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Rabbit Hutch


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41 replies to this topic

#31 Martini

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Posted 17 December 2007 - 07:35 AM

^ Please don't. This thread is to discuss the riddle in the OP, not new geometry problems you'd like to introduce.
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#32 dissatisfied

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Posted 31 December 2007 - 01:13 AM

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.
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#33 dissatisfied

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Posted 31 December 2007 - 01:31 AM

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.
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#34 Deadboy

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Posted 01 February 2008 - 06:05 AM

I've done this one in 11. No overlap and no breaking matchsticks in half.
P11.JPG
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#35 Sharpie357

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Posted 06 February 2008 - 04:44 PM

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.
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#36 zach

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Posted 30 March 2008 - 05:23 AM

After i did the courting of the original solution i found that i used 13 matchsticks:
kraliky.gif
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 :
1st_solution.PNG
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....
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#37 magicalcheese

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Posted 15 June 2008 - 11:43 PM

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.



Spoiler for Solution


easier way, just make a cube...
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#38 int21h

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Posted 07 August 2008 - 07:34 PM

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 :mellow: 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.
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#39 HAL9000

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Posted 12 September 2008 - 04:51 PM

how about only using 7
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#40 MeTed123

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Posted 03 October 2008 - 07:41 PM

This is what I came up with:

my_answer.gif
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