Mr. Fubini

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one person, lets call him Mr. Fubini, walked south one mile, west one mile, and then back north one mile. he then looked around and noticed he was back where he started. where on the earth is this possible?

(note there is a really obvious answer honestly everyone gets in like 2 minutes, and because its so easy, ill just say it. if you havent figured out the answer when you are reading this, you definetly didnt think about the question. so, easy answer is: mr fubini was on the north pole. if you look at the intersections of the meridians, its kind of obvious.

however, there is another solution to the problem (this is the harder part).

if he wasnt on the north pole, then where was he?

btw, i named him Mr. Fubini because he was the guy who made Fubini's theorem, which sounds like one of the most useless theorems out there, so it seemed appropriate for him to get confused this way :rofl: :rofl:
 
I'm working on the equation...but...What about if your starting point was at a distance away from the pole where the circumference of the earth was one mile. The north and south movements cancel each other out and the west one mile would put you exactly were you started.
 
Because 1/2, 1/3, 1/4, 1/5, miles away would also work; As long as you are located on a circle:

1 + 1/(2pi*n)

The west movement would be nullified as well. Assuming n is a positive integer.
 

T@P

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exactly right! nice thinking
 
T@P said:
btw, i named him Mr. Fubini because he was the guy who made Fubini's theorem, which sounds like one of the most useless theorems out there, so it seemed appropriate for him to get confused this way :rofl: :rofl:
Fubini's theorem, useless ? Let's see you go back to integrating over regions without using iterated integrals. :yuck: :biggrin:
 
Say Mr. Fubini's journey is 1000 miles south, 1000 miles west, and 1000 miles north so you have to figure in the curvature of the earth. Then what are the distances from the pole he can start?
 

T@P

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lol i only think that on the surface, his theorem sounds useless. misquoted, it states "no matter HOW you cut the frickin potato, the volume is THE SAME"

or something to that extent. dont sweat over it :)

and BicycleTree, the curvature of the earth is taken into acount all the time. using the second solution, you would find the correct "height" of the earth so that the circumference there is 1000 miles, then simply walk straight up and down. i guess a better way to explain it is if you take a map of the earth, it looks like a watermelon with all the curvy lines. but in the middle, there are two normally "straight" and perpendicular lines (perpendicular on the flat paper too). the position of those lines depends only on where you look from, so really curvature has noting to do with it... i think i made it far more complicated then it should be, but i hope you take my point.
 
Yes, finding the correct "heights" of the earth--and from those heights, the curved distances along the earth's surface so you can say how far he is from the pole--would then be the problem.

I don't understand what you're saying about perpendicular lines.
 
Papa_Rocks said:
I'm working on the equation...but...What about if your starting point was at a distance away from the pole where the circumference of the earth was one mile. The 'north and south' movements cancel each other out and the west one mile would put you exactly were you started.
That's the right reasoning but it forgets that Mr. Fubini is at first going south, hence the diameter he is before going west is then more than 1 mile if he started like you say. In fact this just needs easy trigonometric calculation to find the starting point.
 

T@P

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ok to clear it up a bit: find some circumference (near the south pole whos radius is perpendicular to the north/south pole diameter) on the earth whos length is either 1 km, or anything that divides it (like 1/2 a km, or 1/3 etc.). take any point on that circumference and go north one mile. this is where you start.

starting from that point, its clear that going south lands you on that circumference, west will get you no where, and north back to where you started. to find that point, really its not hard. you know that the radius of the point must be 1/(2*pi) (lets call it r), you know big R (radius of the earth) and from there its a right triangle for the "height". thats it really, no trig or fancy math.
 
Yes...thanks for precisions...

What about this one (I didn't solve it) : Mr. Inibuf achieved to make what he calls a "spherical-planar triangle" : 1km S, 1km W, and Sqrt(2)km NE....At which latitude did he start (near the north pole in degree) ?
 

T@P

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ooof this is stretching my visualization capabilitis...

first of all, if you want to make a triangle you dont need the west/south stuff. but since you have it i will assume that you want to walk around the perimeter of the triangle...
if you do it that way i think it works any place on the globe as long as it isnt too close to the poles (directions get funny there). like imagine starting on the equator. you go south. then west. then north east, thats just like going in a triangle. works on a plane as well as on a sphere. i think im missing your point somehow.

if you want me to calculate how far away you have to be for it to make sense (i.e. going south when you are at the south pole is fuzzy at best), its just stay 1 km away from the south pole, and you should be ok. similarly, stay one km away from the north pole...

maybe you worded it wrong? because the way its written now its a no-brainer to put it bluntly.

or maybe its because i just got a haircut. take your pick
 
Your solution near the south pole is weird : you start 1km north from it...hence you land on it after the first movement...then u get stuck and didnt complete the triangle.

Near the north pole : the problem is that going NE draws a funny shape on the sphere:

-If it's a great circle, then basic spherical trigonometry gives the length of the hypothenuse c, in this triangle :

cos(c)=cos(a)*cos(b)

hence, if it's a great circle, you never get Sqrt(2) for the hypothenuse (just by expanding the cosine you see this, but you get Sqrt(2) for infinitesimal edges).

-I suppose it's not either a great circle (It should be a part of a spiral)...so how do you know the length on that curve has to be Sqrt(2) to reconnect to the first point ? And first of all, do you eventually reconnect to the starting point that way ?

That's all the stuff to be proven....in fact it's not a brain teaser, i suppose it's a huge amount of calculus.
 

T@P

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oh haha sorry i see what you mean now... the length wont be sqrt(2) unless you find some messed up point... yes that will be a vastly unforgiving amount of work. let me think about it... there may be an "easier" way to look at it...
 
Ok...in fact it's as most as easy as you said....going NE gives the condition :

dtheta=sin(theta)dphi....hence :

ds^2=R^2(dtheta^2+sin(theta)^2dphi^2)

=>ds=Sqrt(2)dtheta.

So the length is exactly Sqrt(2) times when going only north...

Then, integrating the first equation gives :

ln(tg(theta))=phi...

This should give the condition for the starting angle, but it's a non-analytical equation, so i give up...
 

T@P

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me too. mr. Inibuf just has to go and ask mr fubini for directions this time...
 

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