Of dogs and men

[bonanova]

An old Sam Lloyd classic that has to be dusted off every now and again. It's deceptively simple, and more fun to solve than to Google.

Enjoy.

A square phalanx of soldiers, 100 feet on a side, marches forward at a constant speed. The company mascot, a small terrier, trots along with them, starting at the center of the rear rank. Since he moves more quickly than the soldiers, he is able to trot at constant speed around the outside of the square, keeping as close as possible to the soldiers at all times. When he returns to his starting position at the rear of the formation, the soldiers have moved forward a distance of 50 feet. How long is the dog's path?

[caike]

I think the answer for the length of the dog's path could be 50*sqrt(65) feet.

[bhramarraj]

Since speeds of the dog and soldiers are not given, it may be assumed that the dog has completed one round of the square formed by the soldiers, and followed the return path same as it took for starting its journey i.e. it has completed 400ft + 50ft + 50ft = 500ft.

[caike]

I think the answer for the length of the dog's path could be 50*sqrt(65) feet.

Sorry, I think I am wrong...

[k-man]

About 402 feet.

When the dog runs along the sides of the phalanx it runs exactly 200 feet (whatever extra distance it has to cover when it runs in the same direction as the phalanx is regained we it runs in the opposite direction.

When the dog runs along the front and back of the phalanx, it runs diagonally. Both front and back are exactly the same, so let's consider one of them and then double the result. The distance is the hypotenuse of the triangle with sides of 100ft and X, where X is the distance travelled by the phalanx in the time it took the dog to run from one corner to another. It's about a quarter of the entire journey, so the phalanx moves about 12.5 feet in that time and the dog runs about 101 feet.

So the total distance is 200 + 101*2 = 402.

It's just an estimate. Need to think some more to come up with the exact answer.

[k-man]

~401.569 feet. The speed of the dog is approximately 8.031 times the speed of the phalanx. Exact answer is too long to post.

[bonanova]

Since speeds of the dog and soldiers are not given, it may be assumed that the dog has completed one round of the square formed by the soldiers, and followed the return path same as it took for starting its journey i.e. it has completed 400ft + 50ft + 50ft = 500ft.

What matters is only the relative speed.

If you played a recording of the event at half speed, the dog would travel the same distance.

The condition that the dog returns to his starting position relative to the phalanx in the same time interval that the soldiers move a given distance, gives a constraint on the dog's speed relative to the cadets. Find the ratio and you've solved the problem.

[bonanova]

I think the answer for the length of the dog's path could be 50*sqrt(65) feet.

Sorry, I think I am wrong...

Not far off, but it's more complex than that.

[bonanova]

~401.569 feet. The speed of the dog is approximately 8.031 times the speed of the phalanx. Exact answer is too long to post.

But the error is not in the 6th decimal place, more like the 3rd.

[bonanova]

Clue:

There is an algebraic solution (closed form) and a trigonometric solution (numerical.)

Let sin (alpha) = soldier speed / dog speed = soldier distance / dog distance. Find alpha.

[TimeSpaceLightForce]

Using graphical solution, the white line is the dog path which measure 40.1548 (scale: 1:10) see below.

position of dog vs. palanx rear center

0 0

5.0389 0.625 from start point

11.25 1.25 from last point

10.077 "

8.7501 "

5.0388 0.625 from last point

40.1548 or 401.548 ft

[bonanova]

k-man and TSLF agree on an answer that differs slightly from mine.
So as it stands I am outvoted.

But I invite a critique of my solution.

• Size of square = x
• cadet speed = v
• distance cadets move = y
• dog speed = c
• distance dog runs = d
• ratios: v/c = y/d = sin(a)

The two ratios above follow from the fact that v and c are constant, so the distances traveled by cadets and dog must have the ratio as their speeds. Using an angle a avoids dealing with all those messy square roots of sums and differences of squares.

Without changing the answer, we can start the dog at the back right corner and have it circle the cadets in a CW direction. There are four legs to the dog's path:

• sideways (mostly)
• forward
• sideways (mostly)
• backward

Consider first, only the forward and backward components of the dog's path.
Call the forward direction the y direction.

y₁ = y₃ = x tan(a)
y₂ = x [c/(c-v)] = x [1/(1-sin(a))]
y₄ = x [c/(c+v)] = x [1/(1+sin(a))]

y₁ + y₃ = 2x tan(a)
y₂ - y₄ = x [2sin(a)/cos²(a)] = 2x tan(a)/cos(a) (y₄ is reverse direction)

y/x = 2 tan(a) [1 + 1/cos(a)]

In our case, this ratio is 50/100 = 0.5
Numerical solution gives:

a = 7.09789^o

Note that this puzzle scales in terms of distance the cadets move.
There is only one equation to solve: a in terms of y/x.

