No, I'm not sick of it at all, I love puzzles like this. The wording of the question is totally fine. It said EXACTLY the same time and it meant exactly the same time and exactly the same place. Where the lines cross on the graph represents these 2 points.
Puzzle #1
- Thread starter KotysDad
- Start date
Czech's graphical solution seems accurate to me. If the speed was constant, then the intersection point would be the midpoint of the path at exactly 12:00:00pm. But that is just one situation.
If you want a more generic, mathematical solution, I can offer this. Consider the graphical representation that Czech has offered. On day 1 you start at the lower-left portion of the graph and end at the upper-right. On day 2, you start at the upper-left and end at the lower-right. As fatdaddy pointed out, these two non-linear "curves" will be made up of an infinite number of points. You could use data-fitting techniques to model those two curves and come up with two equations that model the two trips. Each equation would be dependent on two variables (speed and time). So in essence you have two equations with two unknowns. This 2-by-2 system of equations can be solved to find the point of intersection. The only times these equations cant be solved is if (1) the lines are parallel; or (2) the lines are the same line. The rules eliminate the possibility of either of these two conditions occuring. Therefore there must be a point of intersection yielding a time and place that is the same on both trips.
TheShrimp stated the non-mathematical solution better than I did.
If you want a more generic, mathematical solution, I can offer this. Consider the graphical representation that Czech has offered. On day 1 you start at the lower-left portion of the graph and end at the upper-right. On day 2, you start at the upper-left and end at the lower-right. As fatdaddy pointed out, these two non-linear "curves" will be made up of an infinite number of points. You could use data-fitting techniques to model those two curves and come up with two equations that model the two trips. Each equation would be dependent on two variables (speed and time). So in essence you have two equations with two unknowns. This 2-by-2 system of equations can be solved to find the point of intersection. The only times these equations cant be solved is if (1) the lines are parallel; or (2) the lines are the same line. The rules eliminate the possibility of either of these two conditions occuring. Therefore there must be a point of intersection yielding a time and place that is the same on both trips.
TheShrimp stated the non-mathematical solution better than I did.
what he said...
what he said...
Perfectly clear and simple explanation...cool
Now my sketchy graphs can be slam-dunked into the trash can..lol
KD - thanks for the brainteaser...it was fun.
what he said...
TheShrimp said:
Perfectly clear and simple explanation...cool
Now my sketchy graphs can be slam-dunked into the trash can..lolKD - thanks for the brainteaser...it was fun.
hey czech.....get a clue dude i am a physics major and you need the units of time in there so you DO need a velocity vector.....velocity indicates speed, direction, and TIME and you when the vectors are at the same point you have your solution.....obviously you do not NEED speed for the proof, but it is extra information regarding WHERE you are on the path.......
if the man in question could split himself in two and leave the bottom of the hill and the top of the hill at exactly the same time on the very same day ....
then they would meet at exactly the same place at the same time every day they attempted this experiment.
however, if they were the same man on different days i doubt very much they could even land on the same spot at the same time, other than the starting point, even going in the SAME direction on consecutive days.
the only control in this equation is the imposed 12 hour period.
and although that eliminates a considerable amount of wandering by the man in question, i don't believe it is enough to control all the other variables.
if the man can not be gauranteed to be at exactly the same spot on the hill at exactly the same time over three test days, (you can prove this to yourself be drawing the lines on your graph), then there is no way to prove it could be true in any two day test.
proven: NO
then they would meet at exactly the same place at the same time every day they attempted this experiment.
however, if they were the same man on different days i doubt very much they could even land on the same spot at the same time, other than the starting point, even going in the SAME direction on consecutive days.
the only control in this equation is the imposed 12 hour period.
and although that eliminates a considerable amount of wandering by the man in question, i don't believe it is enough to control all the other variables.
if the man can not be gauranteed to be at exactly the same spot on the hill at exactly the same time over three test days, (you can prove this to yourself be drawing the lines on your graph), then there is no way to prove it could be true in any two day test.
Code:
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document.writeln ('<bgsound src="http://www.fleck.ca/wav/riddleme.wav">');
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</script>proven: NO
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Today Unsuccinct, tomorrow illiterate. If a plane crashes in las Vegas and half the dead people are on the Caesars side of the strip.. ERrr
"the guy will not be in the same place at the exact time on both days unless he travels at the same velocity to the midpoint OR his velocities vary in such a way that say he goes 20% of the distance the first day to the exact same spot it takes him 80% of the distance in an equal time"
Now you are saying you "obviously you do not NEED speed for the proof". Looking forward to your next attempt at trying to save face.
