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Sanya12364 Posts
On December 12 2012 10:30 Iranon wrote:Show nested quote +On December 12 2012 09:36 TanGeng wrote:
Another challenge:
An silk road merchantman needs to buy supplies for his trade caravan and has a circular golden necklace with 79 links. The merchant isn't sure how much he will be spending and needs to be as flexible as possible in payment. He wants to be able to pay with any number of links between 1 and 79. Before heading to the store, the merchant can get his companion to cut some links in the gold chain, but will have to give up a foot of silk for each cut. How many feets of silk does the merchant does the merchant have to give to his friend? Not sure if this is optimal, but eleven is sufficient. + Show Spoiler +Using payments of 1, 2, 4, 8, 16, up to 2^n, you can make any number up to 2^(n+1)-1. So to get up to 79, the merchant can cut the chain up into pieces of length (1, 2, 4, 8, 16, 32), (1, 2, 4, 8), and 1 link left over. That's 79 total, cutting the circular chain into 11 pieces, so the optimal solution is at most 11.
Any price up to 63 can be covered with some combination of the first set of pieces listed above. Any price between 64 and 78 can be covered by the entire first set plus some combination of the second set. And 79 is obviously all 11 pieces.
It's much less than that:
+ Show Spoiler [hint] +
Here's what happens when the friend makes the first cut.
The link is cut, and the merchant pull the link out of the chain. Basically, the link can be disconnected from the chain on both sides. It leaves behind an orphan link of one length and a straight chain of 78 in length.
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On December 12 2012 10:31 TanGeng wrote:Show nested quote +On December 12 2012 10:18 Iranon wrote:
Question: Are there infinitely many "impossible" areas, or will we eventually reach a point where every total area from there on can be covered exactly by drawing a bunch of non-overlapping squares? If the latter, find this point.
+ Show Spoiler +
23
use modulo 4(smallest 2x2 square). 3x3 is equivalent to 1. The biggest "impossible" is therefore the one that requires 3 3x3 squares to complete, and that happens at 27.
(Easier) What if you're allowed to let the squares overlap? So for example, 7 is now possible if you take two 2x2s and have them intersect at a corner... + Show Spoiler +
6
use modulo 3 instead of modulo 4 - and the 2x2 can be used as 3 (equivalent to 0) or 4 (equivalent to 1).
modulo 1 is covered at 4 (one 2x2)
module 2 is covered at 8 (two 2x2)
modulo 3 is covered at 9 (one 3x3)
last one to not be covered is at 6.
+ Show Spoiler +Nice! The last part should be five, not six, though (you can get area six by overlapping two 2x2s along an edge).
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On December 12 2012 10:18 Iranon wrote:Here's one I just made up. I was hunting around for interesting number-theoretic coincidences, so I'm not sure this will actually make a good puzzle, but I think it's neat... Also I'll be pretty impressed if people come up with a solution, since unlike most of the rest of this thread, I doubt you can find a solution via 30 seconds of Google-fu. Take a piece of graph paper, color in a bunch of non-overlapping squares of varying sizes, and keep track of the total area (number of grid sections) you just covered. For example, you might color in a 2x2 square, three 5x5 squares, a 6x6 square, and a 10x10 square, for a total area of 4+3*25+36+100=215. If you're not allowed to use squares of size 1, then clearly some total areas will be impossible to achieve. For example... + Show Spoiler +Total area 1,2,3: obviously impossible
Total area 4: one 2x2
Total area 5,6,7: also obviously impossible
Total area 8: two 2x2s
Total area 9: one 3x3
Total area 10,11: impossible, slightly less obvious
Total area 12: three 2x2s
Total area 13: one 2x2 and one 3x3
Total area 14: impossible, slightly still less obvious
...and so on...
