*resolved* thank you -- unsolvable problem

my friend gave this to me, I cant think of a possible way to solve this:

Quote:

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A space traveler named Rex is dropped off by his team at the bank of a river on a strange planet that has only one river. The river is completely straight and is very long. Rex begins to explore the planet in his dune buggy by driving in a straight line away from the river. But alas, his journey is abruptly ended when a tornado hits his dune buggy, spinning him around and knocking him unconscious so that he has forgotten which direction he was traveling.
Rex's dilemma: Rex does not know how to get back to the river so that his team can pick him up. What he does know is that he has traveled 1000 miles, he has enough gas in his tank to travel an additional 6400 miles. He has an accurate compass and an accurate odometer. He can travel in any direction he desires. What strategy should he employ to make sure that he can get back to the river?
(Any point on the river is OK)

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Note: Rex is driving perpendicular to the river.

I though of a circle with radious 1000mi... so you can walk 1000mi and then drive in a circle with a radius of 1000mi, but that would be more than 6400mi

I cant solve this!!!
see if you can:D

aaah, I think you have to do this to solve it, but I dont know how:


come up with the shortest path that would cross all the tangent lines to a circle of radiuos 1000


here's what I came up with: it is 155 miles longer than the limit. any ideas?:
http://www.vbforums.com/attachment.php?postid=1301387

the black line is supposed to be the path he takes, and the center of the circle is where he starts... the red lines are just to visualize it better.
This will cross all the tangent lines of the circle, but it's still as short as 6400 miles:(

Why not just travel 1000 in a random direction, then go around in a circle at 1000 (1000*pi)

Won't that only be (1000+3140=4140) < 6400 ?????

Quote:

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Originally posted by alkatran
Why not just travel 1000 in a random direction, then go around in a circle at 1000 (1000*pi)

Won't that only be (1000+3140=4140) < 6400 ?????

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no, the circumference of a circle is 2 Pi r, so that would be:

1000+6283 = 7283

Quote:

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Originally posted by si_the_geek
no, the circumference of a circle is 2 Pi r, so that would be:

1000+6283 = 7283

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hehe yeah :)
the way I;ve drawn it in the diagram, it will end up to be 6555 miles and it will cross all the tangent lines to that circle (tangent lines = possibly the river) Still long :(

this is hard! :D

Maybe he should walk for a bit after his tank runs out ;)
Or check his compass before he leaves
Or look for the vegetation that SHOULD be near river
OR FOLLOW HIS TRACKS!!!

All logical solutions that probably don't apply :(

Can he see the river before he reaches it?

Quote:

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Originally posted by alkatran
Maybe he should walk for a bit after his tank runs out ;)
Or check his compass before he leaves
Or look for the vegetation that SHOULD be near river
OR FOLLOW HIS TRACKS!!!

All logical solutions that probably don't apply :(

Can he see the river before he reaches it?

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no no no... hehe, it's NOT a trick question like that. It's more like a math question. No I dont think he can see the river before he reaches it.

If Rex were to go 1000+x miles then go in a circle, he would have the river CROSSING AT 2 POINTS IN THE CIRCLE, I don't have time to find the actual answer, I was never taught how to find circumference of a.. chopped.. circle, and I'm supposed to be sleeping.

So you get to grab it and run. Hope I helped!


I

Quote:

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Originally posted by alkatran
If Rex were to go 1000+x miles then go in a circle, he would have the river CROSSING AT 2 POINTS IN THE CIRCLE, I don't have time to find the actual answer, I was never taught how to find circumference of a.. chopped.. circle, and I'm supposed to be sleeping.

So you get to grab it and run. Hope I helped!


I

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hehe, sorry it's kinda wrong:D
well, I DID thoght of going in a circle, but that totals to 7000 something miles, which is obviously greater than 6400 miles. My solution which I posted above is 6555 miles, which is still more than 6400 miles, but better than any other solution I have though of.


umm, and it wont cross the river twice. Note that the guy is driving PERPENDICULAR to the river, so the circle is gunno touch the river at a single point

Travel a circle out say, 100-300 feet. Find his tracks, then head back!

Although.

Since the circumference of the circle is close the the supposed limit I'd suggest either a brillant, or ignorant answer was in the mind of the creator.

Quote:

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Originally posted by DiGiTaIErRoR
Travel a circle out say, 100-300 feet. Find his tracks, then head back!

