4 congruent equil. triangles with 6 matches in 2-D (my puzzle)

Jun 2014
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191
Earth
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How would you make exactly four congruent equilateral triangles with just
six matches of equal length in two dimensions?


No other triangles may be created when you are done.

Matches may not be bent, torn, or separated into other matches.

Match ends do not necessarily have to join other match ends. Specifically speaking,
certain match ends might be free-standing.

Matches may rest across/intersect other matches.


- - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -- - - -


(You may find it helpful to work with drawing line segments to represent matches.)
 
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mathman

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May 2007
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Star of David. Six triangles surrounding a hexagon.
 
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Denis

Math Team
Oct 2011
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Ottawa Ontario, Canada
Methinks not allowed, because 2 other equilateral triangles
(sides = matches) are formed.
Same thing with a square: works ok, but extra right triangles created.

MMBt s'got a dilly there!
 
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Benit13

Math Team
Apr 2014
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It is dangerous to go alone. Take this *hands over 6 matches*
 
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Denis

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Oct 2011
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Ottawa Ontario, Canada
Skip, methinks your diagram is showing 5 triangles;
includes a larger one side=match.

Benit: you been drinking? :)
 
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skipjack

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Dec 2006
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Thanks, Denis. There is one rather nice, rotationally symmetrical solution (the one with a central triangle surrounded by three more triangles) at that site. I see that skeeter has posted below a version of it that reduces the number of "free" match ends (three such ends is probably the minimum possible).

I like mathman's Star of David idea. It produces six congruent triangles in an elegant, symmetrical configuration, but one can modify it slightly to remove two of those triangles.

In the diagram below, I've done that, and extended the matches slightly so that all the ends of the matches are "free". This makes my diagram very similar to one of the answers at the page that Denis linked to.

Star.png

Obviously, the horizontal match at the upper right of the above diagram can be moved to various other positions to create (after slight movements of other matches) various other solutions, including the solution I mentioned in my opening paragraph above. All of the correct solutions at the site that Denis found can similarly be obtained from the Star of David configuration.
 
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Denis

Math Team
Oct 2011
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Ottawa Ontario, Canada
Hey, real cute Skeeter...the Star of Texas ?

Quick...go get a copyright on it...before MMBt steals it :)
 
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