If I give you a specified total volume to try and cover, like 797, it's pretty difficult to decide whether or not this is possible without writing a program to do it for you. It turns out to be possible (two 2x2s, three 3x3s, three 4x4s, four 6x6s, three 7x7s, a 9x9, a 10x10, and two 11x11s do the trick). Question: Are there infinitely many "impossible" areas, or will we eventually reach a point where every total area from there on can be covered exactly by drawing a bunch of non-overlapping squares? If the latter, find this point. Reworded in terms of pure numbers: find the largest number that cannot be represented as a sum of squares other than 1, or explain why no such number exists. Once you understand things the right way, you can solve this by hand pretty quickly. (Note: there's a ton of information out there about representing numbers as a sum of squares, but excluding the square 1 makes this somewhat tricky. Obviously everything can be represented as a sum of squares if you allow using 1, so the sum-of-squares problem typically includes 1 but restricts the number of squares you're allowed to use. I'm excluding 1 but letting you use as many squares as you want.)Bonus 1: (Easier) What if you're allowed to let the squares overlap? So for example, 7 is now possible if you take two 2x2s and have them intersect at a corner... Bonus 2: (Harder) How does the answer change if instead of using two-dimensional squares and looking for the last "impossible" area, use three-dimensional cubes and look for the last "impossible" volume?
+ Show Spoiler +Nope, there's a very small finite number of impossible areas.
The key to the problem is figuring out that you have an object with an area of 4, so if you ever encounter 4 consecutive numbers you can reach, you will have every possible area. To get 1 above the last of the numbers, add 4 to the first of the series. To get 2 above it, add 4 to the second of the series, and so forth for 3 and 4. And then you can do the same process to get the next 4 afterward. (Hope that was clear, since I don't feel like writing out the PMI proof right now).
The first place with four consecutive reachable areas I encountered was: 24,25,26,27.
edit: Well, that was a while after the answer was posted. I thought I'd refreshed more recently...
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On December 12 2012 10:36 TanGeng wrote:Show nested quote +On December 12 2012 10:30 Iranon wrote:On December 12 2012 09:36 TanGeng wrote:
Another challenge:
An silk road merchantman needs to buy supplies for his trade caravan and has a circular golden necklace with 79 links. The merchant isn't sure how much he will be spending and needs to be as flexible as possible in payment. He wants to be able to pay with any number of links between 1 and 79. Before heading to the store, the merchant can get his companion to cut some links in the gold chain, but will have to give up a foot of silk for each cut. How many feets of silk does the merchant does the merchant have to give to his friend? Not sure if this is optimal, but eleven is sufficient. + Show Spoiler +Using payments of 1, 2, 4, 8, 16, up to 2^n, you can make any number up to 2^(n+1)-1. So to get up to 79, the merchant can cut the chain up into pieces of length (1, 2, 4, 8, 16, 32), (1, 2, 4, 8), and 1 link left over. That's 79 total, cutting the circular chain into 11 pieces, so the optimal solution is at most 11.
Any price up to 63 can be covered with some combination of the first set of pieces listed above. Any price between 64 and 78 can be covered by the entire first set plus some combination of the second set. And 79 is obviously all 11 pieces.
It's much less than that: + Show Spoiler [hint] +
Here's what happens when the friend makes the first cut.
The link is cut, and the merchant pull the link out of the chain. Basically, the link can be disconnected from the chain on both sides. It leaves behind an orphan link of one length and a straight chain of 78 in length.
+ Show Spoiler +Just out of curiosity, the answer is 7, right?
