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How many triangles on the coordinate plane

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by cmlee » Sun Feb 05, 2012 1:07 am
This is how i solved it felt it was simpler than other explanations.

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

Image

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

Image

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 !
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by intenseCK » Thu Feb 16, 2012 12:03 am
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?
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by Brent@GMATPrepNow » Thu Feb 16, 2012 6:55 am
intenseCK wrote: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?
For the OG12 PS#229 question, we cannot use 110C3 for three reasons:
1) We have a condition that says a certain side of the triangle must be parallel to the x-axis (110C3 ignores this condition)
2) We are told that triangle PQR is a right triangle (110C3 ignores this condition)
3) 3 points in a row (i.e., an area of zero) does not constitute a triangle

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by Hirshi » Tue Feb 21, 2012 10:10 am
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
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by Rastis » Wed Feb 22, 2012 5:40 am
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.
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by pavan.mpv » Thu Feb 23, 2012 2:48 am
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
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by spartacus1412 » Mon Mar 19, 2012 11:46 pm
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.
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by ronnie1985 » Wed Mar 21, 2012 9:54 am
Draw the grid to solve the problem. The problem appeared in Manhattan Challenge Problem.
Follow your passion, Success as perceived by others shall follow you
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by [email protected] » Wed Mar 21, 2012 5:08 pm
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.

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by shantanu86 » Tue Apr 24, 2012 2:06 am
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 :) :)
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by Chinmoy » Tue Jun 26, 2012 3:03 am
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.
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by Brent@GMATPrepNow » Tue Jun 26, 2012 6:54 am
Rastis wrote: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 Rastis,

Sorry for the delay.
9C3 is notation that we use for something called "combinations"
Basically, nCr represents that the number of ways we can select r objects from n objects.

So, 9C3 represents the # of ways we can select 3 points (to create a triangle) from a total of 9 points.

If you're interested, I created a free video on calculating combinations in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

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by gopal.prashad » Tue Jul 10, 2012 5:26 pm
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.
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by Shubhu@MBA » Fri Jul 13, 2012 2:24 pm
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 :)
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by jmckenna14 » Thu Aug 16, 2012 8:08 am
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,
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