Line segments of two types

[bonanova]

Six line segments serve to connect pairwise any four points in the plane, no three of which are collinear. It's clear that no placement of the points permits all six to have the same length. How many unique placements permit the segments to have only two distinct lengths?

Example:

Spoiler

The four sides and two diagonals of a square.

 

[Donald Cartmill]

 

I have it ...the 6th solution

1) a square with 4 sides and two diagonals =6 lines 

2) would be a diamond shape created by two equilateral triangles;  5 sides equal and one long diagonal

3)  An equilateral triangle ,3 equal sides ;  The 4th point being in the center equidistant from the other 3 points providing 3 other lines of equal length but obviously shorter  

4)   an equilateral triangle is 3 lines  connecting points "A" ,"B" and "C", Assume  "A" is the vertex and "B" and "C" are the base points.  A 4th point "X"gives us a line  the same length as  a sides of the triangle ,and directly off of vertex point "A" away from the triangle , but perpendicular to the opposing side "B"C";   This arrangement gives us 4 lines of equal length .  Now the lines connecting  the 2 base points  "B"and "C", and point "X" ,are of equal length and longer than the other 4 lines

5)  an equilateral triangle is 3 lines  connecting points "A" ,"B" and "C", Assume  "A" is the vertex and "B" and "C" are the base points.  A 4th point "X"gives us a line  the same length as  a sides of the triangle ,and directly off of vertex point "A",but downward bisecting line BC  = 4 lines of equal length;  2 short lines BX and CX

6) 4 points  A,B,C,D,  lay on two intersecting  circles.  point A is the center of one circle C and D lay on this circle.  D is the center of the 2nd circle with A and B laying on the circle.  The 4 points create a trapezoid in which the distance BC , is the same length as chords AB  and CD  (that is 3 short lengths )  The diagonals AC ,AD and BD are equal all being equal to the radius.

That should be all of them ????

6) 4 points  A,B,C,D,  lay on two intersecting  circles having the same radius.  point A is the center of one circle C and D lay on this circle.  D is the center of the 2nd circle with A and B laying on the circle.  The 4 points create a trapezoid in which the distance BC , is the same length as chords AB  and CD  (that is 3 short lengths )  The diagonals AC ,AD and BD are equal all being equal to the radius.

[Donald Cartmill]

Spoiler

 

I think there are 3 solutions 1) a square with 4 sides and two diagonals =6 lines 

2) would be a diamond shape created by two equilateral triangles;  5 sides equal and one long diagonal

3)  An equilateral triangle ,3 equal sides ;  The 4th point being in the center equidistant from the other 3 points providing 3 other lines of equal length but obviously shorter  

 

 

Edited January 29, 2018 by bonanova
Spolier added

[bonanova]

3 hours ago, Donald Cartmill said:

  Hide contents

I think there are 3 solutions 1) a square with 4 sides and two diagonals =6 lines 

2) would be a diamond shape created by two equilateral triangles;  5 sides equal and one long diagonal

3)  An equilateral triangle ,3 equal sides ;  The 4th point being in the center equidistant from the other 3 points providing 3 other lines of equal length but obviously shorter 

 

Good start, but there are more.

[Donald Cartmill]

Spoiler

 

I think there are 4 solutions 1) a square with 4 sides and two diagonals =6 lines 

2) would be a diamond shape created by two equilateral triangles;  5 sides equal and one long diagonal

3)  An equilateral triangle ,3 equal sides ;  The 4th point being in the center equidistant from the other 3 points providing 3 other lines of equal length but obviously shorter  

4)   an equilateral triangle is 3 lines  connecting points "A" ,"B" and "C", Assume  "A" is the vertex and "B" and "C" are the base points.  A 4th point "X"gives us a line  the same length as  a sides of the triangle ,and directly off of vertex point "A" away from the triangle , but perpendicular to the opposing side "B"C";   This arrangement gives us 4 lines of equal length .  Now the lines connecting  the 2 base points  "B"and "C", and point "X" ,are of equal length and longer than the other 4 lines

 

 

Edited January 31, 2018 by bonanova
Added spoiler

[Donald Cartmill]

Spoiler

 

I think there are 5 solutions 1) a square with 4 sides and two diagonals =6 lines 

2) would be a diamond shape created by two equilateral triangles;  5 sides equal and one long diagonal

3)  An equilateral triangle ,3 equal sides ;  The 4th point being in the center equidistant from the other 3 points providing 3 other lines of equal length but obviously shorter  