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On December 12 2012 10:49 XXGeneration wrote:Show nested quote +On December 12 2012 10:36 TanGeng wrote:On December 12 2012 10:30 Iranon wrote:On December 12 2012 09:36 TanGeng wrote:
Another challenge:
An silk road merchantman needs to buy supplies for his trade caravan and has a circular golden necklace with 79 links. The merchant isn't sure how much he will be spending and needs to be as flexible as possible in payment. He wants to be able to pay with any number of links between 1 and 79. Before heading to the store, the merchant can get his companion to cut some links in the gold chain, but will have to give up a foot of silk for each cut. How many feets of silk does the merchant does the merchant have to give to his friend? Not sure if this is optimal, but eleven is sufficient. + Show Spoiler +Using payments of 1, 2, 4, 8, 16, up to 2^n, you can make any number up to 2^(n+1)-1. So to get up to 79, the merchant can cut the chain up into pieces of length (1, 2, 4, 8, 16, 32), (1, 2, 4, 8), and 1 link left over. That's 79 total, cutting the circular chain into 11 pieces, so the optimal solution is at most 11.
Any price up to 63 can be covered with some combination of the first set of pieces listed above. Any price between 64 and 78 can be covered by the entire first set plus some combination of the second set. And 79 is obviously all 11 pieces.
It's much less than that: + Show Spoiler [hint] +
Here's what happens when the friend makes the first cut.
The link is cut, and the merchant pull the link out of the chain. Basically, the link can be disconnected from the chain on both sides. It leaves behind an orphan link of one length and a straight chain of 78 in length.
+ Show Spoiler +Just out of curiosity, the answer is 7, right?
+ Show Spoiler +I got seven as well, the number of links on left in each piece were 1, 2, 3, 5, 8, 13, 21, 26. Is this the same as what you got?
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On December 11 2012 23:58 Fen2 wrote:Show nested quote +On December 11 2012 20:49 Neino wrote:On December 11 2012 13:06 frogrubdown wrote:On December 11 2012 12:49 ziggurat wrote:
I saw this many years ago, hopefully you find it interesting....
"Last night I went to a dinner party with my wife and four other couples (total of 10 people present). As people arrived, handshakes were exchanged. However, not everyone shook hands. Some did and some didn't. No one shook hands with his or her spouse, and no one shook hands with him or her self. After the party I asked all the others how many hands they shook, and to my surprise each person told me a different number.
The puzzle to solve is: how many hands did my wife shake?" Got it. + Show Spoiler +4. There are 9 possible numbers of hands shaken (0-8) for 9 possible shakers (10 minus you). The person that shakes hands with 8 must be married to the person that shakes hands with 0 because everyone outside of this couple in the group is shaken at least once (by the 8-shaker). Similarly, the 7-shaker must be married to the 1-shaker, since, from the prior sentence, we know that the 1-shaker was shaken only by the 8-shaker, and the 7-shaker must shake hands with everyone else except the 0-shaker and 1-shaker to get to 7. Similar reasoning applies to 6-shaker/2-shaker and 5-shaker/3-shaker. Only possibility left is 4. edit: Not sure how clear the above reasoning was. Maybe this will be better. + Show Spoiler +Every shaker position in {0,...,8} is occupied. The 8-shaker must shake hands with everyone not in spouse-group to reach 8. So everyone outside it has at least one, meaning the spouse is the only candidate for 0-shaker. The 7-shaker must shake hands with everyone not in spouse-group or 0-shaker, meaning everyone not in either of these spouse groups is at least a 2-shaker. So the 7-shaker spouse is the only candidate for 1-shaker. The 6-shaker must shake hands with everyone not in spouse-group or 0-shaker or 1-shaker (we already know the 1 person that shakes hands with this person), meaning everyone not in the first 3 spouse-groups shakes hands with at least 3 people. So the 6-shaker spouse is the only candidate for 2-shaker... that seems less clear if anything. I feel like I'm missing something here, but I just woke up so I got an excuse. How do you know the person telling the riddle isn't the person who shook hands with 8 people, and his wife shook hands with 0? I just missed how you determined which couple the story teller is + Show Spoiler +Due to the line that says he asked and everyone gave a different response.
The groups of couples go
8 0
7 1
6 2
5 3
4 4
As you can see, there are two people who shook 4 peoples hands. As the question asker said that everyone gave different answers, the duplicate must be between him and his wife as he didnt ask himself.
