Problem Solving

This is how i solved it felt it was simpler than other explanations.

the possible integere co-ordinates of (x,y) are :



So all triangles that can be formed by these co-ordinates can be calculated simply by selecting any 3 points out of the 9 = 9C3

But in the question we are asked to find the triangles with positive area and hence we need to remove the triangles with zero area (i.e when all the co-ordinates of the triangle are co-linear) which are included in those 9C3



Such possibilities are shown in the image and they are 8 in number (3 green 3 red 2 blue). These are triangle with area=0 and hence need to subtracted from 9C3=84

So 84-8=76 !

this question is confusing and so is the OG Edition 12 PS#229.
why that question can't be solved using the 110C3 method? that question also doesnt pose any condition also as to the area shd be positive as well?

Ans 1.

Brent@GMATPrepNow wrote:Source: Magoosh Practice Questions

How many triangles with positive area can be drawn on the coordinate plane such that the vertices have integer coordinates (x,y) satisfying 1≤x≤3 and 1≤y≤3?
(A) 72
(B) 76
(C) 78
(D) 80
(E) 84

Can anyone post a better explanation of how to get the answer? I could not bring up the answer from the original GMAT expert bc it was blocked at my work comp. What is 9C3? I cannot follow any of this logic.

Hi guys. I am new to this forum. Maybe I am dumb. I didn't understand why everyone is starting with 9 points. Can any one of you take time and explain me in detail?

Thanks,
Pavan

the range of value that x or y can take are from : 1 to 3.
x:1, 2, 3
y:1, 2, 3
hence integer points available for constructing a triangle are:
(1,1),(1,2),(1,3),(2,1)..(2,3),(3,1)...(3,3)
total number of triangle can be found by selecting any three points from the available sample space - when the points chosen lie on a straight line.

9C3-8(4 edges + 2 diagonals + 2 mid-point connecting lines)
that make sit 84 - 8 =76.

Draw the grid to solve the problem. The problem appeared in Manhattan Challenge Problem.

Brent I tried solving this question the grockit way... There is a similar question for squares and rectangles.

But for the triangles, solving for each and every point is very complicated.

What explanation Brent gave is the correct and concise explanation.

No need to worry abt this...

IMO its 76..

X e [1,3] & Y e [1,3] means we have 9 points in total to choose from.

9C3= 84 total triangles..

but triangles with collinear vertices have 0 area.
Total such triangles- 4 (sides of square) + 2 (diagonals of square) + 2(side bisectors) = 8

Therefore positive area triangles= 84 - 8 = 76

I did the math correct regarding 9C3. But i think we should subtract 9C2 to get rid of all the possible straight lines. Please suggest whether i am correct or not. According to me the answer which i am getting is 9C3-9C2 = 46. Thanks in advance.

AkshayaChandan wrote:I have selected the option E as 84 is highest count of triangle in option. The logic behind it is for the given limits of value we can have infinite number of real number combinations. for eg keeping Y fixed x co-ordinate can have the value between 1 to 3. It isn't a mandatory condition that we should select whole number. Is it?

I just thought the same way.
But if read closely it says integer coordinates. which makes it mandatory condition to select from whole numbers only.

Let's call our triangle as ABC.
Now if we want to select A & B co-ordinates from same line(i.e. x co-ordinate will remain same), we can do it in 3C2 ways.
We can select the 3rd co-ordinate (point C) from second or third line 2 * 3C1

This logic will give us 54 triangles
=> 2(3C2 * 3C1) + 2(C2 * 3C1) + 2(C2 * 3C1) = 54

Second way to form triangles will be by selecting 1 point from each line
3C1 * 3C1 * 3C1 = 27

But,there will be 5 straight lines with this logic.

Thus,total number of triangles = 54+27-5=76

Hence, answer to this question will be 76

I wish I had time to upload an image to explain this

Why did Brent subtract the number of straight lines from the combination?

I understand there are 84 total combinations because of 9 total points and 3 points to a triangle (subtracting redundancies), so 9!/3! = 84. But why do we subtract the number of straight lines? Confused.

@Brent can you explain?

Thanks,