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as I said, it's a math question and not a trick question like that:D

This is killing me, I've been working on it for 2 days now...cant solve it:mad:

Could it be a matter of probability then? Rex travels out, and he then has a 50-50 chance that he'll make it to the river. He travels out 1000 miles and starts moving in a circle.

Edit: yes, I'm aware this has been suggested before, I'm asking about the probability thing.

http://www.vbforums.com/attachment.p...postid=1302706

umm, for the third time, walking in a circle works, and you will find the river, BUT!!! it's gunno take you more than 6400 miles to drive in a circle. you have to drive 1000 miles to get to the center of the circle, and then 6283 miles to drive in the circle, which is 7283 miles total. It has to be less than or equaly to 6400 miles.



umm, btw where did that 1001 come from:D
and the answer has to work 100% of the times... not 50% of the times :(

Quote:

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Originally posted by MrPolite

umm, btw where did that 1001 come from:D
and the answer has to work 100% of the times... not 50% of the times :(

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I know... I wasn't sure about the probability thing.
Give me some time, aqa. Man javob midonam.

Quote:

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Originally posted by mendhak
I know... I wasn't sure about the probability thing.
Give me some time, aqa. Man javob midonam.

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chashm ostad ;)
it's a hard one I think, hehe
btw if anyone want to "visualize" my other solution:
http://www.vbforums.com/attachment.php?postid=1302734
you start by going on that line, away from the center, in any random direction. You will eventually cross the river at some point (you will cross the river, OR, you will be next to it at one point)

Stil thinking about it , but I think you got the best.
6400 miles is not possible unless you use some "realism" like: He can see x-miles while sitting in his car OR .....

Could always poor the petrol over everything and light a GIANT flare!!!!!!!:D :D :D :D


Hmmm its a bummer:D :D :D :D :D

How long is very long?

And at what point did he start on the river?

If he travels in a circle with radius 1100 miles, the circle will cut the river at 2 places. The distance that he will travel before he reaches one of the two intersection will be (2pi * 1100) less the length of arc with an angle of pi/2 plus 1100 to reach the edge of the circle. Therefore :

(2pi * 1100) - (pi/2 * 1100) + 1100

Is this right??? Dont have a calculator or paper with me...

Quote:

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Originally posted by vbud
If he travels in a circle with radius 1100 miles, the circle will cut the river at 2 places. The distance that he will travel before he reaches one of the two intersection will be (2pi * 1100) less the length of arc with an angle of pi/2 plus 1100 to reach the edge of the circle. Therefore :

(2pi * 1100) - (pi/2 * 1100) + 1100

Is this right??? Dont have a calculator or paper with me...

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it = 6283.62

the circumference of 1000 radii circle is 6283.18 weird.

Wrong calculation
Calculate:
Angle in centre of 1100 circle to the intersection of a tangent to a circle of 1000

2*arccos(1000/1100)=49,24


On your solution you have the following:

(2*Pi*1100)* (1-(49,24/360)+1100=7066,16

Yeah i got the same solution using Angle of arc = Cos (inv) 1000/1100 * 2 i.e 49.24.

since length of arc is S=R * angle of arc

S = 1100 * ((Pi/180) * (360-49.24))

Total distance is S + 1100 = 7066.something...

no good....
:(

Ok, I think I was getting close with the bigger circle thing.. now to shave some time off of it..

http://www.vbforums.com/attachment.p...postid=1302887

the pale line is the 1000 mile radius circle, the black line is his path (should probably be rounder/straighter at end ot maximize potential, maybe even be getting smaller the whole time, reaching 1000 miles from center at end) and the blue line is the river. Oh and the red line is what you don't have to drive because you covered it at start.

As you can see I went to 1000 miles then went out at an angle.. covering more ground/angles in less distance, which also made the end point fall back a bit. Is there any other way to optimize??

aclcatran, I don't see any improvement, by using your solution.
In the start you are usingan angle to .

Quote:

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covering more ground/angles in less distance

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, you are not covering more angles, you are covering LESS angles than by going straight.
In the final part, the spiral to the circle is just a waste, because the line continued to the river is shorter than the spiral part!

Sorry, I'm still a bit sleepy.. and I haven't been posting solutions per say, I've been posting ideas, to see if anyone else will do calcs because I don't know how.
HMmm.. I know! threaten your friend and force him to tell you, MrPolite! (or just ask, whatever)

Well you could use half the fuel in one go like in the animation below

:D :D :D :D :D

I really really did!