Ah, of course! Cheers 
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Sanya12364 Posts
On December 12 2012 10:49 XXGeneration wrote:Show nested quote +On December 12 2012 09:36 TanGeng wrote:
Another challenge:
An silk road merchantman needs to buy supplies for his trade caravan and has a circular golden necklace with 79 links. The merchant isn't sure how much he will be spending and needs to be as flexible as possible in payment. He wants to be able to pay with any number of links between 1 and 79. Before heading to the store, the merchant can get his companion to cut some links in the gold chain, but will have to give up a foot of silk for each cut. How many feets of silk does the merchant does the merchant have to give to his friend? + Show Spoiler +Just out of curiosity, the answer is 7, right?
You can do better.
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On December 12 2012 12:12 TanGeng wrote:Show nested quote +On December 12 2012 10:49 XXGeneration wrote:On December 12 2012 09:36 TanGeng wrote:
Another challenge:
An silk road merchantman needs to buy supplies for his trade caravan and has a circular golden necklace with 79 links. The merchant isn't sure how much he will be spending and needs to be as flexible as possible in payment. He wants to be able to pay with any number of links between 1 and 79. Before heading to the store, the merchant can get his companion to cut some links in the gold chain, but will have to give up a foot of silk for each cut. How many feets of silk does the merchant does the merchant have to give to his friend? + Show Spoiler +Just out of curiosity, the answer is 7, right? You can do better.
i take it it's unacceptable to just line up everything such that 1 cut will cut through it all?
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Sanya12364 Posts
On December 12 2012 13:00 thirtyapm wrote:Show nested quote +On December 12 2012 12:12 TanGeng wrote:On December 12 2012 10:49 XXGeneration wrote:On December 12 2012 09:36 TanGeng wrote:
Another challenge:
An silk road merchantman needs to buy supplies for his trade caravan and has a circular golden necklace with 79 links. The merchant isn't sure how much he will be spending and needs to be as flexible as possible in payment. He wants to be able to pay with any number of links between 1 and 79. Before heading to the store, the merchant can get his companion to cut some links in the gold chain, but will have to give up a foot of silk for each cut. How many feets of silk does the merchant does the merchant have to give to his friend? + Show Spoiler +Just out of curiosity, the answer is 7, right? You can do better. i take it it's unacceptable to just line up everything such that 1 cut will cut through it all?
Ha! Will cost 79 feet of silk. The cost by number of cut links and not number of cuts.
Another one: possibly very simple.
A dark priest has a chalice that holds slightly more than 1/2 liter of fluid. The priest is looking to make a sacrificial offering and must pour exactly 1/3 liter of blood into the chalice.
The priest has 10cm x 10cm x 10cm cubic container with an open top. If you know your metric system, the volume of the cube is 1 liter. How does the priest pour out the 1/3 liter of blood to make the offering?
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On December 12 2012 01:55 AmericanUmlaut wrote:Show nested quote +On December 11 2012 22:33 gameguard wrote:On December 11 2012 21:39 eluv wrote:On November 30 2012 00:41 aseq wrote:
Hmm, them having to answer at the same time makes it rather impossible, it seems.
There is a variation to this problem where the gnomes can see all the hats except their own (no front or back watching). They're told to form a line, but have all the white and hats next to each other and all the red hats next to each other. They can't say anything at all (but have a strategy). How do they do that? + Show Spoiler +
Line up "at-random", just so you have some order. Then first in line starts a new line. Each person in the old line then enters the new line as follows. If the new line consists entirely of 1 color, enter it at one end. If the new line has both color hats already, enter at the "split" between the two.
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten? dunno if this is right lol. + Show Spoiler +
if you swim in a zigzag line away from the wolf's starting position, wouldnt the wolf get stuck oscillating back and fourth across his initial position?