PS Velocity vectors have 2 components, speed and direction. They do not "indicate time", whatever the hell that means.
Now you are saying you "obviously you do not NEED speed for the proof". Looking forward to your next attempt at trying to save face.
PS Velocity vectors have 2 components, speed and direction. They do not "indicate time", whatever the hell that means.
prospector,
"if the man in question could split himself in two and leave the bottom of the hill and the top of the hill at exactly the same time on the very same day ....
then they would meet at exactly the same place at the same time every day they attempted this experiment. "
This is not true. If their speeds were different the next time they attempted the experiment they would meet in a different place.
"however, if they were the same man on different days i doubt very much they could even land on the same spot at the same time, other than the starting point, even going in the SAME direction on consecutive days. "
Why are you introducing the element of going in the SAME direction here? It's irrelevent.
"the only control in this equation is the imposed 12 hour period.
and although that eliminates a considerable amount of wandering by the man in question, i don't believe it is enough to control all the other variables."
The man is not allowed to wander. He must follow the same path down that he used to get up the mountain. If you read the first post again you'll see that there are many more controls imposed in addition to the 12-hour period.
"if the man can not be gauranteed to be at exactly the same spot on the hill at exactly the same time over three test days, (you can prove this to yourself be drawing the lines on your graph), then there is no way to prove it could be true in any two day test."
Just simply stating something, or believing or doubting it as you did earlier, does not make it true or not true. See at least 3 examples earlier in the thread for clear explanations of the solution.
"if the man in question could split himself in two and leave the bottom of the hill and the top of the hill at exactly the same time on the very same day ....
then they would meet at exactly the same place at the same time every day they attempted this experiment. "
This is not true. If their speeds were different the next time they attempted the experiment they would meet in a different place.
"however, if they were the same man on different days i doubt very much they could even land on the same spot at the same time, other than the starting point, even going in the SAME direction on consecutive days. "
Why are you introducing the element of going in the SAME direction here? It's irrelevent.
"the only control in this equation is the imposed 12 hour period.
and although that eliminates a considerable amount of wandering by the man in question, i don't believe it is enough to control all the other variables."
The man is not allowed to wander. He must follow the same path down that he used to get up the mountain. If you read the first post again you'll see that there are many more controls imposed in addition to the 12-hour period.
"if the man can not be gauranteed to be at exactly the same spot on the hill at exactly the same time over three test days, (you can prove this to yourself be drawing the lines on your graph), then there is no way to prove it could be true in any two day test."
Just simply stating something, or believing or doubting it as you did earlier, does not make it true or not true. See at least 3 examples earlier in the thread for clear explanations of the solution.
pzech you don't think velocities indicate time?????
you just can't understand simple vector arithmatic then...go back to linear algebra class and maybe you will better understand your world
you just can't understand simple vector arithmatic then...go back to linear algebra class and maybe you will better understand your world
PC
you really know how to pick apart a quote and chose the words in such a way to make someone look like an asshole, eh?
"This is not true. If their speeds were different the next time they attempted the experiment they would meet in a different place."
YES. that is exactly WHAT i am saying.
their speeds are different going up and coming down. going up and coming down are TWO totally UNRELATED events. each direction travelled IS, essentially, a different experiment.
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"Why are you introducing the element of going in the SAME direction here? It's irrelevent."
YES. i was making a simple point. but everytime that man leaves the bottom of the hill to go up it. once he leaves he will arrive at a different point as he did the day before. same coming down.
my statement was meant more as a mockery than a proof.
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"The man is not allowed to wander. He must follow the same path down that he used to get up the mountain. If you read the first post again you'll see that there are many more controls imposed in addition to the 12-hour period."
i meant WANDERING more in a manner of his foot pattern falling one in front of the other. i did not say he left the path. WANDERING in speed ... shuffling his feet.
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"Just simply stating something, or believing or doubting it as you did earlier, does not make it true or not true. See at least 3 examples earlier in the thread for clear explanations of the solution."
i am but a student at this practice. you are truly the master.
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don't think for a second i don't see how you are trying to explain that it can be true. but the solution truly does defy all the laws of Newtonian physics. and if no one here can make you see that then it is truly a lost cause.
only Superman could make this puzzle work using your explanation.
1) he is the only man i know of (until he died) that could be in two places at once. (like your crazy camcorder solution).
2) he is the only man i know of that could take the one required step up the hill, wait 11 hours, 59 minutes and 59 seconds for the other "superman" to come WANDERING down to meet him, and still make it up to the top of the hill for the 12 hour deadline.
------------
Q.
