one

Still wrong

three

getting closer, I have two clues, do you want them?

I have seen this puzzle before and it's far more different from what I've seen. I think we should make our puzzle on our own. Maybe think a unique concept so that more people can enjoy it.

sure

zero!

69 huehue

Four

Still all wrong

Any number between 1 and 10000

well 2sqrt2 is possible with an X formation. less than that? let's try to prove it.

let's assume the square's side is 1.

the minimum is sqrt(2) since a line connecting the opposite corners is this length and any other connection is larger by the triangle's inequality.

so what we're trying to do is to find/disprove a connection whose length is 2sqrt(2)>L>sqrt(2).

let's see. suppose I use FIVE lines; one is of length 0.5 which is parallel to one of the sides and precisely in the middle. four lines go from its ends to the corners. we then have 0.5+4sqrt(0.5^2+0.25^2)=0.5+sqrt(5)<2sqrt(2).

is that the optimal solution tho?


pedit: 5 then.

possible better solution: form 2 equilateral triangles with 2 opposite sides of the square where the third point in both is inside the square. follow earlier solution. let's see what becomes of it.

A(0,0) B(1,0) C(0,1) D(1,1)

E would then be (sqrt(0.75),0.5) and F would be (1-sqrt(0.75),0.5).

that would give us perimeter of...

P=4*sqrt(0.25+1+0.75-2sqrt(0.75))  + 2sqrt(0.75)-1=
=4sqrt(2-2sqrt(0.75))+2sqrt(0.75)-1=~2.8026

whereas 0.5+sqrt(5)=2.73

so no. I can only assume that the 0.5+sqrt(5) solution is the best; can probably differentiate a proven solution if I get a pen, but I don't, so nvm. FIVE

Shos, instead of length 1/2 make that line in the middle length (1 - 1/sqrt(3)). That is the minimum for your method.

SHOS WINS!

The problem is that I thought of that, but when trying to find the minimum I forgot a 2 in the derivative, so I found that function was continuously growing and that 0 should be the minimal length.

that comes out as (7-sqrt(3))/3=~1.75. neat.

could you make some kind of drawing so that non shos and jellsprout can understand?

this was almost uploaded to the featured content, I think

steiner trees

interesting to see that the conjecture is still open!

eh so it's my turn for a puzzle.

you have 13 coins. one of them is fake and weights either more or less than the others - you don't know.

you have a weighing scale that looks like that and it can tell you which side is heavier between the two sides you put on it.

you have precisely THREE weighs, can you find the fake coin? if so, how?

Since the question is that, YES!

fixed.

Divide the coins in 3 piles: two of 4 coins and one of 5. Compare the two piles of 4 coins.
If the two piles are in equilibrium:
|All the 8 coins from the two piles of 4 are true. From the remaining 5 coins take three and compare them to 3 of the true coins.
|If the two piles are in equilibrium:
||One of the remaining two coins must be false. Take one of them and compare it to a true coin. If they are in equilibrium, the remaining coin is false, otherwise the coin on the weight is false.
|Elseif the two piles are not in equilibrium:
The pile of three coins is either lighter or heavier. For now we assume that the pile is heavier, but you can do the exact same deduction if it is lighter.
Take two of the three unknown coins and compare them.
||If they are in equilibrium:
|||The third coin is false.
||Elseif they are not in equlibrium:
|||The heavier coin is false.

Elseif the two piles are not inequilibrium:
|The 5 coins from the remaining pile are true. Let us call the lighter pile L and the heavier pile H. Take 3 coins from pile L and 2 from pile H and compare them to the 5 true coins.
|If there is equilibrium:
||Take one coin from the remainder of pile H and one from the remainder of pile L and compare them to two true coins. If there is equilibrium, the final coin is the false one. If side with the two unknowns is heavier, the coin from pile H is false. If the side with the two unknown is lighter, the coin from pile L is false.
|Elseif the pile of unknowns is lighter:
||Take two of the three unknown coins from pile L and compare them. If there is equilibrium, the third is the false one, if there isn't, the lighter coin is the false one.
Elseif the pile of unknowns is heavier:
||Take two coins from the heavier pile and compare them. The heavier is the false one.

I think that covers all cases.


A similar puzzle. A woodshop has 10 employees. These employees are supposed to cut the wood in pieces of exactly 100 gram. But upon closer inspection, it turns out that one of the employees is instead cutting the wood in pieces of 90 grams. The owner has a scale which outputs the weight in grams, but he can only use it one time. How can he use this to find out which employee is cutting the small pieces of wood?