Once you start moving, the wolf will start moving toward the edge closest to you. If you zig toward any point in the circle ahead of the wolf's position, he'll just keep running that direction. If you zig toward any point in the circle behind his position, he'll turn around when you get closer to that point than to the point you initially swam toward. Every zig and zag will end with the wolf closer to your current position, unless you swim back toward the center, in which case you've gained nothing.
i dunno... im not doing any mathematical calculations but just thinking of a hypothetical situation. Lets say you are at the center and the wolf is at the north shore relative to you (0 degree). You make a small zig southeasterly. This will make the wolf run clockwise since the closest shore to you is somewhere in the 90~180 degree quadrant. Now you make a zag southwesterly. If you think about it, the wolf will only have moved a tiny bit clockwise if your zigzags are small. When you cross over the midline of the circle, suddenly the closest shore ends up at the 180~270 degree quadrant. Assuming the wolf only ran a small distance from its starting point, he would have to double back and run the other way since that is the shortest path towards the closest shore. You keep doing this until you pass the threshold.
My deductions might be way off though. Does this make sense to anyone else? haha
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You start off like this, moving some distance (1) from the center of the lake. It doesn't matter how far. The wolf moves four times as far along the edge toward the bank you're moving toward.
![[image loading]](http://i.imgur.com/dqfqH.png)
Now you're here. The green point is the spot the wolf wants to move toward.
![[image loading]](http://i.imgur.com/gEvOC.png)
Now you zig away.
![[image loading]](http://i.imgur.com/Eng9q.png)
But the dot won't move past the spot 180 degrees across from the wolf and get him to turn around.
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On December 12 2012 12:12 TanGeng wrote:Show nested quote +On December 12 2012 10:49 XXGeneration wrote:On December 12 2012 09:36 TanGeng wrote:
Another challenge:
An silk road merchantman needs to buy supplies for his trade caravan and has a circular golden necklace with 79 links. The merchant isn't sure how much he will be spending and needs to be as flexible as possible in payment. He wants to be able to pay with any number of links between 1 and 79. Before heading to the store, the merchant can get his companion to cut some links in the gold chain, but will have to give up a foot of silk for each cut. How many feets of silk does the merchant does the merchant have to give to his friend? + Show Spoiler +Just out of curiosity, the answer is 7, right? You can do better.
+ Show Spoiler +6.
Cut it into segments of 1, 2, 4, 8, 16, 32 and 47. You can then pay any number between 1 and 79.
I am sure most people here thought of binary, but the little catch is that you only need to be able to make any number between 1 and 40 with your base 2 segments, as you can use the remaining segment to work backwards, if you need to pay more than 63.
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On December 12 2012 09:57 BajaBlood wrote:Show nested quote +On December 12 2012 09:12 TanGeng wrote:On December 11 2012 21:39 eluv wrote:
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten? There is a harder challenge: The wolf runs 4.5 times faster than you can swim with everything else being the same. How do you reach the shore? Yikes, the harder version is too much for me, but I've got an idea for the 4x challenge + Show Spoiler +
If you imagine the pond has radius 'x', then the wolf has to run pi*x units to get to the opposite side. If you are swimming in a circle of radius x/4 or less from the center of the pond, then for you the distance to travel from one side to the other is less than 25% of the amount the wolf has to travel, so you can change angles faster than it.
So therefore if you swim in a circle at a radius a little less than x/4 for long enough, you'll end up across from the wolf. At this point, it's a little over 3/4*x for you to get to the center, whereas the wolf will need to travel pi*x. Even if he is going 4 times faster than you, he'll only make it 3*x of the way to where you are landing, so he won't make it in time.
Now if he's going 4.5 times your speed, this won't work. What a conundrum...
Edit: My instinct is that you would do something similar to what I posted to start, then spiral towards the pond edge to escape, but I have no idea how to show whether that would be better or how much better it would be.
+ Show Spoiler +
Swim 1mm north, stop and wait for the wolf to reach the north point of the lake. Swim directly south.
