A column of soldiers is 25 miles long and they march 25 miles a day One morning a messenger started at the rear of the column with a message for the guy up front. The messenger began to march and gave the message to the guy up front and then returned to his position by the end of the day. Assume that the messenger marched at the same rate of speeed the whole time. How many miles did the messanger march?

Help!

A_________________B-----------C
A is the starting point of the messenger(let messenger be known as A), B is the starting point of the soldier up front(let soldier up front be known as B), point C is where they meet.
Let BC = x, let time to x = t1
Av. Speed = distance/time
therefore for B ---- 25 = x/t1
t1 = x/25......1
let the messenger extra speed = y
therefore for A ---- 25 + y = (25 + x)/t1
sub in 1 25 + y = (625 + 25x)/x.......2
this is all journeys to point x, the messsnger still has to return to his place in the line which will now be point B or a distance of x from where he is
that is 25 + y = x/(1 - t1)
sub in 2 & 1 (625 + 25x)/x = x/(1 - x/25)
solve that to find x = 25/sqrt2
= 25sqrt2/2
total distance the messenger travels = 25 + 2x
= 25 + 25sqrt2 or 60.3553(4dp)miles.
Hope this helps

it must be "at the end of the day." relativity says that all points of reference are correct: in other words, if you consider yourself to be still and everything else to be moving, your measurements will still be accurate.

suppose you are the 2nd to last soldier. At the start of the day, you see the guy behind you run up to reach the front. since his speed is constant, at exactly midday you see him hand a message to the guy at the front. then he comes back. you measure his distance: 50 miles.

but since the question is asked from the viewpoint of someone who is not in the column of soldiers, 25 must be added to your answer to get the "real" answer: 75 miles

It doesn't give the exact speed of the messenger, but we can make some observations about him if he goes an arbitrary speed v_m.

v_a = speed of army (25 miles/day)
v_m = speed of messenger (in miles/day)
d = length of column (25 miles)
t1 = time of first leg (messenger running to front) (in days)
t2 = time of second leg (messenger returning to end) (in days)

first leg:
The messenger runs d plus the distance the front of the column has traveled.

v_a * t1 + d = v_m * t1

Solve for t1 to get:

t1 = d/(v_m - v_a)

second leg:
The messenger and the end of the troop are a distance d apart. The sum of the individual distances each travel is d.

v_a * t2 + v_m * t2 = d

Solve for t2 to get:

t2 = d/(v_m + v_a)

The total distance the messenger travels will be

D = v_m * (t1 + t2)

Plug in the values of t1 and t2, rearrange a bit, and you get

D = 2 * d * 1/(1-(v_a/v_m)^2) (distance equation)

Looking at this equation we can tell a few things. If v_m is 0, D is undefined. This make sense. If the messenger doesn't move, he cannot catch the front of the line. If v_m = v_a, then we have a division by zero again. This is also intuitive because he will stay in his place at the end of the line. If v_m is very large compared to v_a, the distance is near 2*d, also predictable. Most likely, v_m will be pretty close to v_m (v_m > v_a), and D will be defined.

But we didn't consider the fact that he completes the total trip by the end of the day. If he returns to his position just as the day's march ends, then v_m = D (in miles/day). After substituting into the distance equation we get (assuming that v_m > v_a)

D - d^2/D = 2 * d

Again the results are realistic. If D is very large compared to d^2, D will be close to 2*d. If D = 0, there is a division by zero. Assuming D > 0, we can multiply by D and get the quadratic

D^2 - 2 * d * D - d^2 = 0

Under the problem parameter d= 25 miles, we can get a positive solution D = 60.3553. This is the correct answer (in my opinion), and is consistent with answers a few others have given.