4)   an equilateral triangle is 3 lines  connecting points "A" ,"B" and "C", Assume  "A" is the vertex and "B" and "C" are the base points.  A 4th point "X"gives us a line  the same length as  a sides of the triangle ,and directly off of vertex point "A" away from the triangle , but perpendicular to the opposing side "B"C";   This arrangement gives us 4 lines of equal length .  Now the lines connecting  the 2 base points  "B"and "C", and point "X" ,are of equal length and longer than the other 4 lines

5)  an equilateral triangle is 3 lines  connecting points "A" ,"B" and "C", Assume  "A" is the vertex and "B" and "C" are the base points.  A 4th point "X"gives us a line  the same length as  a sides of the triangle ,and directly off of vertex point "A",but downward bisecting line BC  = 4 lines of equal length;  2 short lines BX and CX

 

 

Edited January 31, 2018 by bonanova
spoiler added

[bonanova]

Closer but not all have been found.

[Donald Cartmill]

Spoiler

I think there are 6 solutions 1) a square with 4 sides and two diagonals =6 lines 

2) would be a diamond shape created by two equilateral triangles;  5 sides equal and one long diagonal

3)  An equilateral triangle ,3 equal sides ;  The 4th point being in the center equidistant from the other 3 points providing 3 other lines of equal length but obviously shorter  

4)   an equilateral triangle is 3 lines  connecting points "A" ,"B" and "C", Assume  "A" is the vertex and "B" and "C" are the base points.  A 4th point "X"gives us a line  the same length as  a sides of the triangle ,and directly off of vertex point "A" away from the triangle , but perpendicular to the opposing side "B"C";   This arrangement gives us 4 lines of equal length .  Now the lines connecting  the 2 base points  "B"and "C", and point "X" ,are of equal length and longer than the other 4 lines

5)  an equilateral triangle is 3 lines  connecting points "A" ,"B" and "C", Assume  "A" is the vertex and "B" and "C" are the base points.  A 4th point "X"gives us a line  the same length as  a sides of the triangle ,and directly off of vertex point "A",but downward bisecting line BC  = 4 lines of equal length;  2 short lines BX and CX

6) A circle has a  radius R ; with 3 points  A , B , C lying on the circle , such that the chord AC is equal to the radius R .   The 4th point  X is the center of the circle .Therefore you have  4 lines of the same length; AC ; AX :BX; CX and 2 shorter lines being the chords AB and BC

Spoilers please

[Donald Cartmill]

Spoiler

 

I think there are 7 solutions 1) a square with 4 sides and two diagonals =6 lines 

2) would be a diamond shape created by two equilateral triangles;  5 sides equal and one long diagonal

3)  An equilateral triangle ,3 equal sides ;  The 4th point being in the center equidistant from the other 3 points providing 3 other lines of equal length but obviously shorter  

4)   an equilateral triangle is 3 lines  connecting points "A" ,"B" and "C", Assume  "A" is the vertex and "B" and "C" are the base points.  A 4th point "X"gives us a line  the same length as  a sides of the triangle ,and directly off of vertex point "A" away from the triangle , but perpendicular to the opposing side "B"C";   This arrangement gives us 4 lines of equal length .  Now the lines connecting  the 2 base points  "B"and "C", and point "X" ,are of equal length and longer than the other 4 lines

5)  an equilateral triangle is 3 lines  connecting points "A" ,"B" and "C", Assume  "A" is the vertex and "B" and "C" are the base points.  A 4th point "X"gives us a line  the same length as  a sides of the triangle ,and directly off of vertex point "A",but downward bisecting line BC  = 4 lines of equal length;  2 short lines BX and CX

6) A circle has a  radius R ; with 3 points  A , B , C lying on the circle , such that the chord AC is equal to the radius R .   The 4th point  X is the center of the circle .Therefore you have  4 lines of the same length; AC ; AX :BX; CX and 2 shorter lines being the chords AB and BC

7) A circle with radius R. with 3 points A,B, C , lying on the circle,such that chord AB is equal to the radius R .  The 4th point lies diametrically opposite and perpendicular to the center of chord AB .  This configuration produces 4 lines equal to the radius R,and 2 addition lines AC and BC which are equal in length.