First you head off at 247.5 degrees (south south west), reach the bottom of the 1000m*1000m square. Now go north north west, until you reach the middle left of the square. Go around the circle until you are at the far east, then go straight down. Here's the formula

sqr(1000^2+500^2)*2+(1000*pi)+1000

this adds up to UNDER 6400! (about 33 under it)

http://www.vbforums.com/attachment.p...postid=1303106

Note: The line would pass through circle if you went directly to west side, so I used the 33 extra miles to go on the circle a bit longer.

Maybe I'm studid, but could you explain your formula.
Especially the part the covers from the start until you come to beginning of the half-circle. I guess it'S this part in your formula:

Quote:

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sqr(1000^2+500^2)*2

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The first part is the triangle in the picture, I just took MrPolite's answer and saw that if I Angled the start a bit, it would decrease the total miles. The equation is to find a²+b²=c²

OK
as a start, if you go on 247,5 degr. from the centre you will end up colliding with the side of the red box NOT the bottom (245 will bring you to the corner).

for the formula, you're probably trying to say that the line from the centre and the line to the perimeter are of equal length.
OK for that, but I'm working on a exact solution, where your second leg will be changed to a tangent .
Stand by.

Forget this one, because I made a mistake, see following post
Here you go:
First Leg (go from centre (1000,1000) to Point(500,0) that is 206,57 degrees) Distance 1118,04

Second leg (go on a line that tangents the circle, direction is 323,14) since the triangel that is formed by Leg1 and Leg2 and ack to the centre is equal to that one formed by Leg1, the vertical thru the centre and the bottom line the length of leg2 is 500 (total so far 1618,04)

Third leg (the part on the circle until the 270 degree position; or where the horizontal thru the centre crosses the circle)
Because the triangle from Leg2 has the equivilant toward the vertical, we can compute the degree-position where leg2 hits the circle. It's 2* the angle between the vertical and Leg1 (2*26,57=53,14) On the perimeter we have a way of (53,14/360*2*Pi*1000=927,47) (total so far 2545,51)

The half perimeter is 3141,59 plus the last 1000 gives a total of disapointing 6687,10.
It's lloks to me, that this was not a solution after all.
Unless you can show your

Quote:

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Note: The line would pass through circle if you went directly to west

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in a bit more detail.

Sorry, I made a mistake:

First Leg (go from centre (1000,1000) to Point(500,0) that is 206,57 degrees) Distance 1118,04

Second leg (go on a line that tangents the circle, direction is 323,14) since the triangel that is formed by Leg1 and Leg2 and ack to the centre is equal to that one formed by Leg1, the vertical thru the centre and the bottom line the length of leg2 is 500 (total so far 1618,04)

Third leg (the part on the circle until the 270 degree position; or where the horizontal thru the centre crosses the circle)
Because the triangle from Leg2 has the equivilant toward the vertical, we can compute the degree- position where leg2 hits the circle. It's 90-(2* the angle between the vertical and Leg1) (90-2*26,57=36,86) On the perimeter we have a way of (36,86/360*2*Pi*1000=643,33) (total so far 2251,37)
The half perimeter is 3141,59 plus the last 1000 gives a total of 6392,96. Close but valid, you really did it!

Ah, the problem has been solved. It has been keeping me busy today as well. I was taking the planet's radius into account in my calculations, but (if the radius is exactly the earth's radius, but there's no info on that) I came only 41 miles short...

Calculation (with earth's radius and using MrPolite's path). 1000 miles covers about 14.48 degrees. The distance along the half circle will be 1000 * pi * cos(14.48) = 3042 miles.
Both two horizontal lines are 1000 miles (which probably isn't true, since I neglect earth's curvature here).
The diagonal line is not 1414 miles (1000 * sqrt(2)) but 1399 lines, because of the planet's curvature.

The sum of all this is 1399 + 3042 + 2 * 1000 = 6441 miles. Just a bit too much ;) But if Rex is in for a nice walk, it can be done, although someone really is tired after driving 6400 miles ;)

For a small planet another tactic is possible as well. Just drive a straight line. This works only if the radius of the planet is not more than 2355 miles, so it will work on the moon and (correct me if I'm wrong) even Mars.

To Opus: I was thinking about how to prove I was right all the
way home, guess you did it for me ;)

To MrPolite: Your welcome, and I did use your answer as a base.

To Riis: What if the guy walks/drives around planet parrallel (err spelling) to the river? He'd be walking for A LONG time (until he accidently went in a very large circle/died)

Hi Riis, although I used to do navigation, Ididn't even think about using "those" formulas.

