20100710, 21:50  #1 
"Forget I exist"
Jul 2009
Dumbassville
3×2,797 Posts 
matchstick question
on pg. 230 of number freak 1 to 200 it talks of the smallest 4 regular matchstick graph outside of one technicality I think i can get it down from 104 to 32 (possibly less actually) if one thing is allowed. the attached image is my possible solution the catch I think the x's shown would need either .5 matches or the allowing of matchsticks to cross each other. anyone for a look ?

20100711, 06:13  #2  
Apr 2010
3^{2}·17 Posts 
Quote:
Edit: And presumably the graph should be planar, i.e. no crossings. Last fiddled with by ccorn on 20100711 at 06:15 

20100711, 15:04  #3 
Apr 2010
3^{2}·17 Posts 
Note that four matchsticks with common end point, together defining three adjacent equilateral triangles, by construction enclose three adjacent 60degree angles, i. e. 180 degrees. This means: If the matchsticks must have same length, your ring construction actually is straight. You cannot bend it into a finite polygon. You need to reduce the connectivity of the boundary in order to allow for "gaps" with sub60degree angles. These are needed in order to bend the boundary to some finite perimeter.

20100711, 16:01  #4 
"Forget I exist"
Jul 2009
Dumbassville
8391_{10} Posts 
I think you are wrong about the hexagon shape as it shows the record holder on that page of the book and it's a irregular octagon if my eyes do not deceive me.

20100711, 17:31  #5 
Apr 2010
3^{2}×17 Posts 
OK, scrap my post #2. I have refined what I mean in post #3. In order to clarify things, it could be helpful if you presented the problem with its original specification.

20100711, 19:50  #6  
"Forget I exist"
Jul 2009
Dumbassville
3×2,797 Posts 
Quote:
http://www.ebooks.com/ebooks/book_di...sp?IID=449939#, found it but the page isn't available without buying. Last fiddled with by science_man_88 on 20100711 at 19:56 

20100711, 22:12  #7 
"Forget I exist"
Jul 2009
Dumbassville
8391_{10} Posts 
got my webcam stuff working. that's my best picture.

20100712, 05:41  #8 
Jul 2006
Calgary
5^{2}·17 Posts 

20100712, 18:10  #9 
"Forget I exist"
Jul 2009
Dumbassville
10000011000111_{2} Posts 

20100712, 20:50  #10 
Apr 2010
3^{2}×17 Posts 
Wow. Look at those boundary points that are used as hinges (with lessthan60degree internal angles). You need some of these, otherwise you cannot get a curved boundary. That's what I meant in post #3.
If 3D were allowed, you could simply build an octaeder. 12 matches only Last fiddled with by ccorn on 20100712 at 21:30 
20100712, 21:37  #11 
"Forget I exist"
Jul 2009
Dumbassville
10000011000111_{2} Posts 
the book shows the solution for 3 at each and tells what the solution for 2 is the solution for 2 is a equilateral triangle(though those hinges work as well).
the solution for 3 is 12 matches 2 diamonds split by a match each and then connected into a hexagon(does this work for 4 with the three match solution?). can you draw the octahedron in 2d ? if so it's a possible lowest solution. 