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No. The instantaneous velocity will be different all along the path. Although the average might be the same and the length of trip may be the same there is no guarantee that a particular spot in the path has corresponding "time of arrival". In fact, I would argue that there is an infantesimally small chance that they would match given the day long trip length, distance walked, terrain variation, etc.


Quoted: No. The instantaneous velocity will be different all along the path. Although the average might be the same and the length of trip may be the same there is no guarantee that a particular spot in the path has corresponding "time of arrival". In fact, I would argue that there is an infantesimally small chance that they would match given the day long trip length, distance walked, terrain variation, etc. View Quote Incorrect 

Likely impossible. Though there is a minuscule possibility.
An equal intersection at the exact millisecond. 


Quoted: No. It is a mathematical certainty there is a point he will be at the same time each day. View Quote View All Quotes View All Quotes Quoted: Quoted: Likely impossible. Though there is a minuscule possibility. An equal intersection at the exact millisecond. No. It is a mathematical certainty there is a point he will be at the same time each day. Yeah, uh, no. 

Quoted: Quoted: Assuming his pace to be equal in both directions, he will pass the halfway point at the same time of day in both directions. His speed is irrelevant. Sure, if you're only interested in answering the one question of the riddle. For that purpose, yes, his speed is irrelevant, and he will meet his former self somewhere along the trail, at whatever time and place that happens to be. For the purpose of deeper thinking, however, we are given that his average pace is the same in both directions, and that he is a monk, which means he is highly disciplined and consistent. He is most likely to meet himself in the middle at 2PM. 


Quoted: No. The instantaneous velocity will be different all along the path. Although the average might be the same and the length of trip may be the same there is no guarantee that a particular spot in the path has corresponding "time of arrival". In fact, I would argue that there is an infantesimally small chance that they would match given the day long trip length, distance walked, terrain variation, etc. View Quote You're picking a specific spot and seeing if they will meet there. That's not the right way to look at it. Where the spot is and when the meeting happens is not relevant. Just that they are required to pass each other at some point. Wherever they pass each other is by definition the same place at the same time. 

Quoted: Quoted: Quoted: Likely impossible. Though there is a minuscule possibility. An equal intersection at the exact millisecond. No. It is a mathematical certainty there is a point he will be at the same time each day. Yeah, uh, no. How is it possible to not pass each other? That is the only way to avoid being at the same place at the same time. 

Quoted: You're picking a specific spot and seeing if they will meet there. That's not the right way to look at it. Where the spot is and when the meeting happens is not relevant. Just that they are required to pass each other at some point. Wherever they pass each other is by definition the same place at the same time. View Quote Correct. 

No. Because he's on earth which is hurling through space, so it's impossible for him to be in the exact same spot.


The only way it would be possible is if he hit the midway point at 2pm for both the ascent and descent, assuming his travel speed is the same for both trips
Monks also have morning, mid day, and evening prayer, so that will probably be tied in with the answer. Not going to consider the possibility of transendance because it’s pure fiction 

Quoted: Imagine two monks on the same trail, one climbing and the other descending. Will they ever meet on that one trail, be at the same place at the same time? Yes, there is only one trail. They are both on it heading in different directions. They will meet somewhere along the trail. That time is the same for both travelers  "they meet at the same time of day"  satisfying the riddle. OK now one monk leaves on Monday and the other on Tuesday it doesn't change the result the time is the same. View Quote This is the best and easiest to understand answer. Indisputable. 

Quoted: The only way it would be possible is if he hit the midway point at 2pm for both the ascent and descent, assuming his travel speed is the same for both trips Monks also have morning, mid day, and evening prayer, so that will probably be tied in with the answer View Quote Travel speed and time of the meeting are irrelevant. The monk will occupy the same place, somewhere on that path, at the same time of the day 24 hours apart. 


No. What about the movement of the planet? Travel of the solar system? Expansion of the Universe? If you can believe in the round earth theory.
He wouldn't even be in the same place if he stayed on a treadmill bro. 

I get that questions like this are meant to just ponder over... many a like in math class of yesterday year.
What I absolutely can stand is why the fuck do we give a shit. Same with most problems like that... I understand the problem solving teachings that are behind them but lets at least solve for something that can improve one's life not waste it. 

I will still argue it is not possible. Yes he will pass the same point, but not at the same time unless the speed is the exact same going up and down.
Look at like this, lets say the distance is 4 miles for easy math. Monk on day one climbs 1.9 miles in the first 4 hours, he is 1.9 miles in to the climb at noon. Day two he starts off slower because he knows its an easier walk, at noon he has covered 1.9 miles. day one and two are at the same time but .2 miles apart. So lets change it some. day one he knows he is a long climb and he starts off in a rush, in 5 hours he covered 3 miles. He will be 3 miles up at 1 pm. Day two he starts off with normal pace, or .5 miles an hour. 5 hours in he has passed where he was the day before at 5 hours but is 2.5 miles down the hill. Unless the speeds average exactly the same to the center point they will never meet at the same time and place. Yes they will meet at some point, but not the same time. Time and distance mean nothing when they can be slower or faster at times. Yes on the whole trip they average the same speed, but doing the first half at any different speeds means they covered different distances and therefore are not at the same place. And yes there are possibilities that it happenes, like day one he walks 3 miles and arrives at noon, and day two he slowly walks one mile and arrives at noon. But the odds are slim, even a .01 mile an hour difference screws up where he is at the same place and time. even the internet answer in the solution does not state they arrive at the same time, just the two monks would meet. and that is different that the orginal riddle of passing the same point at the same time. https://creativityboost.net/2017/10/19/themonkandthemountainclimb/ The answer to the riddle per the internet says to look at it like 2 monks, but it is not. Yes two monks would pass at the same time and place, but the one monk has already passed the spot once, so the spot and time is already locked in time and place. Unlike two monks where the spot and time is not locked until they pass. 