And to alkatran, a "straight line" on a planet isn't straight, it's a circle on the surface that that same centreposition as the planet itself. Since both should be true (for the river and the walk) they must meet at two points on the surface of the planet.

This is what I mean.http://www.vbforums.com/attachment.p...postid=1303587

Quote:

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Originally posted by alkatran
I really really did!

First you head off at 247.5 degrees (south south west), reach the bottom of the 1000m*1000m square. Now go north north west, until you reach the middle left of the square. Go around the circle until you are at the far east, then go straight down. Here's the formula

sqr(1000^2+500^2)*2+(1000*pi)+1000

this adds up to UNDER 6400! (about 33 under it)

http://www.vbforums.com/attachment.p...postid=1303106

Note: The line would pass through circle if you went directly to west side, so I used the 33 extra miles to go on the circle a bit longer.

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alright, I understand the problem with my solution now:D

but I dont understand how you got 247.5 degrees, and I dont understand how you know that it wont cross the circle (if your path crosses the circle, then it's gunno miss the tangent line that is parallel to that line)

this is what happens if it crosses the circle:
http://www.vbforums.com/attachment.php?postid=1303733


would this possibly be a solution?
http://www.vbforums.com/attachment.php?postid=1303739


:D

ignore this

The trick is to get the lines to and from bottom to be symetrical, and reach the furthest point possible on circle like that.

Quote:

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Originally posted by alkatran
The trick is to get the lines to and from bottom to be symetrical, and reach the furthest point possible on circle like that.

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I still dotn get it
how do you know that your path wont cross the circle, like the first picture in my last post?

and how do you get that angle :D:D:D

It does, well, it would.

As soon as you reach the circle you start to follow it.

Quote:

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Originally posted by alkatran
It does, well, it would.

As soon as you reach the circle you start to follow it.

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hmm, alright, makes sense :)

but can you show me how you got that angle?

Hi MrPolite,
didin't you read my repy's?
I know, without a drawing you have to READ it!
See the attached circle.bmp.
I computed the solution of alkatran, with a change for the leg that goes inside rthe circle. I divided it in two parts, one is the tangent to the circle and the second is the walk along the perimeter. That way no possible tangent is missed and the total distance is less than 6400.

Quote:

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Originally posted by alkatran
I really really did!

First you head off at 247.5 degrees (south south west), reach the bottom of the 1000m*1000m square. Now go north north west, until you reach the middle left of the square. Go around the circle until you are at the far east, then go straight down. Here's the formula

sqr(1000^2+500^2)*2+(1000*pi)+1000

this adds up to UNDER 6400! (about 33 under it)

http://www.vbforums.com/attachment.p...postid=1303106

Note: The line would pass through circle if you went directly to west side, so I used the 33 extra miles to go on the circle a bit longer.

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The angle length, would be around 1100.
Then to 45 degrees (1000 ft) from circle is 367.7.
8th of the circumference of the circle is 785.4.
half the circle is 3141.6.
going down after that is 1000.

6394.7

Yea. Close....

Here is the COMPUTED solution again:
Look at the circle2.gif to see what I'm talking about.
The start is at the centre A (X;Y are 1000;1000) and goes to point C (500,0). The lenght of this leg can be computed by:
SQR(1000^2 +500^2)=1118,04
To reach B you have to go at an angle (direction) of 206,57 degrees, that makes the angle between AB and AC 26,57.

The second Leg is from C to D. Since the triangle ACB is identicla to ACD (it's just mirrored at AC) the lenght of CD is 500

to compute the part on the circle from D to E we need the direction of the line AD. Since the triangles are identical, the angles in A for both triangles are the same. So covered already 2*26,57 degrees of the circle, to get to E we have to cover additional 36,86 degrees. That is a way of (36,87/360)*2*PI*1000=643,33
Now we have to cover the remaining half circle 3141,59 and the final 1000.
All adds up to 6392,96

Hopefully, everybody reads this one now! The credit goes to alkatran!
And MrPolite, please put a (SOLVED) in the original post!

Forgot the gif. Here it is

Opus, when I sum those numbers, I get a value of 6402.96. So the poor guy still has to walk nearly 3 miles.
I've put the calculations into an Excel sheet, and this returns 6403.128 for a horizontal distance of 500.
However, there are possible solutions for every horizontal distance between roughly 525 and 630.
The minimum is at about 577 miles. (Angle is 209.98 deg.) First part: diagonal line = 1154.525, second part: 577, third part (short parth along circle): 524.1243, fourth part (semicircle): 3141.59, fifth part: 1000. This sums to 6397.242.
Kind regards,

Riis

Blame on me, but my kids are up to % and trig, maybe I need anotherone doing the first class!
BTW: Riis could I get your Excel-sheet?

