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Quoted: I get that, and it seems a bit yo obvious to be the correct answer/response and isn't how I construe the OP when he states 'same time of day'. To me, that means at the same time as shown on a clock. He would have to walk at exactly the same pace but up & down or lead/lag one way or the other while watching his clock to time it correctly. In your explanation, he absolutely crosses his own path. However, there is no guarantee that it is at the same clock time during the day... ...this riddle is intentionally stated to allow for varying ways to construe its intent. Very sneaky View Quote View All Quotes View All Quotes Quoted: Quoted: Quoted: Of course he'd cross his holographic self at some point along the trail, but at the exact same time of the day? Doubtful, unless this is a play on words when 'same time of day' doesn't mean the time shown on the clock... 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. I get that, and it seems a bit yo obvious to be the correct answer/response and isn't how I construe the OP when he states 'same time of day'. To me, that means at the same time as shown on a clock. He would have to walk at exactly the same pace but up & down or lead/lag one way or the other while watching his clock to time it correctly. In your explanation, he absolutely crosses his own path. However, there is no guarantee that it is at the same clock time during the day... ...this riddle is intentionally stated to allow for varying ways to construe its intent. Very sneaky LOL The ghost track answer is correct. If you start the ghost track up at 8 am and the guy starts down at 8am the next day, he passes the ghost track somewhere at the same time no matter what they did coming up or down. No special clock watching required. 


No. Measurement is infinite, so precision is impossible. All things are meaningless. Thus we drift into the void.






Quoted: OK. Show me where the monks are in your example between 1:40 and 2:20. View Quote Okay scaled the distance at 2pm, which is the half way point in time. Assumed the distance is 6 miles like my other chart. Attached File Attached File 

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. View Quote In your first scenario the monks would have to walk up and down the mountain, then both of them reverse their routes and meet at the same exact place at the same exact time on the second day for it to be the same scenario as the O.P.s. 

Quoted: I remember a poster getting banned for welching on a pmag bet not too long ago. View Quote And you have not done what your side of the bet is, show the math. I also upped it to 10 pmags if you just draw a line on the chart and the over lay has them match time and place. But you have ignored that one, so I assume you concede they cant match. 

Quoted: Same time of day, as shown on the clock. They both leave at 8am and will surely cross paths along the way... ...hold up, wait a minute. This scenario gets different results in my head when thinking about (2) people on the sane day walking different directions vs the same guy walking both directions on different days View Quote View All Quotes View All Quotes Quoted: Quoted: Quoted: Quoted: Quoted: Of course he'd cross his holographic self at some point along the trail, but at the exact same time of the day? Doubtful, unless this is a play on words when 'same time of day' doesn't mean the time shown on the clock... 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. I get that, and it seems a bit yo obvious to be the correct answer/response and isn't how I construe the OP when he states 'same time of day'. To me, that means at the same time as shown on a clock. He would have to walk at exactly the same pace but up & down or lead/lag one way or the other while watching his clock to time it correctly. In your explanation, he absolutely crosses his own path. However, there is no guarantee that it is at the same clock time during the day... ...this riddle is intentionally stated to allow for varying ways to construe its intent. Very sneaky Are you arguing that if 2 monks, let's call them A and B, are both on the trail, A going up and B going down, that when they meet each other on the trail it will not be the same time for both of them? Same time of day, as shown on the clock. They both leave at 8am and will surely cross paths along the way... ...hold up, wait a minute. This scenario gets different results in my head when thinking about (2) people on the sane day walking different directions vs the same guy walking both directions on different days Oh, and this thread still sucks... 

Quoted: Not a troll and I know betting here can cause the loss of an account if its welched. I wont welch, he just have to do the blank chart and if they cross same place and time I will send em. View Quote The fact that the lines cross at all proves you wrong. In order for you to not owe pmag(s) you need to draw the graph without the lines intersecting. 

This whole graph thing is best visualized as a line segment with points at each end. The points represent the "2" monks on their common path of travel, the line.
Press play and both points start heading toward one another over the course of 8 hours. Eventually some time between 1 second after starting them 8 hours later the two points are going to converge no matter their respective speed or if they sometimes set idle for hours. They will converge and that point was the common time of day they shared in the same place on the path. 