Here are solutions for three values of y/x

y/x .a . . . c/v = d/y = 1/sin(a)
---+--------+-------------------+
0.5 7.09789 8.09291 <== this puzzle
1.0 13.83756 4.18112
2.0 25.39026 2.33218

We can sum similar expressions for d₁ d₂ d₃ and d₄:

d₁ = d₃ = x/cos(a)

d₂ = x [c/(c-v)] = x [1/(1-sin(a))]
d₄ = x [c/(c+v)] = x [1/(1+sin(a))]

d₁ + d₃ = 2x/cos(a)
d₂ + d₄ = x [2 / cos²(a)] = 2x/cos(a)[1/cos(a)]

d/x = 2/cos(a) [1 + 1/cos(a)]

Which agrees with our ratio: d = y/sin(a)

Either way, once we know that a = 7.09789, we obtain
d = 50/sin(7.09789) = 50*8.09291 = 404.645

Interestingly, k-man and TSLF obtain values close to
d = 50/tan(7.09789) = 50*8.03089 = 401.544.

[bushindo]

k-man and TSLF agree on an answer that differs slightly from mine.

So as it stands I am outvoted.

But I invite a critique of my solution.

• Size of square = x
• cadet speed = v
• distance cadets move = y
• dog speed = c
• distance dog runs = d
• ratios: v/c = y/d = sin(a)

The two ratios above follow from the fact that v and c are constant, so the distances traveled by cadets and dog must have the ratio as their speeds. Using an angle a avoids dealing with all those messy square roots of sums and differences of squares.

Without changing the answer, we can start the dog at the back right corner and have it circle the cadets in a CW direction. There are four legs to the dog's path:

• sideways (mostly)
• forward
• sideways (mostly)
• backward

Consider first, only the forward and backward components of the dog's path.

Call the forward direction the y direction.

y₁ = y₃ = x tan(a)

y₂ = x [c/(c-v)] = x [1/(1-sin(a))]

y₄ = x [c/(c+v)] = x [1/(1+sin(a))]

y₁ + y₃ = 2x tan(a)

y₂ - y₄ = x [2sin(a)/cos²(a)] = 2x tan(a)/cos(a) (y₄ is reverse direction)

y/x = 2 tan(a) [1 + 1/cos(a)]

In our case, this ratio is 50/100 = 0.5

Numerical solution gives:

a = 7.09789^o

Note that this puzzle scales in terms of distance the cadets move.

There is only one equation to solve: a in terms of y/x.

Here are solutions for three values of y/x

y/x .a . . . c/v = d/y = 1/sin(a)

---+--------+-------------------+

0.5 7.09789 8.09291 <== this puzzle

1.0 13.83756 4.18112

2.0 25.39026 2.33218

We can sum similar expressions for d₁ d₂ d₃ and d₄:

d₁ = d₃ = x/cos(a)

d₂ = x [c/(c-v)] = x [1/(1-sin(a))]

d₄ = x [c/(c+v)] = x [1/(1+sin(a))]

d₁ + d₃ = 2x/cos(a)

d₂ + d₄ = x [2 / cos²(a)] = 2x/cos(a)[1/cos(a)]

d/x = 2/cos(a) [1 + 1/cos(a)]

Which agrees with our ratio: d = y/sin(a)

Either way, once we know that a = 7.09789, we obtain

d = 50/sin(7.09789) = 50*8.09291 = 404.645

Interestingly, k-man and TSLF obtain values close to

d = 50/tan(7.09789) = 50*8.03089 = 401.544.

bonanova's work checks out to me. I used k-man and TSLF's answers and worked backward to get a distance of about 50.4 feet moved, so I'm casting a vote for bonanova.

[bushindo]

About 402 feet.

When the dog runs along the sides of the phalanx it runs exactly 200 feet (whatever extra distance it has to cover when it runs in the same direction as the phalanx is regained we it runs in the opposite direction.

When the dog runs along the front and back of the phalanx, it runs diagonally. Both front and back are exactly the same, so let's consider one of them and then double the result. The distance is the hypotenuse of the triangle with sides of 100ft and X, where X is the distance travelled by the phalanx in the time it took the dog to run from one corner to another. It's about a quarter of the entire journey, so the phalanx moves about 12.5 feet in that time and the dog runs about 101 feet.

So the total distance is 200 + 101*2 = 402.

It's just an estimate. Need to think some more to come up with the exact answer.

