ARMY puzzle...
Erol Esen
An army 40 miles long is moving at constant speed. A
horseman takes a message from the rear end of the army
to the general in the front end of the army and returns
back to the rear end of the army (moving with constant
speed). During this time, the army moves a total of 40
miles.
What total distance did the horseman travel?
Gene Miller
This puzzle (or a simple variation) is almost certainly in the FAQ.
SPOILER
This involves solving a quadratic equation.
Let the total elapsed time = 1 hour.
Then the army's speed is 40 mph (pretty fast!)
Let the horseman's speed = h.
And let the time that the horseman reaches the front of the army = t.
Then the net distance covered by the horseman = ht - h(1-t) = 40.
And the distance covered when the horseman reaches the front = ht = 40+40t.
Just solve for h and t, and the total distance covered = ht + h(1-t).
--
Gene Miller Multimedia Communications
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Tim Finkenstadt
> An army 40 miles long is moving at constant speed. A
> to the general in the front end of the army and returns
> back to the rear end of the army (moving with constant
> speed). During this time, the army moves a total of 40
> miles.
> What total distance did the horseman travel?
_
___/
/\ /\
|----------------------------------------|
40 miles
The horseman on his way to the front traveled:
(40 + s) miles. s == The number of miles the army traveled.
On his way back, he traveled:
(40 - s) miles.
Adding the two sums together you get: 80 miles.
Christine Park
> The horseman on his way to the front traveled:
> (40 + s) miles. s == The number of miles the army traveled.
> On his way back, he traveled:
> (40 - s) miles.
> Adding the two sums together you get: 80 miles.
Start # <-horse
********** <-army
|--40mi--|--40mi--|
Horse # <-horse travels 40+s
at ********** <-army
Front |--40mi--|--40mi--|
|--- 40+s ---|
Finish # <-horse travels s
********** <-army
|--40mi--|--40mi--|
|-s-|
I get total distance: 40 + 2s
Am I way off?
^ ^ Christine Park
@ @
o Sandy Eggo, Caliph or Nyeah
u cp...@qualcomm.com
Chris Cole
A harder version of this question is in the rec.puzzles Archive:
==> analysis/dog.p <==
A body of soldiers form a 50m-by-50m square ABCD on the parade ground.
In a unit of time, they march forward 50m in formation to take up the
position DCEF. The army's mascot, a small dog, is standing next to its
handler at location A. When the
B----C----E soldiers start marching, the dog
| | | forward--> begins to run around the moving
A----D----F body in a clockwise direction,
keeping as close to it as possible.
When one unit of time has elapsed, the dog has made one complete
circuit and has got back to its handler, who is now at location D. (We
can assume the dog runs at a constant speed and does not delay when
turning the corners.)
How far does the dog travel?
==> analysis/dog.s <==
Let L be the side of the square, 50m, and let D be the distance the
dog travels.
Let v1 be the soldiers' marching speed and v2 be the speed of the dog.
Then v1 = L / (1 time unit) and v2 = v1*D/L.
Let t1, t2, t3, t4 be the time the dog takes to traverse each side of
the square, in order. Find t1 through t4 in terms of L and D and solve
t1+t2+t3+t4 = 1 time unit.
While the dog runs along the back edge of the square in time t1, the
soldiers advance a distance d=t1*v1, so the dog has to cover a distance
sqrt(L^2 + (t1*v1)^2), which takes a time t1=sqrt(L^2 + (t1*v1)^2)/v2.
Solving for t1 gives t1=L/sqrt(v2^2 - v1^2).
The rest of the times are t2 = L/(v2-v1), t3 = t1, and t4 = L/(v2+v1).
In t1+t2+t3+t4, eliminate v2 by using v2=v1*D/L and eliminate v1 by
using v1=L/(1 time unit), obtaining
2 L (D + sqrt(D^2-L^2)) / (D^2 - L^2) = 1
which can be turned into
D^4 - 4LD^3 - 2L^2D^2 + 4L^3D + 5L^4 = 0
which has a root D = 4.18113L = 209.056m.
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tom schmeling
Erol Esen <es...@CS.ColoState.EDU> wrote:
>
>An army 40 miles long is moving at constant speed. A
>horseman takes a message from the rear end of the army
>to the general in the front end of the army and returns
>back to the rear end of the army (moving with constant
>speed). During this time, the army moves a total of 40
>miles.
>
>What total distance did the horseman travel?
I may be nuts, but how about approximately 96.5 miles?
mile 80 mile 40 Mile 0
*-----|-----------*---------------------*
X
G**********************H (army at start)
GH******************(army at time when H catches G at point X)
G*****************H (army at end)
As I understand the problem, the horseman (H) starts off as the army
starts marching and must complete the round trip to A and arrive at the
rear end of the army at exactly the time that the general (G) of the army
reaches mile 80 and the rear of the army reaches mile 40.
I assume that the requirement of constant speeds for both G and H
means that no one can stop and rest and that H must turn
around immediately on reaching G.