Quoted: And you have not done what your side of the bet is, show the math. I also upped it to 10 pmags if you just draw a line on the chart and the over lay has them match time and place. But you have ignored that one, so I assume you concede they cant match. View Quote The answer has been posted several times. 

Quoted: Okay here is chart with them meeting at the same time, distance traveled is vertical and time is across. https://www.ar15.com/media/mediaFiles/497353/same_time_jpg2965085.JPG View Quote Two different Y axes? Attached File 

StaccatoC2 doesn't get it.
Admiral_Crunch's explanation on page 1 should be enough. The OP says it takes 12 hours to go up and 12 to go down. Based on this, on AVERAGE, they will meet at the halfway point at 2pm. Now, let's say it takes less time to go down than it does to go up. A reasonable assumption. The monk on day 1 will walk per the following equation: mph1 x t1 = d1 For monk2: mph2 x t2 =d2 Since they have to be in the same spot at the same time, t1 = t2. Therefore, mph1/d1 = mph2/d2 We also know that the total distance, D, equals d1+d2. @StaccatoC2, provide values for mph1, mph2, and D and anyone else on the thread will be able to tell you how far each monk walked and at what time they met at the same spot on the trail. Send the other guy a pmag or 10 or make a donation to St. Judes. Now an example: mph1 = 2 mph mph2 = 4 mph D = 12 Plugging in the numbers to the above equation: 2/d1 = 4/d2 12 = d1 + d2 so d1 =12  d2 2/(12d2) = 4/d2 2d2 = 48  4d2 6d2 = 48 d2 = 8 d1 = 12  8 = 4 mph1 x t1 = d1 2 x t1 = 4 t1 = 2 hrs At 10 am, monk1 1 will be 4 miles up the mountain and monk 2 will be 8 miles down. Sure hope I didn't screw up the math after all that. Try it with whatever mph and total distance you want. 

8am8pm is 12 hours. Let's assume it's a 12 mile trip for easy math. The ascending monk travels at 1 mph and the descending monk travels at 3 mph. At 11am (3 hours in), the Monks meet eachother. The descending monk stops at that point to rest until 7 pm then continues. The ascending monk never stops. Both Monks complete their journey at 8pm.
ETA beat by someone who does math shit... In the end, there is a certainty that the monk occupies the same place at one point in the journey. 

Quoted: Okay scaled the distance at 2pm, which is the half way point in time. Assumed the distance is 6 miles like my other chart. https://www.ar15.com/media/mediaFiles/497353/meeting_place_jpg2965093.JPG https://www.ar15.com/media/mediaFiles/497353/monk2_jpg2965094.JPG View Quote Your graph is irrelevant. Your Y axes don't match so you can't compare the two. Suppose the mountain is 1000 meters high. Your descending monk is at, say, 550 meters at 1:40. We know the ascending monk doesn't get to to 550 meters until 2:20. So let's say he's at 500 meters at 1:40. Explain the altitudes of the monks between 1:40 and 2:20. General terms are fine. For what it's worth, what you've done is literally no different than saying one monk is at the bottom at 8 AM, but the other monk doesn't get there until 8 PM. You would need to mathematically prove that for all possible times and altitudes between the start and the finish, they can't be the same. I'll give your a hint  that's not possible. 

Get a piece of string. The path
Put a glass bead on each end. The monks Start a timer, you have 12 hours to slide the beads toward the opposite end they are on. You can slide them in any amount you want but by the end of 12 hours each bead needs to be at the opposite end. Do they collide somewhere on that string in that 12 hour interval? Does not the common clock they share you started at the beginning tell you just one time...that they both share? 

Quoted: 8am8pm is 12 hours. Let's assume it's a 12 mile trip for easy math. The ascending monk travels at 1 mph and the descending monk travels at 3 mph. At 11am (3 hours in), the Monks meet eachother. The descending monk stops at that point to rest until 7 pm then continues. The ascending monk never stops. Both Monks complete their journey at 8pm. ETA beat by someone who does math shit... View Quote Too funny. I didn't want to do it but I couldn't take it any longer. 

