+0  
 
0
2198
6
avatar

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

Best Answer 

 #2
avatar+94110 
+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
avatar+92744 
+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  

-------------------------------------------------------------------------------------------------------------------------

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

 

 

    

CPhill  Mar 29, 2015
 #2
avatar+94110 
+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
avatar+92744 
+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
 #4
avatar
0

Yes. I tried. 516 is the correct solution. No time to explain rn. I am just 100% sure that >516< is the correct solution.

Guest Jul 3, 2016
 #5
avatar
0

It can not be 516 because it has to be less than 4 not 4

Guest Mar 5, 2017
 #6
avatar
0

You just have to do 16 choose 3 to get the answer: 516 which is correct

Guest Mar 10, 2017

16 Online Users

avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.