I think Christine Park was right with the answer 40+2S (where S is the
distance from mile 40 to X). But H's rate of speed relative to G is
fixed. If H travels at 2 times A's speed, she will not reach G until G
reaches mile 80 and will have no time for the return trip. If H
travels at 3 times G's speed (and turns right around) she will be back at
mile 40 before the rear of the army reaches it. (unless H meets the
rear of the army exactly as it reaches mile 40, the condition that the
army advances 40 miles during the H's trip will not be met.)
(Please note that the math below is off by a bit. It's pretty close
I think, but i mostly guessed at the values)
If we set the army's rate of travel at 1, and let it march for time 28.2828..
G is at mile 68.2828.... In the same time, H travels at 2.41456...
times the army's speed for the same time, also reaching mile
68.2828.... G continues on, reaching mile 80 in time (40-28.2828=) 11.71.
In the same time H turns around and travels the 28.2828
(2.41456*11.71 is actually 28.27) miles back to mile 40 and meets the
rear of the army. Thus H has traveled 40 + (28.2828*2)= about 96.5 miles.
If this approach to the problem is right and someone can give a more
precise answer, I'd like to see it.
JRW
>Subject: ARMY puzzle...
>Date: Fri, 17 Sep 1993 16:16:43 GMT
Great puzzle, simple to explain but took me awhile to figure it out.
Let X = distance army travels from start until horseman reaches front.
assuming constant velocity, then ratio of time for the horse to reach
the front of the army to time to return to end is:
(X+40)/X
likewise the ratio of times for the army is
X/(40-X)
These are the same, so setting them equal and solving for X, obtain
X = 20*sqrt(2)
Thus horseman travels 40 + X + X = 40 * (1 + sqrt(2)) or about 96.5 miles.
Initially I tried to solve for velocity of the horse and/or the times. Of
course this won't work since any velocity of the horse with work so long as
it is
(1 + sqrt(2)) times faster than the army's velocity.
Karen Collins
>>>>> I do not do puzzles much but it seems to me that if the army is
40 miles long and the runner goes to the front, that is 40 miles.
Then to the back again is another 40 miles. He has gone 80 miles.
An I anywhere in the ballpark here.
Nuff said,
Ande
Guy Steven
: Regarding Tim Finkenstadt's answer to the army puzzle:
The answer 40 + 2s is correct, the problem is what does 40 + 2s equal.
See below for answer
My guess is 96.57, calculated as follows.
The whole exercise is completed in time T.
The army covers 40 miles, the horse x miles.
=> The speed of the army is 40/T, the speed of the horse x/T.
The horse reaches the head of the army at 40+s miles after time y.
It has covered a distance 40+s.
The army has covered distance s in the same time.
40/T * y = s
r/T * y = 40 + s
x = (40/(x-40)) * T
The army travelled distance s in time (40/(x-40))T. The rider must travel
back this distance in time 1-(40/(x-40))T.
=> 40/T * (40/(x-40))T = x/T * ((x-80)/(x-40))T
=> 1600 = x^2 - 80x
=> x^2 - 80x -1600 = 0
for quadratic ax^2 + bx + c = 0, x = (-b (+/-) sqrt(b^2 - 4ac) )/ 2a
Answer 96.57 miles.
_____________________________________________________________________________
Guy Steven | |
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Christchurch | C\ /\ |
New Zealand | Ampers and Ecks Ltd.| e-mail: r...@etive.equinox.gen.nz
-----------------------------------------------------------------------------
Tony Quinn
>In article <Sep17.161...@yuma.acns.colostate.edu>,
>Erol Esen <es...@CS.ColoState.EDU> wrote:
>>
>>An army 40 miles long is moving at constant speed. A
>>horseman takes a message from the rear end of the army
>>to the general in the front end of the army and returns
>>back to the rear end of the army (moving with constant
>>speed). During this time, the army moves a total of 40
>>miles.
>>
>>What total distance did the horseman travel?
How about from the back to the front (40 miles) and from the front to
the back (40 miles).. no need for fancy equations just consider the
movement relative to each other!
--
-------------------------------------------------------------------------------
Tony Quinn --- The voice of insanity.
Replies to tony...@sixpints.demon.co.uk
-------------------------------------------------------------------------------
David Karr
[...][...]
>>>What total distance did the horseman travel?
>
>How about from the back to the front (40 miles) and from the front to
>the back (40 miles).. no need for fancy equations just consider the
>movement relative to each other!
If you accept that solution, I have an even simpler one. Assuming the
horseman remained mounted (a reasonable assumption since he was
supposed to turn back instantly), then the distance he moves is
exactly ZERO miles---relative to the horse. No need even for simple
arithmetic!
-- David Karr (ka...@cs.cornell.edu)
Wangsawidjaja Mintardjo
>horseman takes a message from the rear end of the army
>to the general in the front end of the army and returns
>back to the rear end of the army (moving with constant
>speed). During this time, the army moves a total of 40
>miles.
>
0. they are in carnival and the parade is circular.
if it is linear, the answer is 40 miles. the messenger travels faster than
the army does to get to the front and slower to get back to the rear.