Quoted: Get a piece of string. The path Put a glass bead on each end. The monks Start a timer, you have 12 hours to slide the beads toward the opposite end they are on. You can slide them in any amount you want but by the end of 8 hours each bead needs to be at the opposite end. Do they collide somewhere on that string in that 12 hour interval? Does not the common clock they share you started at the beginning tell you just one time...that they both share? View Quote Excellent visual. 

View Quote Fine here is a chart that even has them crossing the same time, 4 times with variable speed. It also bases distance traveled and shows how they have covered different distances there fore are not in the same location. Attached File 

Quoted: Fine here is a chart that even has them crossing the same time, 4 times with variable speed. It also bases distance traveled and shows how they have covered different distances there fore are not in the same location. https://www.ar15.com/media/mediaFiles/497353/same_plane_chart_jpg2965126.JPG View Quote I want my Pmags. But your trolling is impressive. Distance traveled is not what you should be graphing. That chart doesn't have to show anything other than they start and end at the same point. Try graphing distance for one, and (6  distance) for the other, and you might find the answer you seek. Because you are looking for altitude to match, not distance traveled. 

Quoted: Fine here is a chart that even has them crossing the same time, 4 times with variable speed. It also bases distance traveled and shows how they have covered different distances there fore are not in the same location. https://www.ar15.com/media/mediaFiles/497353/same_plane_chart_jpg2965126.JPG View Quote Tell ya what. Instead of sending me a pmag, donate $10 to st jude, and post a screen shot of the receipt here, and I’ll consider your bet taken care of. 

Quoted: https://i.ibb.co/mh2vDZJ/monkgraph1.jpg ETA: I still have the tool pulled up. Be more than happy to put in varying rates if you want. ETAA: My graph shows an X axis with time (from 0 to 12, or 8am to 8pm), and the Y axis as the height of the mountain ( from 012, or twelfths of the total height of the path). Your double inverted Y axis showing lines which mean 2 different things at the same time... is hogwash. I'll make 2 graphs, one for each monk, and overlay them to illustrate that, if you'd like. View Quote Ok do the charts where day one monk travels at an average of say 2 miles an hour for the first 3, then slows down to 1.5 for the reminder of the day because he pushed to hard which would make the path 19.5 miles. Then the monk coming down knows its an easy walk and keeps a constant speed of 1.625 mph. @sykkone 

Quoted: Fine here is a chart that even has them crossing the same time, 4 times with variable speed. It also bases distance traveled and shows how they have covered different distances there fore are not in the same location. https://www.ar15.com/media/mediaFiles/497353/same_plane_chart_jpg2965126.JPG View Quote Never mind. 

Quoted: Get a piece of string. The path Put a glass bead on each end. The monks Start a timer, you have 12 hours to slide the beads toward the opposite end they are on. You can slide them in any amount you want but by the end of 12 hours each bead needs to be at the opposite end. Do they collide somewhere on that string in that 12 hour interval? Does not the common clock they share you started at the beginning tell you just one time...that they both share? View Quote Yes but will the distance traveled be the same for both, then that distance is from other ends of the beginning and end the exact same. 

Quoted: Tell ya what. Instead of sending me a pmag, donate $10 to st jude, and post a screen shot of the receipt here, and I’ll consider your bet taken care of. View Quote When I was in high school, my wrestling coach taught "Alternative Learning Center." It was basically just in school suspension for kids who were not quite legally retarded, just prone to trouble because they generally had no attention span. But they were universally pretty dumb. The wrestlers were allowed in the back room there during free periods, where we'd play NES. As the lone smart guy on the team, I ended up trying to tutor these kids. Sometimes it worked, but I will never forget trying to explain algebraic substitution to this one girl. The problem was 3x=27, and no matter how many different ways I tried she didn't get it. I eventually gave up and just said I couldn't help her. 


Quoted: Tell ya what. Instead of sending me a pmag, donate $10 to st jude, and post a screen shot of the receipt here, and I’ll consider your bet taken care of. View Quote Tell me the chart I posted is wrong on locations. Tell me how they are same location, with any kind of variable speed they cant be. Again the two monks walking up and down is not accurate. Once he walks up where and when is locked in time, so the down monk will not run in to the up monk he only has time and places that cant change. 