I do believe that is it ...7 solutions

 

 

 

[bonanova]

2 hours ago, Donald Cartmill said:

  Hide contents

 

I think there are 7 solutions 1) a square with 4 sides and two diagonals =6 lines 

2) would be a diamond shape created by two equilateral triangles;  5 sides equal and one long diagonal

3)  An equilateral triangle ,3 equal sides ;  The 4th point being in the center equidistant from the other 3 points providing 3 other lines of equal length but obviously shorter  

4)   an equilateral triangle is 3 lines  connecting points "A" ,"B" and "C", Assume  "A" is the vertex and "B" and "C" are the base points.  A 4th point "X"gives us a line  the same length as  a sides of the triangle ,and directly off of vertex point "A" away from the triangle , but perpendicular to the opposing side "B"C";   This arrangement gives us 4 lines of equal length .  Now the lines connecting  the 2 base points  "B"and "C", and point "X" ,are of equal length and longer than the other 4 lines

5)  an equilateral triangle is 3 lines  connecting points "A" ,"B" and "C", Assume  "A" is the vertex and "B" and "C" are the base points.  A 4th point "X"gives us a line  the same length as  a sides of the triangle ,and directly off of vertex point "A",but downward bisecting line BC  = 4 lines of equal length;  2 short lines BX and CX

6) A circle has a  radius R ; with 3 points  A , B , C lying on the circle , such that the chord AC is equal to the radius R .   The 4th point  X is the center of the circle .Therefore you have  4 lines of the same length; AC ; AX :BX; CX and 2 shorter lines being the chords AB and BC

7) A circle with radius R. with 3 points A,B, C , lying on the circle,such that chord AB is equal to the radius R .  The 4th point lies diametrically opposite and perpendicular to the center of chord AB .  This configuration produces 4 lines equal to the radius R,and 2 addition lines AC and BC which are equal in length.

I do believe that is it ...7 solutions

 

Spoiler

If I understand your descriptions, case 7 is also case 4; case 6 is also case 5.  Can you check that?

 

[Donald Cartmill]

If I understand your descriptions, case 7 is also case 4; case 6 is also case 5.  Can you check that?

Spoiler

 

I stand corrected 5 solutions . As you pointed out solutions 6 and 7 are merely a different approach  arriving at the same solution as in 4 and 5.                                                1) a square with 4 sides and two diagonals =6 lines 

2) would be a diamond shape created by two equilateral triangles;  5 sides equal and one long diagonal

3)  An equilateral triangle ,3 equal sides ;  The 4th point being in the center equidistant from the other 3 points providing 3 other lines of equal length but obviously shorter  

4)   an equilateral triangle is 3 lines  connecting points "A" ,"B" and "C", Assume  "A" is the vertex and "B" and "C" are the base points.  A 4th point "X"gives us a line  the same length as  a sides of the triangle ,and directly off of vertex point "A" away from the triangle , but perpendicular to the opposing side "B"C";   This arrangement gives us 4 lines of equal length .  Now the lines connecting  the 2 base points  "B"and "C", and point "X" ,are of equal length and longer than the other 4 lines

5)  an equilateral triangle is 3 lines  connecting points "A" ,"B" and "C", Assume  "A" is the vertex and "B" and "C" are the base points.  A 4th point "X"gives us a line  the same length as  a sides of the triangle ,and directly off of vertex point "A",but downward bisecting line BC  = 4 lines of equal length;  2 short lines BX and CX

6) DUPLICATION A circle has a  radius R ; with 3 points  A , B , C lying on the circle , such that the chord AC is equal to the radius R .   The 4th point  X is the center of the circle .Therefore you have  4 lines of the same length; AC ; AX :BX; CX and 2 shorter lines being the chords AB and BC

7) DUPLICATIONA circle with radius R. with 3 points A,B, C , lying on the circle,such that chord AB is equal to the radius R .  The 4th point lies diametrically opposite and perpendicular to the center of chord AB .  This configuration produces 4 lines equal to the radius R,and 2 addition lines AC and BC which are equal in length.

 

 

[bonanova]

Those are still not all of them.