In order to prove this, I would need to use transcendental functions.
But the answer is “Yes”. 

Quoted: I will still argue it is not possible. Yes he will pass the same point, but not at the same time unless the speed is the exact same going up and down. Look at like this, lets say the distance is 4 miles for easy math. Monk on day one climbs 1.9 miles in the first 4 hours, he is 1.9 miles in to the climb at noon. Day two he starts off slower because he knows its an easier walk, at noon he has covered 1.9 miles. day one and two are at the same time but .2 miles apart. So lets change it some. day one he knows he is a long climb and he starts off in a rush, in 5 hours he covered 3 miles. He will be 3 miles up at 1 pm. Day two he starts off with normal pace, or .5 miles an hour. 5 hours in he has passed where he was the day before at 5 hours but is 2.5 miles down the hill. Unless the speeds average exactly the same to the center point they will never meet at the same time and place. Yes they will meet at some point, but not the same time. Time and distance mean nothing when they can be slower or faster at times. Yes on the whole trip they average the same speed, but doing the first half at any different speeds means they covered different distances and therefore are not at the same place. And yes there are possibilities that it happenes, like day one he walks 3 miles and arrives at noon, and day two he slowly walks one mile and arrives at noon. But the odds are slim, even a .01 mile an hour difference screws up where he is at the same place and time. even the internet answer in the solution does not state they arrive at the same time, just the two monks would meet. and that is different that the orginal riddle of passing the same point at the same time. https://creativityboost.net/2017/10/19/themonkandthemountainclimb/ View Quote Not only is it possible, it is a certainty, that there is a point he will be at at the same time on each day. 

Quoted: I will still argue it is not possible. Yes he will pass the same point, but not at the same time unless the speed is the exact same going up and down. ... View Quote Imagine the two monks scenario. One leaves from the bottom, one leaves from the top. At some point in the day, their paths cross. That's "the spot". The first monk arrives at the top of the mountain, and calls it a day. The second monk leaves the story in some way. Probably falls off the mountain or something. The next morning the monk leaves the top of the mountain headed down. At some point in his day, he crosses "the spot". That place where he met the other brother. "The spot" is where he's in the same place at the same time on two different days. It can't be any other way. 



Quoted: Not only is it possible, it is a certainty, that there is a point he will be at at the same time on each day. View Quote Show it by math, using any different speed. You cant, its not possible unless he does the exact same speed to the center point. Bet you a pmag you cant show it math mathematically with different speeds up until the meeting point. Not saying a majority agree with you, that you have to show it mathematically for the bet to valid. My math you quoted shows its impossible. lets take the math further. 4 miles in 8 hours average speed is .5 mph. Monk on day for the first 4 hours averages .51 mph he is 2.04 miles in to the trip at noon. So now the constant is at exactly noon he is exactly 2.04 miles up the hill. Days two he averages exactly .5 mph at noon he is exactly 2 miles. Yes at exactly noon he 2 miles in to the trip, but he is 211.2 feet past where he was the day before. Therefore at the same time he was not at the same point. lets change the math some, you know scientific method means trying to disprove something and if you cant than its correct. day one he averages .49 miles for 4 hours. at exactly noon he is 1.96 miles up the hill. So now the target is he has to be 2.04 miles down on day two at exactly noon. If he walks any speed other than exactly the average of .51 mph he will never be at the the same point at noon. If he walked .5 he is only 2 miles, if he walks .52 mph he is 2.08 miles or 211.2 feet past where he was the day before at exactly noon. 

Quoted: Show it by math, using any different speed. You cant, its not possible unless he does the exact same speed to the center point. Bet you a pmag you cant show it math mathematically with different speeds up until the meeting point. Not saying a majority agree with you, that you have to show it mathematically for the bet to valid. My math you quoted shows its impossible. lets take the math further. 4 miles in 8 hours average speed is .5 mph. Monk on day for the first 4 hours averages .51 mph he is 2.04 miles in to the trip at noon. So now the constant is at exactly noon he is exactly 2.04 miles up the hill. Days two he averages exactly .5 mph at noon he is exactly 2 miles. Yes at exactly noon he 2 miles in to the trip, but he is 211.2 feet past where he was the day before. Therefore at the same time he was not at the same point. lets change the math some, you know scientific method means trying to disprove something and if you cant than its correct. day one he averages .49 miles for 4 hours. at exactly noon he is 1.96 miles up the hill. So now the target is he has to be 2.04 miles down on day two at exactly noon. If he walks any speed other than exactly the average of .51 mph he will never be at the the same point at noon. If he walked .5 he is only 2 miles, if he walks .52 mph he is 2.08 miles or 211.2 feet past where he was the day before at exactly noon. View Quote take the two routes, but instead of imagining them on consecutive days, imagine them on the same day. They both leave at 8am, one going uphill and the other going downhill. They will meet somewhere along the route. Thanks for the pmag. 

