A *lattice point* is a point with integer coordinates such as (2,3).

In how many ways can we pick 3 lattice points such that both coordinates of all three points are nonnegative integers less than 4, and connecting the three points forms a triangle with positive area?

Guest Mar 29, 2015

#2**+10 **

I am not over confident either but I will give it a go

There are 16 points in this region so there are 16C3 = 560 ways of choosing 3 of these points.

Now you have to take away all the sets of collinear points (those that form a line)

There a 4 points in every row and every column. And there are 4 rows and 4 columns

the number of ways that 3 collinear points can be chose from these is 8* 4C3 = 8*4 = 32

Now there are also 4 points in each of the central diagonals so this is 2*4C3 = 2*4=8

now the diagonal next to the cetral diagonal has 3 points and there are 4 of these = 1*4=4

so I think that the answer is

$${\mathtt{560}}{\mathtt{\,-\,}}{\mathtt{32}}{\mathtt{\,-\,}}{\mathtt{8}}{\mathtt{\,-\,}}{\mathtt{4}} = {\mathtt{516}}$$ ways that a triangle can be formed. :)

Melody
Mar 30, 2015

#1**+5 **

I'll give this one a shot......but....I'm not overly confident.....!!!

Notice the follwing patterns......

For a 1 x 1 square, we have two possible values fo x and y, 0 and 1. And we have two ways to pick an apex point and a base of 1. But, we can orient these two triangles in 4 ways......apex points at the "top," apex points to the "right," apeax points to the "left" and apex points on the "bottom"....and each will have a base of 1...so that's eight possible triangles

1x 1 = (2 apex points)*(1 base) * [4 orientations] = 8 possible triangles

For a 2 x 2 square, we have three possible values for x and y .... 0, 1, 2. And we have three possible apex points, three possible bases - two of one unit and one of two units, and three possible heights - two of one unit and one of two units. And these can be oriented in 4 different ways, as before.

So the possible triangles in a 2 x 2 configuration are....

2 x 2 = [3 apex points]*[3 possible bases]* [3 possible heights]* [4 possible orientations] = 108 possible triangles

And for our situation, we have a 3 x 3 square.

We have four possible values for x and y - 0,1,2,3. And we have four possible apex points, 6 possible bases and heights- three of one unit, two of two units and three of one unit.... and four possible orientations, as before.

So....the possible triangles are

3 x 3 = [4 apex points]* [6 bases] *[6 heights]*[4 possible orientations ] = 576 possible triangles

It appears that the pattern for an n x n arrangement is [n + 1]*[C(n+1, 2)]^2 * 4

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I hope this is correct........can somebody else check my reasoning ????

CPhill
Mar 29, 2015

#2**+10 **

Best Answer

I am not over confident either but I will give it a go

There are 16 points in this region so there are 16C3 = 560 ways of choosing 3 of these points.

Now you have to take away all the sets of collinear points (those that form a line)

There a 4 points in every row and every column. And there are 4 rows and 4 columns

the number of ways that 3 collinear points can be chose from these is 8* 4C3 = 8*4 = 32

Now there are also 4 points in each of the central diagonals so this is 2*4C3 = 2*4=8

now the diagonal next to the cetral diagonal has 3 points and there are 4 of these = 1*4=4

so I think that the answer is

$${\mathtt{560}}{\mathtt{\,-\,}}{\mathtt{32}}{\mathtt{\,-\,}}{\mathtt{8}}{\mathtt{\,-\,}}{\mathtt{4}} = {\mathtt{516}}$$ ways that a triangle can be formed. :)

Melody
Mar 30, 2015

#3**+5 **

Yep.....after reconsidering....I don't like my answer at all..I realize there might be some double - if not "triple" counting in mine....I'll look at Melody's answer in more detail......but.....just not right now.....

CPhill
Mar 30, 2015