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On December 13 2012 01:35 hzflank wrote:Show nested quote +On December 12 2012 09:57 BajaBlood wrote:On December 12 2012 09:12 TanGeng wrote:On December 11 2012 21:39 eluv wrote:
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten? There is a harder challenge: The wolf runs 4.5 times faster than you can swim with everything else being the same. How do you reach the shore? Yikes, the harder version is too much for me, but I've got an idea for the 4x challenge + Show Spoiler +
If you imagine the pond has radius 'x', then the wolf has to run pi*x units to get to the opposite side. If you are swimming in a circle of radius x/4 or less from the center of the pond, then for you the distance to travel from one side to the other is less than 25% of the amount the wolf has to travel, so you can change angles faster than it.
So therefore if you swim in a circle at a radius a little less than x/4 for long enough, you'll end up across from the wolf. At this point, it's a little over 3/4*x for you to get to the center, whereas the wolf will need to travel pi*x. Even if he is going 4 times faster than you, he'll only make it 3*x of the way to where you are landing, so he won't make it in time.
Now if he's going 4.5 times your speed, this won't work. What a conundrum...
Edit: My instinct is that you would do something similar to what I posted to start, then spiral towards the pond edge to escape, but I have no idea how to show whether that would be better or how much better it would be.
+ Show Spoiler +
Swim 1mm north, stop and wait for the wolf to reach the north point of the lake. Swim directly south.
+ Show Spoiler +That obviously doesn't work. The wolf then needs to cover pi distance, the girl 1. And the wolf is 4x faster. I get the solution up to 4.14, I can't get the solution for 4.59 though.
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On December 13 2012 01:42 aseq wrote:Show nested quote +On December 13 2012 01:35 hzflank wrote:On December 12 2012 09:57 BajaBlood wrote:On December 12 2012 09:12 TanGeng wrote:On December 11 2012 21:39 eluv wrote:
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten? There is a harder challenge: The wolf runs 4.5 times faster than you can swim with everything else being the same. How do you reach the shore? Yikes, the harder version is too much for me, but I've got an idea for the 4x challenge + Show Spoiler +
If you imagine the pond has radius 'x', then the wolf has to run pi*x units to get to the opposite side. If you are swimming in a circle of radius x/4 or less from the center of the pond, then for you the distance to travel from one side to the other is less than 25% of the amount the wolf has to travel, so you can change angles faster than it.
So therefore if you swim in a circle at a radius a little less than x/4 for long enough, you'll end up across from the wolf. At this point, it's a little over 3/4*x for you to get to the center, whereas the wolf will need to travel pi*x. Even if he is going 4 times faster than you, he'll only make it 3*x of the way to where you are landing, so he won't make it in time.
Now if he's going 4.5 times your speed, this won't work. What a conundrum...
Edit: My instinct is that you would do something similar to what I posted to start, then spiral towards the pond edge to escape, but I have no idea how to show whether that would be better or how much better it would be.
+ Show Spoiler +
Swim 1mm north, stop and wait for the wolf to reach the north point of the lake. Swim directly south.
+ Show Spoiler +That obviously doesn't work. The wolf then needs to cover pi distance, the girl 1. And the wolf is 4x faster. I get the solution up to 4.14, I can't get the solution for 4.59 though.
Sorry, was thinking 2 pi, have not slept in too long :p I can see the theory, but I don't think I have the mental capacity to calculate it.
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On December 12 2012 13:31 TanGeng wrote:
Another one: possibly very simple.
A dark priest has a chalice that holds slightly more than 1/2 liter of fluid. The priest is looking to make a sacrificial offering and must pour exactly 1/3 liter of blood into the chalice.
The priest has 10cm x 10cm x 10cm cubic container with an open top. If you know your metric system, the volume of the cube is 1 liter. How does the priest pour out the 1/3 liter of blood to make the offering?
gave this a try, thought i found something interesting and then realized my math sucked. Still, might be someone sees a use for my stuff so will post it here in a spoiler.