Quoted: Imagine the two monks scenario. One leaves from the bottom, one leaves from the top. At some point in the day, their paths cross. That's "the spot". The first monk arrives at the top of the mountain, and calls it a day. The second monk leaves the story in some way. Probably falls off the mountain or something. The next morning the monk leaves the top of the mountain headed down. At some point in his day, he crosses "the spot". That place where he met the other brother. "The spot" is where he's in the same place at the same time on two different days. It can't be any other way. View Quote Again same bet at I gave to low country, 1 pmag if you can prove it mathematically. Not everyone agrees, you need to prove it by math. And you cant plan a speed for day two, day one is now set in time for ever. Thats the problem with the two monks way of solving it, its variable until they meet. One monk its not on day one place and time is locked. He cant be in the same place and time with variable average speed on day two except in the one math problem where we solve for speed on day two and distance on day one. Yes I will admit by sheer luck he walks the perfect speed on day two than it works. But it is not certain. 

Quoted: take the two routes, but instead of imagining them on consecutive days, imagine them on the same day. They both leave at 8am, one going uphill and the other going downhill. They will meet somewhere along the route. Thanks for the pmag. View Quote now you are changing the riddle. its not two monks its one. That is what you dont get, again day one speed place and time is no longer variable, he was at some exact spot at some exact time. For you to be right he needs to be at the exact same place and time not random meeting his ghost from the day before. The bet was one pmag, for you to show it by math. By responding I think you agreed to the bet, now show the math or PM me for my mailing address. 

Quoted: now you are changing the riddle. its not two monks its one. That is what you dont get, again day one speed place and time is no longer variable, he was at some exact spot at some exact time. For you to be right he needs to be at the exact same place and time not random meeting his ghost from the day before. The bet was one pmag, for you to show it by math. By responding I think you agreed to the bet, now show the math or PM me for my mailing address. View Quote I’m sorry you can’t wrap your head around this one, it fools many people. But I’m correct, and you are incorrect. It is a certainty that there is a point on the path where he will be at the same time on each day. https://condor.depaul.edu/dallbrit/extra/360/420_22MONK.HTM https://creativityboost.net/2017/10/19/themonkandthemountainclimb/ https://gizmodo.com/thisweekspuzzlehasaverysimplesolutioncanyouf1685948256 

Absolutely yes. Imagine two monks, one at the top heading down, one at the bottom heading up. Each sets out on the trail at 8:am. At any point will they meet on the trail?




The earth is never in the same point in space, ever. The sun is always moving through space, dragging the planets with it. Therefor no, the monk is never in the same place at any given time.


You don't need to use math. Imagine it's a graph.
Attached File Red is the ascending monk. He starts at the beginning of the day at the bottom of the mountain and ends at the end of the day at the top of the mountain. Green is the descending monk. He starts at the beginning of the day at the top of the mountain and ends at the end of the day at the bottom of the mountain. Those are the known constraints. Given those, it is NOT POSSIBLE to draw lines of ANY slope, changing, constant, negative, positive, or otherwise, such that they don't intersect at at least one point. This is among the easiest problems and I am dumbfounded how people can't. 



Quoted: You don't need to use math. Imagine it's a graph. https://www.ar15.com/media/mediaFiles/267086/graph_png2964958.JPG Red is the ascending monk. He starts at the beginning of the day at the bottom of the mountain and ends at the end of the day at the top of the mountain. Green is the descending monk. He starts at the beginning of the day at the top of the mountain and ends at the end of the day at the bottom of the mountain. Those are the known constraints. Given those, it is NOT POSSIBLE to draw lines of ANY slope, changing, constant, negative, positive, or otherwise, such that they don't intersect at at least one point. This is among the easiest problems and I am dumbfounded how people can't. View Quote Yes they intersect but not nessessarly at the same point and place in time or distance. 





I will take that bet. I believe I can prove it mathematically.
Is it open to me? Edit: it has already been shown in graph form. 



Quoted: You don't need to use math. Imagine it's a graph. https://www.ar15.com/media/mediaFiles/267086/graph_png2964958.JPG Red is the ascending monk. He starts at the beginning of the day at the bottom of the mountain and ends at the end of the day at the top of the mountain. Green is the descending monk. He starts at the beginning of the day at the top of the mountain and ends at the end of the day at the bottom of the mountain. Those are the known constraints. Given those, it is NOT POSSIBLE to draw lines of ANY slope, changing, constant, negative, positive, or otherwise, such that they don't intersect at at least one point. This is among the easiest problems and I am dumbfounded how people can't. View Quote True for the place but not time. It’s a not on the X Y plane. It’s not tied directly to either value. 



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