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How many triangles on the coordinate plane

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by tomada » Sat Dec 18, 2010 1:35 pm
The question specifies integer coordinates
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?
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by Koala » Sun Dec 19, 2010 5:31 am
hitmis wrote:Why should we not have to also subtract combinations of (x1,y1) = (x2,y2)=(x3,y3) , 9 such combinations, which would also not create a triangle.?
When you consider 9C3, that means you want to calculate the number of sets of 3 different points among 9 different points.
Then, the combination you're talking about is not included in the 9C3.
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by Koala » Sun Dec 19, 2010 5:35 am
wydadi wrote:There are 9 points on the limited plane.

Each point can form 8 couple of points with the 8 remaining points.

If we consider the couple (A,B), we have 7 points left to form an (A,B,C) group. This group is a triangle only if C is not on the same line as A and B ==> means we have 6 points left (since there is one point on the line (AB) that is in the limited plane).

Then we have 9*8*6 groupe (A,B,C). Each group correspond to a triangle on the limited plane.

And since order is not important (triangle ABC is the same as triangle BAC), so each triangle (ABC) is repeated 6 times.

The result is (9*8*6)/6=9*8=72.
Here the error is in the blodface sentence. After selecting two points among the nine available, we don't always have 6 points left to form a triangle. We could have even 7 points to consider, since the first two points chosen are not always on "the same line" ==> for example, points A(1,1) and B(3,2) are not "on the same line" since we have 7 points left to form a triangle with.

The answer of Brent is the most accurate.
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by kumarkislay » Fri Dec 24, 2010 11:01 am
Koala wrote:
wydadi wrote:There are 9 points on the limited plane.

Each point can form 8 couple of points with the 8 remaining points.

If we consider the couple (A,B), we have 7 points left to form an (A,B,C) group. This group is a triangle only if C is not on the same line as A and B ==> means we have 6 points left (since there is one point on the line (AB) that is in the limited plane).

Then we have 9*8*6 groupe (A,B,C). Each group correspond to a triangle on the limited plane.

And since order is not important (triangle ABC is the same as triangle BAC), so each triangle (ABC) is repeated 6 times.

The result is (9*8*6)/6=9*8=72.
Here the error is in the blodface sentence. After selecting two points among the nine available, we don't always have 6 points left to form a triangle. We could have even 7 points to consider, since the first two points chosen are not always on "the same line" ==> for example, points A(1,1) and B(3,2) are not "on the same line" since we have 7 points left to form a triangle with.

The answer of Brent is the most accurate.
Agreed. And since there would be 4 such cases when 7 points would be available - [], [], [], [] => total number of possible triangles would be 72 + 4 = 76. This also matches with 9C3 - 3 (horiz lines) - 3 (vert lines)- 2 (diagonals) = 76.
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by hungpham » Sun Feb 13, 2011 4:58 am
The answer is C: 78

There are 9 points

Total = (9 x 8 x 7)/6 = 84

Not triangles = 6 including [(0,1) (0,2) (0,3)]; [(1,0) (2,0) (3,0)]; [(1,2) (2,1) (3,0)]; [(0,3) (1,2) (2,1)]; [(0,3) (1,2) (3,0)]; [(0,3) (2,1) (3,0)]

Result = 84 - 6 = 78

Question: If each vertex has to satisfy 1<=x<=3 and 1<=y<=3, there are only three points and there is only one triangle.
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by m.abdulk » Tue Mar 22, 2011 8:05 am
nice one... loved it.. thugh i was unable to solve it...
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by krishnakumar.ks » Thu Apr 07, 2011 11:28 pm
76 is definitely not the correct answer as the point made by hitmis is a valid one.
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by krishnakumar.ks » Thu Apr 07, 2011 11:30 pm
Y did 0 come into picture when 1<=x<=3 and 1<=y<=3
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by kruthika » Fri Apr 08, 2011 5:24 am
Could someone confirm the correct answer to this question? This is definitely confusing! I arrived at 76, but some of them say 78..
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by Hillel » Sat Apr 09, 2011 10:52 am
I got (B) 76 :

Actually, there are 9 available points/vertices.
Every triangle needs 3 vertices.
so choose 3 out of 9, this is:
9!
-----
(9-3)!3!

which gives 84.

But we have not to take into account all vertices on the same line

so we have 3 lines of x, 3 lines of y + 2 diagnals.

84-8 = 76
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by quantskillsgmat » Thu Apr 14, 2011 9:05 pm
i think ur answer is wrong.
my answer is 54.
becz the reason that u r giving 9c3-4.there are more three point sets that will nt form traingle.3 verical lines 3 horizontal lines and 2 dignols.
my approach is for any set of two points that lie on same line six traingle are possible.so for 9 points 54 traingle are possible.
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by quantskillsgmat » Thu Apr 14, 2011 10:26 pm
now i got ryt approach
answer is 76, this is how
9c3-8=76 there are 8 set of collinear point which will nt form traingles.u can consider a square 4 sides and two digonals.they cant form traingle becz they r collinear.
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by telberrak » Wed Apr 20, 2011 3:13 am
we can choose 9C3 = 84 combinations of 3 points (A,B,C) from the 9 points.
the number of combination of (A,B,C) that are on the same line are 8 (3 horizontals, 3 verticals, 2 diagonals)

the answer is 84 - 8 = 76 triangles
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by gmatscrewed » Mon May 09, 2011 10:18 am
i do not get it at all! How do you know that there are 9 points how do u count them? can somebody help me i feel very dumb right now.
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by pemdas » Mon May 09, 2011 10:47 am
the points can be easily plotted on x-y abscess, the points must be integer values (if we count them 9 only)
C(9,3)=9!/(3!*6!)=84
less the solid line combination sets (horizontals, verticals, diagonals) 8
76 triangles
Brent@GMATPrepNow wrote:Source: Beat The GMAT 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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