6/(1-5/7) = 21
I'm off for the day so anyone may start a new puzzle
Tricky Puzzles
I've got a puzzle, so may I go next? I've got one question about it though. It's more like a yes/no question puzzle, so how will I go about getting it out?
Skype wrote:[7:23:42 AM] Darx: Click here to give me an internet!
[7:23:57 AM] 'a'a: *clicks here*
Yay! Thankyou!
Skype wrote:[7:23:42 AM] Darx: Click here to give me an internet!
[7:23:57 AM] 'a'a: *clicks here*
Here's an easy one, that can everyone do. 
A man has a swimmingpool with a square shape and on each corner a tree. After several years he wants the pool to be twice as big, but still square shapy. And he definitaly doesn't want to cut the trees. How does he do that?
A man has a swimmingpool with a square shape and on each corner a tree. After several years he wants the pool to be twice as big, but still square shapy. And he definitaly doesn't want to cut the trees. How does he do that?
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Last edited by Marinus on Fri Aug 08, 2008 1:19 pm, edited 1 time in total.
Yes, I will! 
I uploaded a new attachment and that shows with the thin lines hoe the pool is gonna be. Both shapes are supposed to be squares
When you draw 2 squares exactly that way and the big square with the thin line is also the shape of the drawing paper, then you can flip the 4 parts which are outside the thick line to the inside, and then those 4 together are exactly as big as the inside square. So the new square is 2 * as big as the old one, so the length of a side of the new one is square-root 2 (1.41) * as big.
This is one of the cleverest puzzles in Death Gate adventure game, reworded to not to make it too easy while still giving all neccesary information:
You are in dog's body (don't ask, it makes sense in context) and need to bring an antidote, to your shackled, poisoned human self. You look around the room and notice a stand of bottles right in front of a color-patterned curtain. You know this stand has the antidote, and that all the bottles are clear, but only the antidote itself is clear, the wine and the poison bottles are coloured. But: there's one sticky situation.
How will you differentiate the COLOR of the liquids when you're a DOG with a black-and-white vision?
You are in dog's body (don't ask, it makes sense in context) and need to bring an antidote, to your shackled, poisoned human self. You look around the room and notice a stand of bottles right in front of a color-patterned curtain. You know this stand has the antidote, and that all the bottles are clear, but only the antidote itself is clear, the wine and the poison bottles are coloured. But: there's one sticky situation.
How will you differentiate the COLOR of the liquids when you're a DOG with a black-and-white vision?
Rest in peace, Kym. I hardly knew ya.
Rest in peace, Marinus. A bright star, you were ahead of me on my own tracks of thought. I miss you.
Rest in peace, Marinus. A bright star, you were ahead of me on my own tracks of thought. I miss you.
First I think the coloured liqiud is always darker when you look through it then the non-coloured. Beside that the coloured liqiud may have an different breaking index or barking index But that's not your intended solution
When you look through the red wine to the green parts of the curtain it will seem black or at least much darker. The same with the green poison and the red parts of the curtain. Through the non-coloured liqiud there will be no difference.
But that's not your intended solution When you look through the red wine to the green parts of the curtain it will seem black or at least much darker. The same with the green poison and the red parts of the curtain. Through the non-coloured liqiud there will be no difference.
barking index
Yep, I couldn't tell what is impossible. The game required you to watch at the parts of the curtain that are covered by the bottles, and the one that doesn't change the colors of the pattern is the clear one.
Death Gate is actually a tricky puzzle on a tricky puzzle. Getting down a huge pit safely, dealing with your magical double (near the end) and the one where you meet Sang-Drax for the first time, and he threatens to kill-you-off-for-real.
Rest in peace, Kym. I hardly knew ya.
Rest in peace, Marinus. A bright star, you were ahead of me on my own tracks of thought. I miss you.
Rest in peace, Marinus. A bright star, you were ahead of me on my own tracks of thought. I miss you.
Your uncle Peegue calls you on the phone and asks for 40 green spheres, for he's gonna make a party in the city hall of Wondertown. By coincidence you find a box where exactly 40 sphere's fit in, like the picture. At the moment you're ready to go, he calls again and asks for also one red sphere. But you have no other boxes. How do you solve this problem? 
