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April 4th, 2016, 07:41 AM  #1 
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  4 congruent equil. triangles with 6 matches in 2D (my puzzle)
. 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 freestanding. Matches may rest across/intersect other matches.                                                     (You may find it helpful to work with drawing line segments to represent matches.) 
April 4th, 2016, 05:06 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,759 Thanks: 696 
Star of David. Six triangles surrounding a hexagon.

April 4th, 2016, 05:29 PM  #3 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,415 Thanks: 1025 
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! 
April 5th, 2016, 02:12 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,634 Thanks: 2080 
Here's a rough sketch: Matches.PNG 
April 5th, 2016, 02:41 AM  #5 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,155 Thanks: 731 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
It is dangerous to go alone. Take this *hands over 6 matches* 
April 5th, 2016, 06:44 AM  #6 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,415 Thanks: 1025 
Skip, methinks your diagram is showing 5 triangles; includes a larger one side=match. Benit: you been drinking? 
April 5th, 2016, 09:49 AM  #7 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,415 Thanks: 1025  
April 5th, 2016, 04:11 PM  #8 
Global Moderator Joined: Dec 2006 Posts: 20,634 Thanks: 2080 
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. Last edited by skipjack; April 5th, 2016 at 04:36 PM. 
April 5th, 2016, 04:21 PM  #9 
Math Team Joined: Jul 2011 From: Texas Posts: 2,924 Thanks: 1521 
...

April 5th, 2016, 05:02 PM  #10 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,415 Thanks: 1025 
Hey, real cute Skeeter...the Star of Texas ? Quick...go get a copyright on it...before MMBt steals it 

Tags 
congruent, equil, matches, puzzle, triangles 
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