Input prompt "Enter ratio of messenger speed to column speed":msr
Let cs= 25 ! Column speed
Let cl= 25 ! Cloumn length
Do
Let ti= 0 ! Time to travel to column head
Let t2= 0 ! Time to travel to column rear
Let d1= 0 ! Distance messinger travels to column head
Let d2= 0 ! Distance messinger then travels to column rear
Let td =0 ! Total distance messinger travels
Let n+n+1
Let t1= 1/(msr-1)
Let d1= t1*msr*25
Let d2= d1-25
Let td= d1+d2
Let t2= d2/(msr*25)
Let tt= t1+t2
If tt<1 then let mrs=msr-(1-tt)
Print td
If n>20 then stop
loop

end

You're driving down the road. There's a car about 50 feet in front of you. That car is going 70mph, and you're going 60. You'll never catch him. If he's going 70, and you're going 70, you'll still never catch him, you'll stay equidistantly behind him. If he's going 70-mph, and you're going 80, then you have to make up that 50 feet. That's time, in which the car in front of you is continuing to travel.

There's no way to tell. d=rt is very important here. But you just have a "d". "r" and "t" are very important. You need one of those 2 to solve the problem. Jeremy

Assumming that the messanger will return from point C , he will travel

2 * < 25 + 12.5 > = 75 kms

A MORE INVOLVED ANSWER ASSUMES THAT THE MESSENGER WILL COMPLETE HIS JOURNEY AT THE END OF THE DAY ALONG WITH THE SOLDIERS.
THAT SOLUTION REQUIRES APPLICATION OF A LITTLE COMMON SENSE. FOR EXAMPLE,HIS SPEED MUST BE GREATER THAN THE .SOLDIERS.IT ALSO REQUIRES A FORMULA.
RATE X TIME EQUALS DISTANCE OR TIME EQUALS DISTANCE DIVIDED BY RATE
RATE WOULD EQUAL 1 PLUS (THE SQUARE ROOT OF 2).DISTANCE KNOWN IS 25,THEREFORE 25 TIMES 1 PLUS(THE SQUARE ROOT OF 2) OR APPROX. 60 MILES IS YOUR ANSWER.

THE ANSWER: 25
CAUSE THE MASSENGER NEEDS TO PASS THE MESSAGE TO
HIS FRONT (NEIGHBOUR) AND HE SHOULD PASS IT TO THE FOLLOWING
FRONT (NEIGHBOUR) AND SON... AND HENCE THE ARMY MARCH 25 MILES
A DAY, THUS EACH SOLDIER DOES SO.
IF IT IS THE RIGHT ANSWER PLEASE CONTACT ME:
dreamcty@emirates.net.ae

The messenger starts to march with the 25 mile of soldiers (he starts marching and he does it at the same speed that the soldiers, then he wont pass any soldier and when they finish marching.

He would be at the rear as he started, and at this point he has covered 25 miles) then he has to go all the way to the first soldier and give him the message (this takes another 25 miles route) then he has to RETURN to the first begining point (it is now 50 miles away) then just do the math: 25 miles marching + 25 miles to get to the first soldier + 50 miles getting all the way to his original possition = 100 miles!

HE TRAVELED 100 Miles (I know that this is a lot but, thats the answer.

I think he marched 75 miles since

If the soldiers weren't moving he would he marched 25 miles to the front and then 25 miles to the back. Then you need to add on the fact that the soldiers have covered 25 miles to get a total of 75 miles.

This is a fun puzzle since many puzzles of this type utilize
some sort of quick logic to get an easy solution. That is what
has led many people to assume the answer must be 50 or 75 miles.
Here the answer is straightforward, but actually requires some
algebra, which has some people bugged.

We can certainly quickly show that 75 is an upper bound. Suppose
the column is laid out as:
A------25miles--------B
where A is the messenger and B is the person in the lead. Then
at the end of the day:
A------25miles------B
so B is at most 50 miles ahead of A's starting point, and then A
must walk at most 25 miles back to return to the end of the column.

Also, 25 miles is an easy lower bound, since A ends up at least
25 miles past where A started.

Other people have done the algebra to show that true answer is
around 60.36, so I'll do the algebra in slightly different
fashion. We know that A has speed at least 25 miles per day, so
I'll denote the speed of A as:
25 + x miles/day,
where if I solve correctly x must be a positive number.
When I'm done solving for x, note that the distance A marches is just
(1 day)(25 + x miles/day) = 25 + x miles
so that will give me the distance marched as well.

