Ok, this one’s tough. I know there’s a bunch of you out there who love the math type puzzles, and this one’s going to be hard to crack:
The 400 soldiers of the 7th company are ready to march.
They form a square of 20 meters x 20 meters.
Their mascot, a german shepherd dog, stands in the middle of the front line.
The 7th company starts to march at constant speed, and the mascot starts to walk the perimeter of the company clockwise at constant speed at the same time.
The dog is trained in such a way that, when the company has marched 20 meters, the mascot has walked the perimeter completely and is back in the middle of the front line.
The soldiers have marched 20 meters but, what is the distance the dog has walked?
Much Thanks to Bilbao for this one (a regular contributor and untouchable 11 solver)
If you can figure this one out, feel free to submit your answers in the comment section below. Will wait several days before unmasking what the math champs can come up with : )
When I worked it out I wound up with a 6th degree equation, which aren’t solvable algebraically in general, so I typed it into an equation solver and it gave me an answer of about 83.6225 meters. That answer seems low to me but I double checked my numbers and that’s what I’m getting.
jajaja I can be wrong but if the dog walk in the perimeter of the company( that is a square of 20X20)for every 20 mts the company advance, and the company had advanced 20 mts, the dog had advanced the perimeter of the square= 20X4= 80mts right or maybe not jajaja but I think is 80 mts, cool.
I say 60 meters
I believe the dog travels 82 meters. The speed up and down the sides of the square average out to be the same as the distance of the sides, however the dog would have to walk at an angle when at the back or front of the formation, adding another 2 meters total. Of course the 2 came from off the cuff calculation so it may be in error.
I get 83.62m for the dog.
I’m sure there is a more clever and elegant solution, but I assumed a dog/soldier speed ratio of 4x (which would be correct if the motions of the soldiers and the dog were independent of one another) as a starting point, and let Pythagoras do the rest. As expected, at 4x speed the soldiers had marched too far by the time the dog got back around. I then let Excel find the right speed, which turned out to be about 4.18x.
With y=dog and x=soldiers, the final equations would look like this:
1. To the first corner,
y/x=4.18, and x^2+10^2=y^2
x=2.464, y=10.299
2. To the second corner,
y/x=4.18, and y+x=20 “against the tide”
x=3.861, y=16.139
3. To the third corner,
y/x=4.18, and x^2+20^2=y^2
x=4.928, y=20.598
4. To the fourth corner,
y/x=4.18, yx=20 “with the tide”
x=6.289, y=26.289
5. To the starting point,
y/x=4.18, x^2+10^2=y^2 (same as 1.)
x=2.464, y=10.299
sum(x) = 20.006m
sum(y) = 83.624m
Should be 40+5*SquareRoot(17) meters.
Not sure on how to write out the process.
The dog has marched about 83.62 meters.
Ok, drumroll please.
The answer is…
20 meters!
The dog has in effect marched 20 meters straight ahead with the group!
Ok that was my smart alleck answer for the day. I’ll work some more on the real problem and get back with you.
I think it is a trick question. The dog could not “walk” the distance and be back in the front of the line before the company marched 20 meters.
This is a good little problem to sink your teeth into… thanks Bilbao!
I visualized the dog moving as a right triangle to the top right corner of the group. The distance along the “xaxis” in this case is 10 meters. So I let the distance the dog traveled here equal to the hypotenuse and called that number X/2. The distance traveled along the “yaxis” I named Y.
Ignoring the distance the dog travels straight up and down the “yaxis” on the right and left sides of the group for a moment, let’s consider the other 2 traingles formed. The big one as the dog goes across the bottom of the group, and the other one at the top left of the group. The top 2 triangles it turns out are equal, and the bottom one is similar. So the distance the dog travels in the front and back of the group would be 2X after some careful logical deduction.