?
--
----------------------------------------------------------------
Wangsawidjaja, Mintardjo Oregon State University.
Groups: student Domain: CS.ORST.EDU
Internet: jacobs.cs.orst.edu!wangsawm Phone: (503) 737-9000
Tony Quinn
3 weeks????? or is it a lot more complex than it seems???
Mintardjo W.
>>What total distance did the horseman travel?
>
> if it is linear, the answer is 40 miles. the messenger travels faster than
> the army does to get to the front and slower to get back to the rear.
>
Disregard my last answer. It was incorrect.
After taking pencil and paper, the distance should be 40 x (1 + sqrt(2))
What's about these:
A giant frog leaps 5 times as far as normal frog does. If a normal frog
leaps 1/100 of the radius of a round pool, how many times does the
giant frog leap to cross it?
A boy along with his cow arrived on a round lake. On the center of the lake
is a round island. On the center of the island is a very tall tree. The boy
successfully tied his cow to the tree on the island without ever wetting
himself. how?
Matthew T. Russotto
}
} What's about these:
}
} is a round island. On the center of the island is a very tall tree. The boy
} successfully tied his cow to the tree on the island without ever wetting
} himself. how?
Given the phrasing "on a lake", perhaps the lake was frozen. On the
other hand, with a really long rope and a lot of walking, he could
have done it without such tricks.
--
Matthew T. Russotto russ...@eng.umd.edu
Some news readers expect "Disclaimer:" here.
Just say NO to police searches and seizures. Make them use force.
(not responsible for bodily harm resulting from following above advice)
Howard Pinsley
> horseman takes a message from the rear end of the army
> to the general in the front end of the army and returns
> back to the rear end of the army (moving with constant
> speed). During this time, the army moves a total of 40
> miles.
>
> What total distance did the horseman travel?
I found this problem quite interesting and challenging. I posted my
solution a few days ago, but my internet connection was having news
problems. Anyway, here's a general algebraic solution to the
problem.
Let: Va = speed of the army
Vh = speed of the horseman
t = total time for the round trip
t1 = time riding to the front
t2 = time riding back
L = length of the army
d = distance the army marched during the
horseman's trip.
the relevant equations are:
(1) t1 + t2 = t (obviously)
(2) Va * t = d (army rate times time)
(3) (Vh - Va)t1 = L (the relative speed of the
horseman w/respect to the army
is reduced by Va since they are
both going forward. The distance
traveled by the horseman w/repect
to the army is the length of the
army).
(4) (Vh + Va)t2 = L (on the return trip, the logic is
reversed, and the horseman's
relative speed is increased by Va).
from (3) and (4)
t1 = L / (Vh - Va) and t2 = L / (Vh + Va)
Now plug these results into (1) to get:
L L
-------- + --------- = t
(Vh - Va) (Vh + Va)
... and simplifying this yields:
2L * Vh
---------- = t
2 2
Vh - Va
Now from (2) we know that t = d / Va so we get:
2L * Vh d
---------- = ----
2 2 Va
Vh - Va
crossmultiplying yields:
2 2
(2L*Va)Vh = (d)Vh - d*Va
or in a quadratic form in terms of Vh:
2 2
(d)Vh - (2L*Va)Vh - d*Va
Solving using the quadratic equation yields:
--------------------
2L*Va + / 2 2 2 2
/ 4L Va + 4d Va
Vh = -----------------------------
2d
or
------------
2L*Va + 2*Va / 2 2
/ L + d
------------------------------
2d
which equals:
Va * [L + sqrt(L^2 + d^2)]
Vh = --------------------------
d
but from (2) we know that Va = d / t
giving us finally:
Vh * t = L + sqrt(L^2 + d^2)
Note that the LHS (Vh * t) is the distance the horseman traveled.
The problem as stated gave the length of the army, L, and the
distance traveled by the army, d, to be 40. This would yield
40 + sqrt(40^2 + 40^2)
= 40 + 40 * sqrt(2)
= 40 (1 + sqrt(2)) as others have mentioned.
Note that if the army is stationary (i.e. d = 0), then the
horseman's distance becomes:
L + sqrt(L^2 + 0)
= 2L as we would expect
and also, trivally, if the army was of zero length, but did move
distance d, the horseman would move
0 + sqrt(0^2 + d^2)
= d
or the same distance as the army.
All in all, a very interesting puzzle. Thanks!
--
Howard W. Pinsley
(212) 836-8322 Fax: (212) 836-8689
he...@dorsai.dorsai.org
--
Howard W. Pinsley
(212) 836-8322 Fax: (212) 836-8689
he...@dorsai.dorsai.org
Eric Arnett
> 40 miles long and the runner goes to the front, that is 40 miles.
> Then to the back again is another 40 miles. He has gone 80 miles.
>
> An I anywhere in the ballpark here.
>
> Nuff said,
> Ande
>
forward so he doesnot go 40 miles backward
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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EA7...@conrad.appstate.edu \ - "Where does it go ?"
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