Now at some time t days A catches up with B. A has covered distance
(t days)(25 + x miles/day)
in this time, and B has covered
(t days)(25 miles/day)
distance in this time. Since B started out 25 miles ahead of A:
t(25 + x miles) = t(25 miles) + 25 miles
Removing the 25 t miles term from both sides:
x t = 25.

Now this is interesting. We know that t is between 0 and 1 and
so x is between 25 and infinity. That means that our messenger
travels *at least* 25 + 25 = 50 miles/day. That means that 50
miles is the least amount of distance that the messenger will
have to travel. In fact, t < 1, and so x > 25, and the messenger
must actually travel a farther distance.

B travels at 25 miles/day, and so after t days, B is
25 + 25 t miles
ahead of A's starting point. This means that the total amount of
distance A covers in a day is 25 + 25 t + 25 t miles because A has
to walk balk to the end of the column at the end of day, an extra
distance of 25 t miles. Hence A travels
25 + 50 t miles
altogether.

So now we know A travels 25 + 50 t miles, but t = 25 / x, so A
travels 25 + 1250 / x miles in a day. We also know that A travels
25 + x miles in a day, and so we have 1250 / x = x, x^2 = 1250
and x = sqrt(1250) miles, which is about 35.36. Hence the total
distance A walks is 25 + sqrt(1250) ~ 60.36 miles

So here is a new approach... using geometry. It goes beyond middle school maths, but hey, so does most of the algebra so far (at least, in the Brit'sh curriculum). As a compensation I have included more detail than I normally would.

The problem may be reconsidered as a problem in the plane, taking as the two dimensions distance along the road and time elapsed. We scale the plane such that 25 miles and one day (the time for which the column marches in one day) each equate to one unit of distance. Let the rear of the column at the start of the day in question be the origin, O. I shall use co-ordinates with the distance co-ordinate before that of time. So, for example, (0.6, 0.5) is the point a distance of 15 miles along the road after 1/2 a day. i shall refer to hte x and y axes in the usual way. x is distance, and y is time. The angle of a line is the angle, measured anticlockwise, from the x axis to that line. So, for example, the line through (0, 0) and (1, 1) has angle 45 degrees.

A person moving with constant velocity has locus a straight line in this representation.
The person at the end of the column is at O at the start of the day and at P = (1, 1) at the end of the day. The person at the front of the column is at A = (1, 0) at the start of the day, and at B = (2, 1) at the end. The messenger begins at O, passes on his message at some point Q on AB and ends up at P. The locus of the messenger therefore consists of the lines OQ and QP. Since we are told that his speed is constant, we know that the gradient of QP is minus that of OQ. Finally, let X be the intersection of AP with OQ. Since the messenger has gone 25 miles when arriving at X, and his speed is constant, the total distance he travels is 25(OQ + QP)/OX miles.

Let C be the circumcentre of OQP. Let l be a line through C of angle 0 (that is, parallel to the x axis). Let Q' be the reflection of Q in l. Then Q' lies on the circumcircle of OQP, and QQ' has angle 90 degrees which implies that the angles OQQ' and Q'QP are equal. Using the converse theorem to 'the angles in the same segment are equal' we obtain that the seqments OQ' and Q'P are equal. So Q' lies on the perpendicular bisector of OP. As C also lies on this line, we see that CQ' is this bisector, which has angle -45 degrees. So CQ, the reflection of this line in l, has angle 45. That is, C lies on the line through Q with angle 45 degrees, which happens to be the line AB. C must also lie on the perpendicular bisector of OP. This uniquely identifies C to be the point of intersection of this line, which happens to be A. That is, A is the circumcenter of OQP.

We deduce that AQ = AP = 1. As AQ and OP are parallel and X is the intersection of OQ with AP, AQX and POX are similar. So XQ/OX = AQ/OP = 1 / sqrt(2). Now QX = QP and OQ = OX + XQ so (OQ + QP)/OX = (OX + XQ + XQ)/OX = OX/OX + 2 * XQ/OX = 1 + 2/(sqrt(2)) = 1 + sqrt(2)

Finally we arrive at our result: The total distance travelled is 25(1 + sqrt(2)) miles

At some point, I may include a diagram of the above.

Thankyou

sfwc
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