My puzzle involving triangles formed by line segments of equal length

L

lookagain

Elite Member
Joined
Aug 22, 2010
Messages
3,230
Suppose you have six line segments, all in the same plane, and all of the same length.

You may want to approximate making formations by using six unused pencils.

You want to place the six line segments in the manners below so as to form the
maximum total of number of overlapping and non-overlapping triangles:


1) No line segments may cross any others.

2) No line segment can share more than one point with another line segment.

3) Each line segment must be in contact with at least one other line segment.
(no free-standing line segments)

4) Included above, two or more line segments may be placed so as to form
a longer line segment.




(Clarifications can be asked for.)
 
Hello, lookagain!

This reminds me of similar problem ... with a sneaky answer.

Suppose you have six line segments, all in the same plane, and all of the same length.

You may want to approximate making formations by using six unused pencils.

You want to place the six line segments in the manners below so as to form the
maximum total of number of overlapping and non-overlapping triangles:


1) No line segments may cross any others.

2) No line segment can share more than one point with another line segment.

3) Each line segment must be in contact with at least one other line segment.
. . (No free-standing line segments.)

4) Included above, two or more line segments may be placed
. . so as to form a longer line segment.
Click to expand...

The problem I know goes like this.
. . You have six identical matchsticks.
. . Use them to form four equilateral triangles.

(That phrase in blue is omitted.)

Solution: form a tetrahedron.
 
Kaylaemarx said:
I suppose both are same questions with a
> > > slight difference < < < isn't it?
Click to expand...

That phrase has a relative meaning in this context. Working in
(allowing) three dimensions can give more opportunities to solve the problem.
I want the problem to be more restrictive, and so I limited it to involving two
dimensions (one plane).

Also, the tetrahedron solution for that other question doesn't involve any
overlapping triangles (it doesn't have any), as opposed to my own
question.
 
Last edited:
Kaylaemarx said:
Okay!!! Then what is the solution for your answer? Please let me know.
Click to expand...

I'm going to show one of the ways here now, because it has been several days since I posted the problem.


Picture an equilateral triangle in the x-y plane with vertex coordinates of (0, 0), (2, 0), and (1, sqrt(3)).
(These are formed with three of the line segments.)


Place a fourth line segment, beginning at (1, 0), and extending up to and through (1, sqrt(3))
outside of the equilateral triangle. This will be referred to as a vertical line segment for this diagram.

Place a fifth line segment which will have an endpoint, call it P, on the vertical line segment.
It will extend back through the vertex at (0, 0) and outside of the equilateral triangle.

Place a sixth line segment, which will also have an endpoint at P, and extending back through
the vertex at (2, 0). It will extend back through that vertex, (2, 0), and outside of the
equilateral triangle.


Now, count up the total of non-overlapping triangles and overlapping triangles
for the whole diagram.


What total do you get?


Do you think there is another way involving a different placement of the six line segments
to get a larger total of triangles? (I do not believe there is one.)
 
You must log in or register to reply here.
Top