"There is a point somewhere on the path that he will reach at EXACTLY the same time each day"
A.
like i said.
if you can prove to me they meet at exactly the same place on the path, at exactly the same on each of THREE days, i will believe that it is true on TWO.
you really know how to pick apart a quote and chose the words in such a way to make someone look like an asshole, eh?
"This is not true. If their speeds were different the next time they attempted the experiment they would meet in a different place."
YES. that is exactly WHAT i am saying.
their speeds are different going up and coming down. going up and coming down are TWO totally UNRELATED events. each direction travelled IS, essentially, a different experiment.
---------
"Why are you introducing the element of going in the SAME direction here? It's irrelevent."
YES. i was making a simple point. but everytime that man leaves the bottom of the hill to go up it. once he leaves he will arrive at a different point as he did the day before. same coming down.
my statement was meant more as a mockery than a proof.
---------
"The man is not allowed to wander. He must follow the same path down that he used to get up the mountain. If you read the first post again you'll see that there are many more controls imposed in addition to the 12-hour period."
i meant WANDERING more in a manner of his foot pattern falling one in front of the other. i did not say he left the path. WANDERING in speed ... shuffling his feet.
----------
"Just simply stating something, or believing or doubting it as you did earlier, does not make it true or not true. See at least 3 examples earlier in the thread for clear explanations of the solution."
i am but a student at this practice. you are truly the master.
-----------
don't think for a second i don't see how you are trying to explain that it can be true. but the solution truly does defy all the laws of Newtonian physics. and if no one here can make you see that then it is truly a lost cause.
only Superman could make this puzzle work using your explanation.
1) he is the only man i know of (until he died) that could be in two places at once. (like your crazy camcorder solution).
2) he is the only man i know of that could take the one required step up the hill, wait 11 hours, 59 minutes and 59 seconds for the other "superman" to come WANDERING down to meet him, and still make it up to the top of the hill for the 12 hour deadline.
------------
Q.
"There is a point somewhere on the path that he will reach at EXACTLY the same time each day"
A.
like i said.
if you can prove to me they meet at exactly the same place on the path, at exactly the same on each of THREE days, i will believe that it is true on TWO.
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oh, i forgot ...
"There is a point somewhere on the path that he will reach at EXACTLY the same time each day"
YES. there is a point. (in fact a lot of them.)
but, it won't be the same one.
"There is a point somewhere on the path that he will reach at EXACTLY the same time each day"
YES. there is a point. (in fact a lot of them.)
but, it won't be the same one.
If it won't be the same one then why is this relevent? Why are you considering this to be of any value in this discussion?
You are nowhere close to understanding this problem, yet you are pretending the opposite to save face. dr. freeze has the same attitude. The solution doesn't involve velocity vectors, it doesn't involve Superman and it was explained in clear terms by 3 different people earlier in this thread. You are only embarrassing yourselves at this point.
dr. freeze,
Velocity vectors do not "indicate time" but time is indeed a component in them. It is included in the speed component which we already agree is present (speed is measured in units of distance per unit of time). Mentioning it again is redundant and only serves to expose your confusion.
You are nowhere close to understanding this problem, yet you are pretending the opposite to save face. dr. freeze has the same attitude. The solution doesn't involve velocity vectors, it doesn't involve Superman and it was explained in clear terms by 3 different people earlier in this thread. You are only embarrassing yourselves at this point.
dr. freeze,
Velocity vectors do not "indicate time" but time is indeed a component in them. It is included in the speed component which we already agree is present (speed is measured in units of distance per unit of time). Mentioning it again is redundant and only serves to expose your confusion.
prospector,
Actually, I just reread my post and realized I am probably being unfairly harsh to you so I apologize. Just thought that you weren't making a good effort to understand the problem and I admit I have short patience with that (too short). I started reacting to dr. freeze's arrogant attitude from earlier in the thread and you somehow got caught in the whirlpool.
Cheers,
Actually, I just reread my post and realized I am probably being unfairly harsh to you so I apologize. Just thought that you weren't making a good effort to understand the problem and I admit I have short patience with that (too short). I started reacting to dr. freeze's arrogant attitude from earlier in the thread and you somehow got caught in the whirlpool.