Quoted: Yes but will the distance traveled be the same for both, then that distance is from other ends of the beginning and end the exact same. View Quote View All Quotes View All Quotes Quoted: Quoted: Get a piece of string. The path Put a glass bead on each end. The monks Start a timer, you have 12 hours to slide the beads toward the opposite end they are on. You can slide them in any amount you want but by the end of 12 hours each bead needs to be at the opposite end. Do they collide somewhere on that string in that 12 hour interval? Does not the common clock they share you started at the beginning tell you just one time...that they both share? Yes but will the distance traveled be the same for both, then that distance is from other ends of the beginning and end the exact same. The distances traveled will be the same by the end of the 12 hours but not necessarily where they meet. The path is the same length for both. If you believe the monk doesn't pass his ghost at some shared time and place on the path then you DO believe 2 glass beads can get to the opposite ends on a string without hitting. 

Quoted: Tell me the chart I posted is wrong on locations. Tell me how they are same location, with any kind of variable speed they cant be. Again the two monks walking up and down is not accurate. Once he walks up where and when is locked in time, so the down monk will not run in to the up monk he only has time and places that cant change. View Quote Go back and tell me altitudes for the monks between 1:40 and 2:20. 

Quoted: Tell me the chart I posted is wrong on locations. Tell me how they are same location, with any kind of variable speed they cant be. Again the two monks walking up and down is not accurate. Once he walks up where and when is locked in time, so the down monk will not run in to the up monk he only has time and places that cant change. View Quote I’m not wasting more time on your troll game. You offered a bet, which you lost. Refusing to pay up makes you a welcher. 

Quoted: Can you go back to the altitude of the monks between 1:40 and 2:20? Choose your own numbers if you don't like mine. View Quote Ok if I understand what your asking, I scaled it below. Its close. at 140 for each day one monk is 2.8 miles up, and day two monk is 3.1 ( probably marked a little far past the point so closer to 3 but resolution to scale is not great ) Attached File 

Quoted: Quoted: The distances traveled will be the same by the end of the 12 hours but not necessarily where they meet. The path is the same length for both. If you believe the monk doesn't pass his ghost at some shared time and place on the path then you DO believe 2 glass beads can get to the opposite ends on a string without hitting. Checkmate https://i.postimg.cc/xCgMZRtT/Path.jpg FUUUUUUUCK 

Quoted: The distances traveled will be the same by the end of the 12 hours but not necessarily where they meet. The path is the same length for both. If you believe the monk doesn't pass his ghost at some shared time and place on the path then you DO believe 2 glass beads can get to the opposite ends on a string without hitting. View Quote Ok let me rephrase that. Have they traveled to the same point from the start and finish in the same amount of time? 

Quoted: Ok if I understand what your asking, I scaled it below. Its close. at 140 for each day one monk is 2.8 miles up, and day two monk is 3.1 ( probably marked a little far past the point so closer to 3 but resolution to scale is not great ) View Quote Great! We might be getting somewhere. Monk one has to go from 2.8 to 3.1. Time taken is irrelevant. If monk two doesn't move, then he's still sitting at 3.1 at 2:20. If he moves, then he's going down as the other is going up. Does this help illustrate that they have to cross paths? At 0.3 miles apart, they might even be able to see each other. Surely you acknowledge they cross? 

Quoted: Fine here is a chart that even has them crossing the same time, 4 times with variable speed. It also bases distance traveled and shows how they have covered different distances there fore are not in the same location. https://www.ar15.com/media/mediaFiles/497353/same_plane_chart_jpg2965126.JPG View Quote View All Quotes View All Quotes Quoted: Fine here is a chart that even has them crossing the same time, 4 times with variable speed. It also bases distance traveled and shows how they have covered different distances there fore are not in the same location. https://www.ar15.com/media/mediaFiles/497353/same_plane_chart_jpg2965126.JPG Cool, your monk discovered time travel and teleportation! 