+ Show Spoiler +chalice=x; container=c
Fill the container and then, from the container fill the chalise. The container now contains c-x.
empty the chalise, refill it from the container. Since the chalise can hold more than ½ of the container, it wont completely fill, so the difference beteen competely filled (x) and what it is now will be called d. (x-d=current state of chalice)
Refill the container completely and from the container, start filling the chalice again.
Once the chalice is full, the container holds c-(x-d), lets call this z. Now empty the chalise again, and fill it from the container. This means the container holds z-x now, or c-2x+d.
So if we empty the chalice again, we have an empty chalice (obviously) and a container that's filled as c-2x+d.
we know c=1L, x is "slightly" more than ½L and d is the difference between them.
Should we fill the chalice from the container, we're not gonna manage to competely fill the chalice (just insert .51 as x and calculate, it wont work, if it would fit the outcome would be x again). So we have an almost full chalice, call it a, and an empty container. Refill the container, fill the chalise with it, the container now holds c-a. a was x-(what was left in c; c-2x+d) => the container holds c-(x-(c-2x+d))=c-(x-c+2x-d)=c-x+c-2x+d=2c-3x+d.
at this point i gave up cuz it got pretty complicated to do it on my screen without writing it down, and i had no idea if im heading the right direction. Ofc i tried 2c-3x+d=1/3c (working it out is 6c-9x+3d=c, 5c-9x+3d=0, 5L=9x-3d and i didnt see what that would be in an instant so left it there.
if someone thinks he can use (part of) it, feel free, hope i was going the right way so i didnt get a headache for nothing :D
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On December 12 2012 13:31 TanGeng wrote:
Another one: possibly very simple.
A dark priest has a chalice that holds slightly more than 1/2 liter of fluid. The priest is looking to make a sacrificial offering and must pour exactly 1/3 liter of blood into the chalice.
The priest has 10cm x 10cm x 10cm cubic container with an open top. If you know your metric system, the volume of the cube is 1 liter. How does the priest pour out the 1/3 liter of blood to make the offering?
+ Show Spoiler +Using the cubic container, if you look at one of the base corners and the 3 corners adjacent to it, this forms a pyramid shape, which has a volume of 1/6 liter. You can tilt the container onto the base corner and fill it up to the height of the 3 other corners. Repeat twice to get 1/3 liter.
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Sanya12364 Posts
On December 13 2012 03:58 Slithe wrote:Show nested quote +On December 12 2012 13:31 TanGeng wrote:
Another one: possibly very simple.
A dark priest has a chalice that holds slightly more than 1/2 liter of fluid. The priest is looking to make a sacrificial offering and must pour exactly 1/3 liter of blood into the chalice.
The priest has 10cm x 10cm x 10cm cubic container with an open top. If you know your metric system, the volume of the cube is 1 liter. How does the priest pour out the 1/3 liter of blood to make the offering?
+ Show Spoiler +Using the cubic container, if you look at one of the base corners and the 3 corners adjacent to it, this forms a pyramid shape, which has a volume of 1/6 liter. You can tilt the container onto the base corner and fill it up to the height of the 3 other corners. Repeat twice to get 1/3 liter.
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Sanya12364 Posts
On December 12 2012 09:12 TanGeng wrote:Show nested quote +On December 11 2012 21:39 eluv wrote:
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten? There is a harder challenge: The wolf runs 4.5 times faster than you can swim with everything else being the same. How do you reach the shore?
+ Show Spoiler [answer] +It takes a bit of thought to arrive at the starting point. Then after that it is provable that the tangent swim-out of the inner circle is the optimal strategy. ![[image loading]](http://i.imgur.com/iXGLi.png) Plugging in the all the numbers it's possible to show that velocity ratios larger than 4 and as high as 4.6 are possible.
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EPT
