Guest Posted December 16, 2007 Report Share Posted December 16, 2007 Look also on my solution with scales it is alternatve to your solution. Link to comment Share on other sites More sharing options...

rookie1ja Posted December 16, 2007 Author Report Share Posted December 16, 2007 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 Link to comment Share on other sites More sharing options...

Guest Posted December 16, 2007 Report Share Posted December 16, 2007 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) Link to comment Share on other sites More sharing options...

rookie1ja Posted December 16, 2007 Author Report Share Posted December 16, 2007 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 Link to comment Share on other sites More sharing options...

Guest Posted December 16, 2007 Report Share Posted December 16, 2007 This is so easy dont think just try ;-) I know the solution and I will give it in next week bye. Link to comment Share on other sites More sharing options...

Guest Posted December 17, 2007 Report Share Posted December 17, 2007 ^ Please don't. This thread is to discuss the riddle in the OP, not new geometry problems you'd like to introduce. Link to comment Share on other sites More sharing options...

Guest Posted December 31, 2007 Report Share Posted December 31, 2007 The dispute Kro-G offers is incorrect, the administrator's solution is correct. For one, there is a congruence shortcut in triangles stating that if three sides in one triangle are congruent to three corresponding sides in another triangle, then the two triangles are congruent. Each of the six triangles in teh larger hexagon is an equilateral triangle, all with the same side length. Therefore, according to this congruence shortcut, the triangles are the same side (see Euclid's Elements I) Moreover, the area calculation was incorrect also. You do not find the area of a triangle by multiplying the base and an arbitrary leg by one half (in this case, a * a, since both are the same length in an equillateral triangle). The area is found by multiplying a leg (a) by the perpendicular height (which lends itself to the Pythagorean theorem resulting in an altitude of a*sqrt(3)) to the vertext opposite your chosen leg, and then multiplying that by one half. Link to comment Share on other sites More sharing options...

Guest Posted December 31, 2007 Report Share Posted December 31, 2007 The dispute Kro-G offers is incorrect, the administrator's solution is correct. For one, there is a congruence shortcut in triangles stating that if three sides in one triangle are congruent to three corresponding sides in another triangle, then the two triangles are congruent. Each of the six triangles in teh larger hexagon is an equilateral triangle, all with the same side length. Therefore, according to this congruence shortcut, the triangles are the same side (see Euclid's Elements I) Moreover, the area calculation was incorrect also. You do not find the area of a triangle by multiplying the base and an arbitrary leg by one half (in this case, a * a, since both are the same length in an equillateral triangle). The area is found by multiplying a leg (a) by the perpendicular height (which lends itself to the Pythagorean theorem resulting in an altitude of a*sqrt(3)) to the vertext opposite your chosen leg, and then multiplying that by one half. My apologies, the altitude is .5*a*sqrt(3). And this has already been addressed by teh administrator anyway. Link to comment Share on other sites More sharing options...

Guest Posted February 1, 2008 Report Share Posted February 1, 2008 I've done this one in 11. No overlap and no breaking matchsticks in half. Link to comment Share on other sites More sharing options...

Guest Posted February 6, 2008 Report Share Posted February 6, 2008 Ever wonder why matchsticks are used and not poles or anything else? If you are to make a rabbit pin with 12 matchsticks, the rabbits would be awfully small. Plus, matchsticks catch on fire easier than regular sticks. Link to comment Share on other sites More sharing options...

Guest Posted March 30, 2008 Report Share Posted March 30, 2008 After i did the courting of the original solution i found that i used 13 matchsticks: So the first thing flashed into my mind is if i can shorten just one single matchsticks then i can be done.so there for my 1st solution is : But since building up a dwelling for those rabbits is what we concerted, then perhaps we are not supposed to put rabbits in the air and a two-dimensional solution should be better.... Link to comment Share on other sites More sharing options...

Guest Posted June 15, 2008 Report Share Posted June 15, 2008 Rabbit Hutch - Back to the Matchstick Puzzles In the picture there are little flats for 6 rabbits. Can you build a dwelling for these 6 rabbits with only 12 matches. Each rabbit must have an equally big space. Edit: 6 separate flats are needed for the rabbits. Rabbit Hutch - solution Picture added as per below hint. easier way, just make a cube... Link to comment Share on other sites More sharing options...

Guest Posted August 7, 2008 Report Share Posted August 7, 2008 In general this problem can be resolved for as: N matches, with K spaces, to determine K spaces with to N-1 matches, where N is 12. You can then try N=9, N=5, and get N-1 edges to create the same amount of closures as in the initial state Therefore by resolving this problem for 5 edges (matches), 2 spaces; then 4 spaces, and 9 edges, and 6 spaces, 12 edges (matches), you get the trick to resolved it for all. A triangle -> hexagon figures could be combined with all the matches. Link to comment Share on other sites More sharing options...

Guest Posted September 12, 2008 Report Share Posted September 12, 2008 how about only using 7 _____________________ | | | | | | |--------------|--------------| | | | | | | | | | |--------------|--------------| | | | |__________|__________| Link to comment Share on other sites More sharing options...

Guest Posted October 3, 2008 Report Share Posted October 3, 2008 This is what I came up with: Link to comment Share on other sites More sharing options...

andromeda Posted October 24, 2008 Report Share Posted October 24, 2008 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? Sure... Link to comment Share on other sites More sharing options...

Guest Posted January 13, 2009 Report Share Posted January 13, 2009 heres the solution i came up with. i was impressed by everyone's creativity on this one Link to comment Share on other sites More sharing options...

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