Quoted: Great! We might be getting somewhere. Time taken is irrelevant. View Quote View All Quotes View All Quotes No time is relevant the op said meet at the same place at the same time he was there the day before. Quoted: Great! We might be getting somewhere. At 0.3 miles apart, they might even be able to see each other. Surely you acknowledge they cross? Yes the paths will cross, but wwould they be at the same time, no because 140 being the time, they are not at the same place. 

Quoted: Great! We might be getting somewhere. Monk one has to go from 2.8 to 3.1. Time taken is irrelevant. If monk two doesn't move, then he's still sitting at 3.1 at 2:20. If he moves, then he's going down as the other is going up. Does this help illustrate that they have to cross paths? At 0.3 miles apart, they might even be able to see each other. Surely you acknowledge they cross? View Quote Ok I also see what your saying, but we are using my chart based on time he did move. 

Its interesting that everyone advocating for 'yes, they do cross paths' uses a chart/graph with opposite lines of charting (x pattern). Of course an x pattern crosses somewhere...
...but the monk is walking the same path in different directions (parallel of sorts path), which means that his timesrelativelocation can definitely never cross between start n finish... 

Quoted: Ok let me rephrase that. Have they traveled to the same point from the start and finish in the same amount of time? View Quote View All Quotes View All Quotes Quoted: Quoted: The distances traveled will be the same by the end of the 12 hours but not necessarily where they meet. The path is the same length for both. If you believe the monk doesn't pass his ghost at some shared time and place on the path then you DO believe 2 glass beads can get to the opposite ends on a string without hitting. Ok let me rephrase that. Have they traveled to the same point from the start and finish in the same amount of time? Absolutely. Think about the beads on a string. If you were tasked with pushing the 2 beads to opposite ends over the course of 12 hours you'd eventually hit a point in time where they would be in each other's way of getting past. It could be near the center or near either end depending on how you moved the pieces and how far each time but they WILL get in the way before you can have them at opposite ends within 12 hours. This is the point that the monk passes his ghost track. You're getting side tracked trying to map out each beads exact location on the string at what time and distance from respective start points but it doesn't matter given the circumstances. If each bead will be at the other end in 12 hours they will collide sometime between 1 second after go and 11hrs 59seconds or so. 

Quoted: Ok I also see what your saying, but we are using my chart based on time he did move. View Quote So he did move. OK. So say at 1:50 Monk 1 is at 2.9. Monk 2 is at 3.0. At 1:55 Monk 1 is resting. Monk 2 has moved to 2.95. They both start walking again. For Monk 1 to pass Monk 2's position, which is only decreasing, he has to occupy the same space and time at least once. In other words, if Monk 2 is going from 3.1 to 2.8 and Monk 1 is going from 2.8 to 3.1, they cannot get there unless they occupy the same precise height. They will do so by definition at the same time of day. I think you are getting confused on the time taken, which is different. It might take one 4 hours and one 2 hours, but they *will* get to the same place and time. 

This time traveling monk is gonna fuck up space time occupying the same time and place as himself.


Quoted: Its interesting that everyone advocating for 'yes, they do cross paths' uses a chart/graph with opposite lines of charting (x pattern). Of course an x pattern crosses somewhere... ...but the monk is walking the same path in different directions (parallel of sorts path), which means that his timesrelativelocation can definitely never cross between start n finish... View Quote That's why I go with the shared line as the path. The end points just drift toward the opposite ends over the course of 12 hours at random rates and distances until they cross. That's the common time and place. 

Draw two lines, both of which originate/finish at the same time/location. You can choose to draw them all squiggly so they do touch atleast once, or you can choose to draw them so they never touch except at the ends...
...its the same path in different directions, not differing paths that cross each other 

Quoted: Its interesting that everyone advocating for 'yes, they do cross paths' uses a chart/graph with opposite lines of charting (x pattern). Of course an x pattern crosses somewhere... ...but the monk is walking the same path in different directions (parallel of sorts path), which means that his timesrelativelocation can definitely never cross between start n finish... View Quote Xpattern is accurate, one can visualize this with the top of the chart being the summit and the bottom being the base. It is a certainty that the monk will occupy a space at the same time of the day regardless of speed, as long as start and finish are constrained within the specified period. ETA they're not parallel paths, they're literally opposites. 

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