiment has marched 20 meters, the mascot has walked the perimeter completely and is back in the middle of the front line.

Question: The soldiers have marched 20 meters but, what is the distance the dog has walked?

Edited July 17, 2009 by DeeGee

[pdqkemp]

Cool puzzle. I don't have what it takes to really do the math right, but I'll try my answer/guess here anyway for fun:

I can visualize that when the dog starts off to go past the first 10 meters of front line that he needs to actually walk more than 10 meters. Then once he makes the first right turn, this will be the only time he needs to walk LESS than the distance of one of the sides because the men are walking opposite of the way the dog is. So after his first right turn, the dog walks LESS than 20 meters. While walking behind the back side of the square, he'll have to go more than 20 meters. While walking back to the front of the square, he'll walk the longest distance after he makes his 3rd right turn. Finally, in getting back past the 10 meters of men to get back to the beginning, he'll walk more than 10 meters.

Again - I don't quite have what it takes for the math. But logically it seems that the dog will have walked the entire perimeter (80 meters), PLUS a total of 20 meters more when all is said and done (and calculated correctly).

So 100 meters total.

I look forward to seeing the actual answer.

[Guest]

Are they marching in a straight line?

If they are marching in a straight line, then the distance will be equal to the perimeter, since the long walk (with the flow of the soldiers) will be cancelled out by the short walk (with the flow of the soldiers). So my guess would be 80 meters...

EDIT: as pdqkemp mentioned, the dog will cover a long distance on 3 sides.. so my answer is off, time to get a pen and paper

On the other hand, if the soldiers are not walking in a straight line then the distances wouldn't necessarily cancel out.

Then again, I have probably missed something (I normally do!!)

Ok, I did a little math and came up with 40 (the 2 sides that cancel each other out) + 41.23 (the top and bottom as it were)

So I reckon its ~81.23 meters

Edited July 17, 2009 by d31b0y

[bonanova]

83.6225 m

4.1811 times the soldier's speed

13.837^o

[Guest]

80 + 20*sqrt(2) = 108.28m

Edited July 17, 2009 by bonanova
Spoiler added.

[Guest]

83.6225 m

4.1811 times the soldier's speed

13.837^o

All correct!

Mind showing how you got this. When I was solving this, I ended up with an equation containing a 2x, an x²-1 and a 2*sq root of x²-1. (x was the ratio of their speeds)

I had to use a solver to get the results. Your solution may be a more elegant way of doing this problem.

[Guest]

Soldiers of a regiment are ready to march.

They form a square of 20 meters x 20 meters.

The commander's dog, stands in the middle of the front line.

The regiment starts to march at constant speed, and the dog starts to walk the perimeter of the regiment clockwise at constant speed at the same time.

The dog is trained in such a way that, when the regiment has marched 20 meters, the mascot has walked the perimeter completely and is back in the middle of the front line.

Question: The soldiers have marched 20 meters but, what is the distance the dog has walked? The dog walks 100 meters. 20 meters around all 4 sides of the regiments perimeter and the 20 meters the regiment walked as well if the dog stayed around the regiment

[Guest]

Ity Bit, d31b0y and pdqKemp - try the dog's path along the perimeter while the soldiers are moving. The 20 meters forward that the dog moved is not exclusive from its going around the perimeter.

Kiran Nag: How do you get that? 108 seems a very long distance!

And yes, the soldiers are marching in a straight line

Edited July 17, 2009 by DeeGee

[Guest]

Ity Bit, d31b0y and pdqKemp - try the dog's path along the perimeter while the soldiers are moving. The 20 meters forward that the dog moved is not exclusive from its going around the perimeter.

Kiran Nag: How do you get that? 108 seems a very long distance!

And yes, the soldiers are marching in a straight line

Dog has traversed the entire perimeter of a 20X20 square, which is 80m.

In addition, it as traversed the hypotenuese of the 20X20 square which is 20*sqrt(2)

So, total is around 108m. Am I missing something here?

[Guest]

Dog has traversed the entire perimeter of a 20X20 square, which is 80m.

In addition, it as traversed the hypotenuese of the 20X20 square which is 20*sqrt(2)

So, total is around 108m. Am I missing something here?

