[k-man]

How many distinct triangles can you make by placing 10 points on a flat surface and connecting them with line segments? Line segments cannot cross.

[curr3nt]

18

[Guest]

[spoiler = more]32 in concentric circles of 3 with middle point.

[plainglazed]

36

[dark_magician_92]

Hey poliwrath

max. triangles possible = 10C3 = 120

[Guest]

My guess is 64:

I built a simple triangle with 3 dots, then placed 7 dots on a straight line below the apex of the triangle.

Each new dot gives 2 more new triangles than the previous one.

the count of triangles is 1+3+5+7+9+11+13+15=64

Edited July 27, 2011 by mulikar

[Guest]

10c3=120 so 120 maximum triangle can be made

[Guest]

10c3=120 so 120 maximum triangle can be made

Or as line should not cross the answer can be 9

[Guest]

Bit difficult to describe this in text but here goes, I counted at least 25 triangles but then my drawing on my lunch hour napkin started to disintegrate...

If you put 3 points to make an equilateral triangle, then a mid point bang in the centre of that triangle. Then half way between the mid point and each of the corners, put another dot. So you should end up with two concentric and similar triangles with a single dot right in the centre of these.

I have assumed that you can make a triangle that will have other lines in the middle of it but no lines cross.

firstly join up the outside edges, 1 big triangle and repeat for the smaller one (that's two) The next lines were radially out from the centre dot through the mid points and onto the outer points.

That bit should be fairly straight forward despite my description.

The are three other lines you can make that is the difficult bit to describe, but I think if you do it this way it will make more triangles than say doing it to resemble a spiral.

If your original triangle has its point at the top and base at the bottom, then draw from the top most point to both of the base points on the smaller triangle. and lastly join either of the base points of the centre triangle to the opposite base point on the larger triangle.

I got to at least 25 triangles before the napkin shredded but I think there may be more, anyone thinking along similar lines? ...

This is just with 7 dots so feel free to put a third triangle around these two and count away...

Edited July 27, 2011 by RESHAW

[Guest]

i can get at least 47. make an equilateral traingle with points at the mid point of each line. create lines going from each point to every other point. where the lines cross, place a point. this gives 10 points all together. this figure has 47 triangles. i suspect more is possible.

[Guest]

22

Edited July 27, 2011 by sks

[BobbyGo]

I counted 63.

sixty_three.bmp

Edited July 27, 2011 by BobbyGo

[k-man]

Mulikar got it! Good job.

BobbyGo was pretty close and even posted a picture of how to arrange the dots, but lost one triangle by aligning the bottom 3 dots horizontally.

Move the middle dot up a little and you get 64 triangles.

My guess is 64:

I built a simple triangle with 3 dots, then placed 7 dots on a straight line below the apex of the triangle.

Each new dot gives 2 more new triangles than the previous one.

the count of triangles is 1+3+5+7+9+11+13+15=64

[k-man]

Hey poliwrath

max. triangles possible = 10C3 = 120

If the lines were allowed to cross then this answer would be correct, but the lines cannot cross.

[dark_magician_92]

If the lines were allowed to cross then this answer would be correct, but the lines cannot cross.

yes, ur right i dint think much that time.

[k-man]

Here is an additional challenge:

Prove that for any number k (k>2) of points placed on the flat surface the maximum number of triangles that can be formed without crossing the lines is always (k-2)²

The proof is trivial for k=3 and k=4, but can you provide the general proof for any k>4?

Edited July 28, 2011 by k-man

[Guest]

Here is an additional challenge:

Prove that for any number k (k>2) of points placed on the flat surface the maximum number of triangles that can be formed without crossing the lines is always (k-2)²

The proof is trivial for k=3 and k=4, but can you provide the general proof for any k>4?

Going by mulikar's solution to original problem, it's a series of odd numbers, with number of terms = k-2. So sum would obviously be (k-2)*(k-2).

[k-man]

Going by mulikar's solution to original problem, it's a series of odd numbers, with number of terms = k-2. So sum would obviously be (k-2)*(k-2).

Mulikar showed that for any number of points k, there is a method of arranging the points on the surface that creates (k-2)² triangles. But nobody has proven that this method always produces the maximum possible number of triangles. What if there is a different method of placing points that can produce more triangles? What if this method produces the same number of triangles for k<x, but produces more triangles for k>x?

That's what this new challenge is - to prove that (k-2)² is the maximum for any k.