[Donald Cartmill]

6 solutions with number 6 being only theoretically correct

Spoiler

 

 1) a square with 4 sides and two diagonals =6 lines 

2) would be a diamond shape created by two equilateral triangles;  5 sides equal and one long diagonal

3)  An equilateral triangle ,3 equal sides ;  The 4th point being in the center equidistant from the other 3 points providing 3 other lines of equal length but obviously shorter  

4)   an equilateral triangle is 3 lines  connecting points "A" ,"B" and "C", Assume  "A" is the vertex and "B" and "C" are the base points.  A 4th point "X"gives us a line  the same length as  a sides of the triangle ,and directly off of vertex point "A" away from the triangle , but perpendicular to the opposing side "B"C";   This arrangement gives us 4 lines of equal length .  Now the lines connecting  the 2 base points  "B"and "C", and point "X" ,are of equal length and longer than the other 4 lines

5)  an equilateral triangle is 3 lines  connecting points "A" ,"B" and "C", Assume  "A" is the vertex and "B" and "C" are the base points.  A 4th point "X"gives us a line  the same length as  a sides of the triangle ,and directly off of vertex point "A",but downward bisecting line BC  = 4 lines of equal length;  2 short lines BX and CX

6) 4 points representing  2 parallel lines infinitesimally close and stretching to infinity.  The long diagonals would be the same length as the two lines and the you would have         2   infinitesimally short lines connecting the two at each end

 

 

[bonanova]

You can fit all the answers on a sheet of paper. Nice thinking tho.

[bonanova]

Bumping this puzzle and offering a clue - unless someone's actively searching for the final case.

[Molly Mae]

 

I'm confident there can be at most 8 cases based on the way points would be related to each other.

plot point 1
plot point 2 at distance called A
plot point 3, possibilities: AA, AB, BA, BB
plot point 4, possibilities:  AAAA, AAAB, AABA, AABB, ABAA, ABAB, ABBA, ABBB

Maybe I'm over-generalizing though.  Maybe some of these would be colinear.  I'll put pen to paper when I can and see if they're viable.

EDIT: Now I'm doubting myself.  I think I missed a distance.

Edited February 7, 2018 by Molly Mae

[bonanova]

On 2/8/2018 at 1:48 AM, Donald Cartmill said:

 

I have it ...the 6th solution

1) a square with 4 sides and two diagonals =6 lines 

2) would be a diamond shape created by two equilateral triangles;  5 sides equal and one long diagonal

3)  An equilateral triangle ,3 equal sides ;  The 4th point being in the center equidistant from the other 3 points providing 3 other lines of equal length but obviously shorter  

4)   an equilateral triangle is 3 lines  connecting points "A" ,"B" and "C", Assume  "A" is the vertex and "B" and "C" are the base points.  A 4th point "X"gives us a line  the same length as  a sides of the triangle ,and directly off of vertex point "A" away from the triangle , but perpendicular to the opposing side "B"C";   This arrangement gives us 4 lines of equal length .  Now the lines connecting  the 2 base points  "B"and "C", and point "X" ,are of equal length and longer than the other 4 lines

5)  an equilateral triangle is 3 lines  connecting points "A" ,"B" and "C", Assume  "A" is the vertex and "B" and "C" are the base points.  A 4th point "X"gives us a line  the same length as  a sides of the triangle ,and directly off of vertex point "A",but downward bisecting line BC  = 4 lines of equal length;  2 short lines BX and CX

6) 4 points  A,B,C,D,  lay on two intersecting  circles.  point A is the center of one circle C and D lay on this circle.  D is the center of the 2nd circle with A and B laying on the circle.  The 4 points create a trapezoid in which the distance BC , is the same length as chords AB  and CD  (that is 3 short lengths )  The diagonals AC ,AD and BD are equal all being equal to the radius.

That should be all of them ????

6) 4 points  A,B,C,D,  lay on two intersecting  circles having the same radius.  point A is the center of one circle C and D lay on this circle.  D is the center of the 2nd circle with A and B laying on the circle.  The 4 points create a trapezoid in which the distance BC , is the same length as chords AB  and CD  (that is 3 short lengths )  The diagonals AC ,AD and BD are equal all being equal to the radius.

I'm not sure I recognize your solution 6 from its description of how it's constructed, but your final description of it seems right. Nice solve.

Spoiler

Another way to describe it is to take a 5-pointed star (which fulfills the 2- length criterion) and remove one of the points.

 

[Donald Cartmill]

I'm not sure I recognize your solution 6 from its description of how it's constructed, but your final description of it seems right. Nice solve.    Let me attempt to clarify

6) 4 points  A,B,C,D,  lay on two intersecting  circles having the same radius.  point A is the center of one circle  with C and D laying on its circumference ,    D is the center of the 2nd circle with A and B laying on its circumference.  The 4 points create a trapezoid in which   The diagonals AC , & BD  plus the  base AD are equal all being  a radius.   Points B and C lay on their respective circumferences at points where the chords   AB =  CD =  equals the top BC   I don't know if that is any more clear