Try to follow the path the dog would follow. Do you think the dog would go at 45 degrees (for hypotenuese)? Second, if the dog goes at some angle, would that not include a side (full or partial)... You seem to be double counting the sides (front line and last line). (once at side of square and once as "hypotenuese"

[Guest]

Taking the direction of travel of the regiment as "y", and their right to be "x"...

The "y" component of the dog's velocity on his first path (vy1) is equal to the velocity of the soldiers (vs).

vy1 = vs

1/8th of the total time is spent on the first path of 10 meters "x" distance. Therefore,

vx1 = 10m/(t/8)

t = 20m/vs

==> vx1 = 4vs

The resultant velocity, which will be the constant speed all around, is

vr = sqrt( (vx1)2 + (vy1)2 )

vr = sqrt( (4vs)2 + (vs)2 )

vr = vs x sqrt(17)

or 4.123 times the speed of the soldiers.

To get the total distance, 20 meters times sqrt(17) = 82.462 meters

[Guest]

Taking the direction of travel of the regiment as "y", and their right to be "x"...

The "y" component of the dog's velocity on his first path (vy1) is equal to the velocity of the soldiers (vs).

vy1 = vs

1/8th of the total time is spent on the first path of 10 meters "x" distance. Therefore,

vx1 = 10m/(t/8)

t = 20m/vs

==> vx1 = 4vs

The resultant velocity, which will be the constant speed all around, is

vr = sqrt( (vx1)2 + (vy1)2 )

vr = sqrt( (4vs)2 + (vs)2 )

vr = vs x sqrt(17)

or 4.123 times the speed of the soldiers.

To get the total distance, 20 meters times sqrt(17) = 82.462 meters

Your assumption that 1/8th of the time is spent in the first leg is not right.

[bonanova]

All correct!

Mind showing how you got this. When I was solving this, I ended up with an equation containing a 2x, an x²-1 and a 2*sq root of x²-1. (x was the ratio of their speeds)

I had to use a solver to get the results. Your solution may be a more elegant way of doing this problem.

The dogs's journey to the four corners of the formation then to his starting point are the five distances that need to be summed.

The first, third and fifth of them are at an angle a to the path the dog would take if the soldiers were not moving;

the second and fourth are parallel to the formation. Call the distances d_i, i=1, ..., 5.

As in the dog/tiger problem, let the dog's speed ratio be s = s_dog/s_soldier = 1/sin(a).

d₁ = d₅ = .5 d₃ = 10/cos(a); these distances sum to 40/cos(a).

d₂ is 20m foreshortened by the motion of the soldiers: d₂ = 20 s_dog/(s_dog + s_soldier) = 20 s/(s + 1) = 20/[1 + sin(a)].

d₄ is 20m lengthened by the motion of the soldiers: d₄= 20 s_dog/(s_dog - s_soldier) = 20 s/(s - 1) = 20/[1 - sin(a)].

Summing all five, d_dog = 40/cos(a) + 20/[1+sin(a)] + 20/[1-sin(a)] = ... = 20 [2/cos(a) + 2/cos²(a)].

During the time the dog moved this distance, the soldiers moved 20 m, so d_dog = 20s = 20/sin(a).

Equating and simplifying, cot(a) - 2/cos(a) = 2.

Here you need a solver, or a short program to home in on a good value for a.

So a = 13.837^o, s = 1/sin(a) = 4.18, and d_dog = 20s = 83.62 m.

[Guest]

Equating and simplifying, cot(a) - 2/cos(a) = 2.

Here you need a solver, or a short program to home in on a good value for a.

Interesting method. I did it taking x as the ratio of speeds of the dog and the soldiers

Got and equation like:

2x + 2*root(x²-1) = x² - 1 or something like that

and needed a solver to go further. At least your final equation had only 2 terms to care of comapred to 3 in mine!

[bonanova]

Interesting method. I did it taking x as the ratio of speeds of the dog and the soldiers

Got and equation like:

2x + 2*root(x²-1) = x² - 1 or something like that

and needed a solver to go further. At least your final equation had only 2 terms to care of comapred to 3 in mine!

You can do pretty much the same math using distances and Pythagoras [x² + y² = z²], or angles and trig identities [sin² + cos² = 1].

In this case the angle a had an identifiable significance - the angle of paths 1, 3 and 5 - so I went with angles.

Your x is my 1/sin(a).