## Puzzle of 7 points and 6 straight lines – 2nd solution

This is the second solution to this puzzle: Construct a geometric shape with 7 points such that there are 6 straight lines, and each line must pass through 3 points. The first solution was already discussed last week.

And here’s the construction:

The 7 points are labelled A to G. The 6 lines are ADB, AFC, AGE, DGF, DEC and FEB. No, I didn’t label the points so I’ll have AGE and the short forms of the months December and February. It just happened that way…

The construction starts with point A, and you draw two lines down to get points B and C. The lines AB and AC must be of the same length. Then from point B, draw a perpendicular line to meet line AC. That meeting point is F.

From this construction, angles AFB, AFE, CFB and CFE are right angles (90 degrees).

Do the same thing from point C and draw a perpendicular line to meet line AB, and you’ll get point D. Similarly, angles ADC, ADE, BDC and BDE are right angles.

The point E is formed from the cross point of lines DC and FB that was just formed.

Draw a line joining D and F. Draw another line joining A and E. The cross point of lines DF and AE is point G.

The first solution focused on getting the points right, and then forming lines to fit them. This solution focused on constructing the lines, and the required points magically appear.

On hindsight, we didn’t need the right angles to be there. As long as D and F meet the lines AB and AC respectively in the same ratio, the solution is still valid. There are 2 criteria to meet:

• Lines AB and AC must be of the same length. This allows symmetry.
• The length ratios AD:AB and AF:AC must be equal. This is dependent on the previous criteria.

[Update] Yes, that is one heck of a correction. 3 criteria:

• Points B, A, and C don’t form a straight line (they’re not collinear)
• D is somewhere on the line AB (and D not equal to A nor B)
• F is somewhere on the line AC (and F not equal to A nor C)

Then follow similar construction steps for points E and G and as Eric puts it, the rest just happens. Thank you Eric for pointing this out.

## Puzzle of 7 points and 6 straight lines

My friend presented me with a puzzle: Construct a geometric shape with 7 points such that there are 6 straight lines, and each line must pass through 3 points.

My friend came up with a solution, but he wasn’t sure, so he consulted me. Actually there was a “model” answer, but he wanted to find another solution. So we had 2 solutions. I’ll present that model answer here so you can understand the puzzle better.

Points A, B and C form an equilateral triangle. Points D, E and F are the midpoints of lines AB, AC and BC respectively. Point G is the centre of the triangle.

Note: 3 points are collinear if they lie on a straight line.

The 7 points are A to G. The 6 lines are formed by ADB, AEC, BFC, AGF, BGE and CGD.

[I’m not going to go too much into math theories and proofs. Besides, I’m not even sure if I can recall them after losing track for so long…]

The first three lines should be easy to understand. D is the midpoint of AB, so ADB is a straight line. So it is for AEC and BFC.

E is the midpoint of AC, and BE is the perpendicular bisector (remember ABC is an equilateral triangle). And G lies on the line BE, since G is the centre of the triangle. So BGE is a straight line.

Using this reasoning, AGF and CGD are also straight lines.

I know, the explanation I gave isn’t quite backed up with math proofs, more with intuitive reasoning. But I’m sure they are (I just don’t know exactly which theorems to cite…).

So I have 2 favours:

• Send me a more proper explanation of the solution given above
• Or send me another solution (remember I have another one?)

I know there are at least 2 solutions. Just wondering if there are more. I will post my other solution, together with any submissions from you next Monday. Or earlier, depending on the number of submissions. You’re welcome to do both. *smile*

Since this is a geometric construction, a picture together with some explanations is appreciated. You can host the image on a site like Flickr, then add a comment to this post with a link to that image and a brief explanation.

Or you can send me your solution via email to vincent at polymathprogrammer dot com. Let me know if you want to remain anonymous (why?), and I’ll just post your solution only.

Your construction doesn’t have to be exact in proportions. Meaning if there’s a right angle, it doesn’t have to be 90 degrees exactly, so long as it looks like it’s 90 degrees. Just add some explanation to supplement the drawing. Try Paint.NET if you don’t have any image editing software.

Have fun!