41 replies to this topic

[Kro-G]

Little house for 6 rabbits (just 12 matches)

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[mustavarest]

I had assumed that the requirement was that the floor area of each new area should be the same as previouslly. It's a long time since I studied geometry but I think the solution given does this whereas most (if not all) of the alternatives offered do not!

[Kro-G]

The solution of author is wrong.
He wrote the same area?

the cell area on the first picture is P=1/2a*a where a is lenght of match for ex=10
so area of 1 cell will be 50

on his picture we have equila teraltriangle so P=(a*a*square root(3))/4 where a is lenght of match for ex=10
so are of 1 cell will be 43,30

on my solution
right-angled triangle so P=a*a/2 where a is lenght of match for ex=10

so area of 1 cell will be 50

[rookie1ja]

The solution of author is wrong.
He wrote the same area?

the cell area on the first picture is P=1/2a*a where a is lenght of match for ex=10
so area of 1 cell will be 50

on his picture we have equila teraltriangle so P=(a*a*square root(3))/4 where a is lenght of match for ex=10
so are of 1 cell will be 43,30

on my solution
right-angled triangle so P=a*a/2 where a is lenght of match for ex=10

so area of 1 cell will be 50

so you claim that my solution:
1. does not offer the same (equally big) area for 6 rabbits
2. made of 12 matchsticks

???

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[Kro-G]

Yes If it is in 2d your solutions is wrong.
Maybe rabbits will have enough space ;-) but it will be less than on the begining.

[Kro-G]

Look also on my solution with scales it is alternatve to your solution.

[rookie1ja]

Yes If it is in 2d your solutions is wrong.
Maybe rabbits will have enough space ;-) but it will be less than on the begining.

I have not mentioned anything about enough space ... less does not mean not enough ... btw, have you seen a rabbit that would fit into your or my image?

ps: have you noticed that your solution is not made of matchsticks of the same size? check the hypotenuses

[Kro-G]

If you are so smart give my solution of this problem triangle A1A2A3 is make one plane A and is formated on platen plane Z, triangle B1B2B3 is make the second plane B and is formated on plated plane x , plane Z is perpendicular to plane X finde the cuting edge of Tr A1A2A3 & B1B2B3 (all is in 3d of course)

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[rookie1ja]

If you are so smart give my solution of this problem triangle A1A2A3 is make one plane A and is formated on platen plane Z, triangle B1B2B3 is make the second plane B and is formated on plated plane x , plane Z is perpendicular to plane X finde the cuting edge of Tr A1A2A3 & B1B2B3 (all is in 3d of course)

good luck with your homework

[Kro-G]

This is so easy dont think just try ;-)
I know the solution and I will give it in next week bye.