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Okay, so here's another math puzzle, probably quite easy, too:
What are the four numbers of which you can get any number between 1 and 40 only by adding or substracting? Every number may only be used once to get a result, so 1+1=2 is for example not allowed. However, you don't have to use every number for each result (which also means for example 16=16 is a valid solution).
What are the four numbers of which you can get any number between 1 and 40 only by adding or substracting? Every number may only be used once to get a result, so 1+1=2 is for example not allowed. However, you don't have to use every number for each result (which also means for example 16=16 is a valid solution).
I had really no idea how to start 
This is about a little game, in Dutch called: butter, cheese and eggs, don't ask me why, 
and, if I'm right in English: tic tac toe. It's a 3*3 square and it's played by two players who write one by one a X or O in one of the squares. One of them uses X and the other O. A player has won if he/she has three X or O on one line, horizontal, vertical or diagonal.
Just in the situation of this riddle there are two rules:
1 If a player has a possibillity to get three on a line he/she must do that.
2 If a player has a possibillity to prevent the other player to get three on a line, he/she must do that.
Of course rule 1 is before 2.
In this situation each player has done three turns, this way:
O O .
O X .
X X .
It is quite obvious that the player who's turn it is will win. The question is, who's turn it is, and why.
and, if I'm right in English: tic tac toe. It's a 3*3 square and it's played by two players who write one by one a X or O in one of the squares. One of them uses X and the other O. A player has won if he/she has three X or O on one line, horizontal, vertical or diagonal. Just in the situation of this riddle there are two rules:
1 If a player has a possibillity to get three on a line he/she must do that.
2 If a player has a possibillity to prevent the other player to get three on a line, he/she must do that.
Of course rule 1 is before 2.
In this situation each player has done three turns, this way:
O O .
O X .
X X .
It is quite obvious that the player who's turn it is will win. The question is, who's turn it is, and why.
If I'm right, then in your riddle the O player must begin first, which is against the rules. Ignoring that meddling bit, it's O's turn, since the only way a three-in-the-line can occur is if a player manages to make a double threat -- since you can only close off one at a time...
Rest in peace, Kym. I hardly knew ya.
Rest in peace, Marinus. A bright star, you were ahead of me on my own tracks of thought. I miss you.
Rest in peace, Marinus. A bright star, you were ahead of me on my own tracks of thought. I miss you.
It's O's turn.
It's pretty logical to me, but still I find it hard to explain. I'll try anyway:
Both players threaten (I don't know if that's the correct word, but I'll use it in the explanation) to complete a line. Because of that, it's obious that the last move had to prevent a possibility; otherwise it wouldn't have been possible for one player to make a threat while another one already exists.
And after figuring out that, you only have to notice that the last move of O couldn't have prevented any line from being completed. The only line where a O prevents something like that is the second column. However, that would mean that the previous move of X would have either been R2C2 or R3C2. If that was the case, however, either of them would have already existed. Both would lead to a threat combined with R3C1, either on the third column or on one of the diagonals which O would have had to prevent.
Altogether, this means that X's last move was R3C1, where he parried the threat on the 1 column, but wasn't able to parry the one on the first row.
EDIT: D'oh, posted at the same time again! Muzo's explanation is certainly much more compact...
It's pretty logical to me, but still I find it hard to explain. I'll try anyway:
Both players threaten (I don't know if that's the correct word, but I'll use it in the explanation) to complete a line. Because of that, it's obious that the last move had to prevent a possibility; otherwise it wouldn't have been possible for one player to make a threat while another one already exists.
And after figuring out that, you only have to notice that the last move of O couldn't have prevented any line from being completed. The only line where a O prevents something like that is the second column. However, that would mean that the previous move of X would have either been R2C2 or R3C2. If that was the case, however, either of them would have already existed. Both would lead to a threat combined with R3C1, either on the third column or on one of the diagonals which O would have had to prevent.
Altogether, this means that X's last move was R3C1, where he parried the threat on the 1 column, but wasn't able to parry the one on the first row.
EDIT: D'oh, posted at the same time again! Muzo's explanation is certainly much more compact...