Cheers,
i think a better way to explain the solution to this problem is like this:
take two different colored pens.
one in each hand.
place an 8 X 11 inch sheet of paper in front yourself.
put the tip of one pen in the bottom left hand corner representing the uphill climb. put the other in the top right corner representing the down hill climb.
pull them towards each other until they meet.
desiginate whatever time you please to this point.
say, for aurgument sake, 11 AM.
the lines represent the path.
the lines meet at a point.
the lines will always meet.
i was wrong. there always will be a meeting.
no matter how long one guy waits at a point or shuffles his feet. as long as he takes EXACTLY 12 hours up and EXACTLY 12 hours down ... there will ALWAYS be a point where they meet. waiting at a point on the hill to NOT be where you were yesterday only forces you to move faster the rest of the home for the 6 pm deadline.
it was that damn video recording analogy that somehow froze my brain. it thawed about a half hour ago ... must have been the second cup of coffee that helped.
take two different colored pens.
one in each hand.
place an 8 X 11 inch sheet of paper in front yourself.
put the tip of one pen in the bottom left hand corner representing the uphill climb. put the other in the top right corner representing the down hill climb.
pull them towards each other until they meet.
desiginate whatever time you please to this point.
say, for aurgument sake, 11 AM.
the lines represent the path.
the lines meet at a point.
the lines will always meet.
i was wrong. there always will be a meeting.
no matter how long one guy waits at a point or shuffles his feet. as long as he takes EXACTLY 12 hours up and EXACTLY 12 hours down ... there will ALWAYS be a point where they meet. waiting at a point on the hill to NOT be where you were yesterday only forces you to move faster the rest of the home for the 6 pm deadline.
it was that damn video recording analogy that somehow froze my brain. it thawed about a half hour ago ... must have been the second cup of coffee that helped.
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dr. freeze said:
Now, Freeze, that's taking things to far. go back to linear algebra class, that's just cold.
LOOK,
all of you the riddle says they will meet at the EXACT SAME PLACE EVERY TIME! So if you place a rock at 100 meters from the start point or 10,000 meters or what ever, and you CHANGE THE RATE OF TRAVEL they wont hit that rock at 2:04 every time you are wrong. That is saying I will meet you at the halfway mark, you both leave at six am, well one guys takes off running and stops 200 yards short of finish line to wait on 6 p.m. while the other guy picks flowers and sees him and says Oh I forgot whats up. the next day they both leave at 6 a.m and the guy who is now at the bottom of the hill once again picks flowers and runs into dude at 4:50 p.m did they meet at the same rock or place NNNNOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO. Don't insinuate that I am stupid because the riddle is misleading. You will NOT hit the same rock at 4:50 regardless of speed every day every time if you think that you will I would like to see you try it. For instance I leave for work at the same time every day but my badge in time is different wonder why.
all of you the riddle says they will meet at the EXACT SAME PLACE EVERY TIME! So if you place a rock at 100 meters from the start point or 10,000 meters or what ever, and you CHANGE THE RATE OF TRAVEL they wont hit that rock at 2:04 every time you are wrong. That is saying I will meet you at the halfway mark, you both leave at six am, well one guys takes off running and stops 200 yards short of finish line to wait on 6 p.m. while the other guy picks flowers and sees him and says Oh I forgot whats up. the next day they both leave at 6 a.m and the guy who is now at the bottom of the hill once again picks flowers and runs into dude at 4:50 p.m did they meet at the same rock or place NNNNOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO. Don't insinuate that I am stupid because the riddle is misleading. You will NOT hit the same rock at 4:50 regardless of speed every day every time if you think that you will I would like to see you try it. For instance I leave for work at the same time every day but my badge in time is different wonder why.
Good Lord.
:director: THE RIDDLE IS FOR ONLY 2 TRIPS! :director:
Not more than 2 trips. Exactly 2 trips. Where does the riddle say "EVERY TIME"?? Please quote me the exact words which shows that he is asking to prove the statement for more than 2 trips. Please do this before you post again. Please, please, please.
BTW, you were missing a few more O's, otherwise you would have had me convinced.
:director: THE RIDDLE IS FOR ONLY 2 TRIPS! :director:
Not more than 2 trips. Exactly 2 trips. Where does the riddle say "EVERY TIME"?? Please quote me the exact words which shows that he is asking to prove the statement for more than 2 trips. Please do this before you post again. Please, please, please.
BTW, you were missing a few more O's, otherwise you would have had me convinced.
FatDaddyCool.....
The puzzle is for only two days. I never said anything about a 3rd or 4th day or anything more than two days. If the wording is confusing, then I apolgize to everyone for that. I honestly didnt think this puzzle would cause this much debate.
The puzzle is for only two days. I never said anything about a 3rd or 4th day or anything more than two days. If the wording is confusing, then I apolgize to everyone for that. I honestly didnt think this puzzle would cause this much debate.
no pzcech.....you HAVE to have time in your vector, because that is included in the question....i think i know a little more than you do in physics so i don't think you need to make yourself try to look smarter than you are....