I think there's an error here (see part highlighted in red above)

To make the notation consistent with bonova's, let

c = dog speed

v = men marching speed

x = side of square = 100 feet

y2 = side where dog is running in the same direction as the men

y4 = side where dog is running in the opposite direction

If the total distance is 401.569 feet, then the speed of the dog is approximately 8.031 times the speed of the phalanx, or c/v = 8.031. For simplicity, let's assume v = 1, and c = 8.031.

y2 = [ x/( c - v ) ] * c = 114.3495

y4 = [ x/( c + v )] * c = 88.85034

y2 + y4 = 203.1998

So the total distance that the dog runs along the two sides of the phalanx (y2 + y4) is not precisely 200.

[k-man]

About 402 feet.

When the dog runs along the sides of the phalanx it runs exactly 200 feet (whatever extra distance it has to cover when it runs in the same direction as the phalanx is regained we it runs in the opposite direction.

When the dog runs along the front and back of the phalanx, it runs diagonally. Both front and back are exactly the same, so let's consider one of them and then double the result. The distance is the hypotenuse of the triangle with sides of 100ft and X, where X is the distance travelled by the phalanx in the time it took the dog to run from one corner to another. It's about a quarter of the entire journey, so the phalanx moves about 12.5 feet in that time and the dog runs about 101 feet.

So the total distance is 200 + 101*2 = 402.

It's just an estimate. Need to think some more to come up with the exact answer.

I think there's an error here (see part highlighted in red above)

To make the notation consistent with bonova's, let

c = dog speed

v = men marching speed

x = side of square = 100 feet

y2 = side where dog is running in the same direction as the men

y4 = side where dog is running in the opposite direction

If the total distance is 401.569 feet, then the speed of the dog is approximately 8.031 times the speed of the phalanx, or c/v = 8.031. For simplicity, let's assume v = 1, and c = 8.031.

y2 = [ x/( c - v ) ] * c = 114.3495

y4 = [ x/( c + v )] * c = 88.85034

y2 + y4 = 203.1998

So the total distance that the dog runs along the two sides of the phalanx (y2 + y4) is not precisely 200.

You're right. I made a wrong assumption here and that's what amounts to the difference in the answers. I'm convinced that Bonanova's solution is correct.

[TimeSpaceLightForce]

i think there is a difference between constant speed relative to soldier and constant speed relative to ground, this is apparent when the dog is moving negative direction, if relative to soldier, the dog must have been slower ...

[bonanova]

i think there is a difference between constant speed relative to soldier and constant speed relative to ground, this is apparent when the dog is moving negative direction, if relative to soldier, the dog must have been slower ...

Was that the assumption you made in your graphical solution?

[TimeSpaceLightForce]

Assumed from OP, it looks like :The dog is moving along the square perimeter
at constant speed with respect to the moving men, not with respect to the ground
or a stationary officer who observer the dog to have run from rear center to 1st corner
then dashed to next corner, then run again to the next, but trotted to the last corner and
finally runs to its starting point as the phalanx halt. That is 3 constant speeds for him but
1 constant speed for the soldiers.

It is easier for the dog to maintain speed relative to formation when in reference to the
uniformly spaced sodiers than to maintain a single ground speed while keeping near the soldier.


Any soldier in group measure the dogs time as follows:
1/2 T to first corner , T to the next , T to the next ,T to the next ,1/2T back to start = 4T
while with constant speed to the ground, correspondingly he measure the advanced distance of corners to be:
6.25 12.5 12.5 12.5 6.25 = 50ft

The officer with yardstick measures the distance of dogs tracks to be 401.548 ft

[bonanova]

OK I see what you did.

It seems more natural to assume the same thing for the dog that one assumes for the men.

The men march at a constant speed; the dog trots at a constant speed.

The men's feet are in contact with the ground; the dog's feet are in contact with the ground.

If we were talking about an airplane, then both airspeed and ground speed are of importance.

If we were talking about a boat, then both water speed and "geographical" speed are of importance.

If the dog's feet were somehow in contact with the men, a "soldier speed" would have import.

That might happen as follows:

Imagine the soldiers are carrying a 100' x 100' platform on their shoulders.

A dog runs at constant speed around the perimeter of the platform.

Now constant speed would mean with respect to the soldiers.

What is the distance traveled by the dog?

With respect to the platform, exactly 400'

With respect to the ground, something like your answer.

[TimeSpaceLightForce]

There are no absolute speeds only relative..In this case , there were 3 inertial frames involved: the ground , the phalanx and the dog.

they are all have a change in distance per time and whether they touches or not it does not matter..the dog in the platform will find it

easier to run around the outside of the square but still relative to the observer on the ground the elevated dog has 3 variable speeds

according to his own reference frame. Even if he is on a steady hot air balloon 1 m above the ground. Thus,any constant speed should

be referenced..

I see that you computed it correctly .. using the constant speed relative to the ground and now its clear why K-man and my graph data differs.

another good puzzle!