2 more distances, RS the short distance traveled on the right, and LS, the longer distance traveled on the left is added to 2X to obtain the dog’s total distance traveled. The dog makes this distance in the same time the group marches 20 meters. Just set the dog’s speed at some simple speed, in this case, 1 meter/s, and solve for the rest. Let Dog’s speed =DS.
so (2X + RS + LS)/DS = 20 / 1
ie. Time = Time
Solving RS and LS for DS we have…
RS = 20DS/(DS+1)
LS = 20DS/(DS1)
The simplest form I could obtain X in was
X= 10DS – (20DS^2/(DS^21))
Also we can visually see Y, for 20 is the sum of
20 = Y – RS + 2Y + LS + Y
So
Y = (20 – LS + RS)/4
From here I could not directly solve for X, DS, or Y as the powers of DS were all in my numerators and denominators and was driving me bananas!
So I just put these equations in Excel(like the greatest program ever or what) and developed an algorthim to find the approximate distance.
83.6225 should be accurate to at least the first 3 decimal places.
It’s always dissapointing when an exact solution eludes you, but the algorithim could be used to find the solution to as many decimal places as desired.
Hope this made sense.
I have a typo in my post. I meant to say set the group of 400 soldiers’ speed to 1 m/s and solve for the dog’s speed.
Distance travelled straight ( in the direction of march ) = 20 meters
Distance travelled around = perimeter = 80 meters
At first, the answer seemed simple (10rad17 + 40 = 81.23105626m) but then I realized it wasn’t.
Then, with a little algebra and a graphing calculator I got an answer of 83.62250891
I assumed the troop moved a 1m/s
I’m getting a tad under 81.2 meters. I won’t bother to explain, in case my answer is wrong, but it probably helps if you move the start position of the dog to a corner of the square (it shouldn’t really make a difference where the dog starts, as long as it has returned to the starting position after the soldiers have traversed 20 meters).
The dog goes at 4.181x the soldiers’ speed, so he walks 83.62m.
I say the dog walked 84.16 meters.
This is the dog’s run, starting from the left period and ending at the right period.
______
/ .20m.\
/__\
vc: company’s speed
vd: dog’s speed
I identify 5 different partial lines
1st, 3rd and 5th stage vc = vd * sin (a)
2nd stage v’ = vc + vd
4th stage v’ = vd – vc
I started adding partial distances and times
D = D1 + D2 + D3 + D4 + D5
20m = D1 * sin (a) + D2 + D3 * sin (a) + D4 + D5 * sin (a)
T = T1 + T2 + T3 + T4 + T5
And
vc = 20m / T
vd = D / T
vd > vc (the dog runs faster than the company to catch the front line)
Operating with trigonometry I arrived to
2 / tan (a) – 1 / (sin (a) + cos (a)) + 1 / (1 – cos (a)) – 1 = 0
I don’a hay a math software, so excel helped me to graphic this and find a = 76°.
Calculating I arrived to 84.16 meters.
Now I’m trying to find the easy way…
Sorry about the drawing, I can’t make it look fine
______
/ .20m.\
/__\
this one was kind of tough; MichaelC and Shawn provide very nice detailed solutions.
Bizarette18, Someone, Jimmy Anders, Bobo the Bear got the right answer, several of you were close.
Here’s a picture which may help too:
http://smartkit.com/wpconten.....lution.jpg
Nice answer, I thought it was a trick question not really hard and overlooked the problem, and that really was 80 mts the answer jajaja, very cool explanation by Shawn and MichaelC.
Congratulations!!! I thought this one would resist your assaults a bit longer.
This was a tricky one as there are 2 cases to consider, depending on whether the distance the company advances while the dog is going backwards (stage 2) is less or greater than 20m. In this case, it is less.
To solve for that case:
K = Vd/Vc (speed of dog / speed of company)
Going through laborious equations leads to:
K^2 – 2K 1 = 2 * sqrt(K^2 – 1)
Solving for K (numerically) > K=~4.181125
distance traveled by dog = 20 * K =~ 83.6225 meters
Note that this analysis makes no assumptions whatsoever about Vc and Vd except that they are constant.
I like this idea, but as most of these kinds of problems go, it is not practical. The dog would have to run out a bit from the soldiers so as not to get run over on the front lines and not to trip on the feet in the back lines. It would be very difficult for him to make steps in a diagonal fashion not being able to see how far his hind quarters are from being kicked. Also, I think the dog would have to run rather than walk, unless the Company was crawling.