Here is the excel sheet, zipped since it is about 160k large.

First column, x, is horizontal distance (BC and CD)
Second column, a, is diagonal distance (AC)
Third column, b, is 'short' path over circle's edge (DE)

Fourth column, total, is sum of AC, CD, DE, semi-circle and 1000


I'm wondering if the graph is exactly symmetrical. I didn't do any research to it. Might be a nice question for this forum :)

the optimal solution would be, when the first checkpoint is at -1000,1000/sqrt(3)
if x=distance to tangent from first checkpoint then you have the first checkpoint distance sqrt(r^2+x^2), the second x, and the third r by the angle between the normal to the tangent trough origo and the other tangent, which is derived trough the angle to the first checkpoint, pi/2-2*arcsin(x/sqrt(r^2+x^2)), the minimum of this function can be found by testing when its derivative is 0.
D[x+sqrt(r^2+x^2)+r*(pi/2-2*arcsin(x/sqrt(r^2+x^2)))]=
1+x/sqrt(r^2+x^2)-(2r(-(x^2/(r^2+x^2)^(3/2))+1/sqrt(r^2+x^2)))/sqrt(1-x^2/(r^2+x^2))=0
of which 2 are imaginary and one is negative and the positive is:
x=r/sqrt(3)

Thanks Kedaman for the derivation ( maybe I had derivation of trig'S in school, but only maybe)

BTW Your Avatar looks nice, should I try to hunt those sub'S done their in the med? I'll sure try using my application

Quote:

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Originally posted by kedaman
the optimal solution would be, when the first checkpoint is at -1000,1000/sqrt(3)
if x=distance to tangent from first checkpoint then you have the first checkpoint distance sqrt(r^2+x^2), the second x, and the third r by the angle between the normal to the tangent trough origo and the other tangent, which is derived trough the angle to the first checkpoint, pi/2-2*arcsin(x/sqrt(r^2+x^2)), the minimum of this function can be found by testing when its derivative is 0.
D[x+sqrt(r^2+x^2)+r*(pi/2-2*arcsin(x/sqrt(r^2+x^2)))]=
1+x/sqrt(r^2+x^2)-(2r(-(x^2/(r^2+x^2)^(3/2))+1/sqrt(r^2+x^2)))/sqrt(1-x^2/(r^2+x^2))=0
of which 2 are imaginary and one is negative and the positive is:
x=r/sqrt(3)

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sorry I'm asking this over and over :(


I understood what you said about the first checkpoint, but nothing after it. Here;s what I do understand:

http://www.vbforums.com/attachment.php?postid=1304869


but I dont know how you get pi/2 - 2teta..... and I dont understand what angle that is.
And in the derivitive, why do you have an X in there? -> D[ x + ...]

I will really appriciate if you show me how you got those...
I tried the other solution that opus was saying before, and I was getting 6403 miles. So I guess your solution is better

Thanks for the help everyone... this is an interesting problem, but I still dont understand kedaman's solution:D(maybe I'm dumb:D:D)

AND: is my diagram right? is x where it's supposed to be :confused:

(I've been sleeping at 3 am all days this week, and umm...ahem... my brain isnt working very well at the moment:p sorry if I make a dumb mistake)

aah alright, did I do this right?:D


http://www.vbforums.com/attachment.php?postid=1305061

Yup :)

Quote:

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Originally posted by riis
Yup :)

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alright I think I'm being annoying... one more thing:D:D:D:D:D:D


why are both sides X? why are they equal?

Short answer: Yup

Long answer:
Theyare equal, since they are both start at the end a line coming from the centre of the circle and both are tangents to that circle. So they have to be identical in length.

Quote:

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Originally posted by opus
Short answer: Yup

Long answer:
Theyare equal, since they are both start at the end a line coming from the centre of the circle and both are tangents to that circle. So they have to be identical in length.

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aah got it! :) thanks :)

A good ending to your story
http://www.math.yorku.ca/Who/Faculty.../Stewart1.html

one thing still confuses me, if he didn't for once look at the compass during his long trip, how did he manage to go in a straight line?

thanks opus, my av inspired by the diplomacy game i'm currently playing