## Welcome to BrainDen.com - Brain Teasers Forum

 Welcome to BrainDen.com - Brain Teasers Forum. Like most online communities you must register to post in our community, but don't worry this is a simple free process. To be a part of BrainDen Forums you may create a new account or sign in if you already have an account. As a member you could start new topics, reply to others, subscribe to topics/forums to get automatic updates, get your own profile and make new friends. Of course, you can also enjoy our collection of amazing optical illusions and cool math games. If you like our site, you may support us by simply clicking Google "+1" or Facebook "Like" buttons at the top. If you have a website, we would appreciate a little link to BrainDen. Thanks and enjoy the Den :-)
Guest Message by DevFuse

# Rabbit Hutch

41 replies to this topic

### #31 Martini

Martini

Senior Member

• Members
• 770 posts

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.
• 0

### #32 dissatisfied

dissatisfied

Newbie

• Members
• 10 posts

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.
• 0

### #33 dissatisfied

dissatisfied

Newbie

• Members
• 10 posts

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.

• 0

Newbie

• Members
• 1 posts

Posted 01 February 2008 - 06:05 AM

I've done this one in 11. No overlap and no breaking matchsticks in half.

• 0

### #35 Sharpie357

Sharpie357

Junior Member

• Members
• 32 posts

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.
• 0

### #36 zach

zach

Newbie

• Members
• 15 posts

Posted 30 March 2008 - 05:23 AM

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....
• 0

### #37 magicalcheese

magicalcheese

Newbie

• Members
• 12 posts

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...
• 0

### #38 int21h

int21h

Newbie

• Members
• 1 posts

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 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.
• 0

### #39 HAL9000

HAL9000

Newbie

• Members
• 1 posts

Posted 12 September 2008 - 04:51 PM

_____________________
| | |
| | |
|--------------|--------------|
| | |
| | |
| | |
|--------------|--------------|
| | |
|__________|__________|
• 0

### #40 MeTed123

MeTed123

Newbie

• Members
• 2 posts

Posted 03 October 2008 - 07:41 PM

This is what I came up with:

• 0

#### 0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users