Also the solution picture is slightly misleading. The second leg should go in the opposite direction to show that his orientation is now N to S rather than S to N. This is not necessary to the solution, but is more correct for visual interpretation of the complete situation.
Copy and paste the below into an editor with fixed fonts like “courier new”. This isn’t drawn to scale or angle, but it shows the orientation of the lines and the dog’s travel. As a reference, the actual vertical distance (S to N) between ‘a’Start and ‘f’Finish should be 20m.
/ f
/
/
/
/
/
/
/
/
e/























\ d  b
\ /
\ / 
\ / 
\ / 
\ / 
\ / 
\ / 
\ / 
\ / 
X 
a \ 
\ 
\ 
\ 
\ 
\ 
\ 
\ 
\ c
In textual discription: leg 1 (line ab) is W to E having to travel somewhat N because the Company is moving S to N. When the dog reaches leg 2 (line bc), he now must travel N to S along the right flank of the Company (in a direction opposite of the Company), for leg 3 (line cd), the dog must now “catch up” with each person he passes E to W so he is again moving in a somewhat N direction. He has to do this twice as far (20m total) in the E to W direction as leg 1 did in the W to E direction. Next, for leg 4 (line de), he must travel S to N along the left flank of the Company (in the same direction as the Company). Finally, for leg 5 (line ef), he must travel W to E somewhat N because the company is still moving S to N. This final leg will be 10m W to E.
All this frames the math which must be done, but it lays the problem out correctly to pursue it.
Please correct me if I am wrong. Thanks!
…arrrggh!!! The post cleans up all spaces here it it again with underscore placeholders…ignore any appearance of horizontal lines. These are not part of the solution, there are only vertical and diagonal lines.
__________/_f
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________/
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e/























\_d__________________b
_\_________________/
__\_______________/_
___\_____________/__
____\___________/___
_____\_________/____
______\_______/_____
_______\_____/______
________\___/_______
_________\_/________
__________X_________
_________a_\________
____________\_______
_____________\______
______________\_____
_______________\____
________________\___
_________________\__
__________________\_
___________________\ c
P.S. I’d like to see a Mythbusters episode on this one.
I’m pretty new to the site and not sure how to tell which answer was correct. But I think i have a fairly exact answer without assumptions that take away from the solution
To greatly simplify things the problem can be broken up into two sections
1) when the dog is running vertically along either side of the men
2) when the dog is running infront of and bahind the men.
1) As the speeds of both man and dog are constant, the dog runs 40 meters during this section. (The dog runs less than 20 meters on one side, and more than 20 meters on the other. But as it is shorter and longer by the same amount it can be simplified to 40 meters total. (better explained by michaelc))
from this
Velocity of the dog
V_d = 40 / t1 (where t1 is the time section 1 takes)
Velocity of the men
V_m = x / t1 (where is the dist traveled by the men as we dont know how far the men got in the time it took the dog to run the 40m)
2) as both velocities are constant this section can be simplified to one large right angle triangle.
Base = 40
hight = 20 – x (how far the men walk in total minus what they walked in section 1 is how far they walk in section 2)
hyp = d (distance run by dog)
Therefore the distance run by the dog in section 2 is
sqrt ( 40^2 + (20x)^2)
total distance run by the dog is section 1 and 2 added together
d = 40 + sqrt ( 40^2 + (20x)^2)
now to find x
v_d = dist total / time total
= (40 + sqrt ( 40^2 + (20x)^2))/t
v_m = 20 / t
make t subject and sub together gets
(40 + sqrt ( 40^2 + (20x)^2))/v_d = 20 / v_m
sub in v_d and v_m from section 1
(40 + sqrt ( 40^2 + (20x)^2))/(40/t1) = 20 / (x/t1)
t1 cancels out leaving only x
solv for x
x = 9.84
sub this back into d to get
d = 81.270m