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How many triangles with positive area can be drawn on the
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29 Jul 2010, 20:09
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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?
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
Thanks for your kind help!
It would be better if you draw it while reading this explanation. With the restriction given (1≤x≤3 and 1≤y≤3) we get 9 points, from which we can form the triangle: (1,1), (1,2), (1,3), (2,1)...
From this 9 points any three (9C3) will form the triangle BUT THE SETS of three points which are collinear.
We'll have 8 sets of collinear points of three: 3 horizontal {(1,1),(2,1),(3,1)}; {(1,2)(2,2)(3,2)}; ... 3 vertical 2 diagonal {(1,1)(2,2)(3,3)}; {(1,3)(2,2)(3,1)};
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
Thanks for your kind help!
It would be better if you draw it while reading this explanation. With the restriction given (1≤x≤3 and 1≤y≤3) we get 9 points, from which we can form the triangle: (1,1), (1,2), (1,3), (2,1)...
From this 9 points any three (9C3) will form the triangle BUT THE SETS of three points which are collinear.
We'll have 8 sets of collinear points of three: 3 horizontal {(1,1),(2,1),(3,1)}; {(1,2)(2,2)(3,2)}; ... 3 vertical 2 diagonal {(1,1)(2,2)(3,3)}; {(1,3)(2,2)(3,1)};
For the first vertex you get 9 options. For the second one 8 options. For the third 6 options.
9 x 8 x 6
But since the order of vertices is not import, so we have to divide by 3!
So, (9x8x6)/3! = 72
Where's the mistake in my reasoning?
Thank you.
As I understand in 9*8*6 by the last 6 you try to get rid of collinear points for the first two chosen ones. But it's not always true, consider the following:
*** *** ***
If you choose red and blue for the first two dots then for the third one you'll have 7 choices, not 6. So there are cases when you have 9*8*7 and cases when you have 9*8*6. That is why you'll get incorrect answer.
Bunuel, is there a quicker way to figure out the collinear points
Points are collinear if they lie on a straight line. So I think it should be quite obvious that 3 vertical points of three, 3 horizontal points of three and 2 diagonal points of three are collinear. Total of 3+3+2=8 collinear points of three.
_________________
Bunuel, is there a quicker way to figure out the collinear points
Points are collinear if they lie on a straight line. So I think it should be quite obvious that 3 vertical points of three, 3 horizontal points of three and 2 diagonal points of three are collinear. Total of 3+3+2=8 collinear points of three.
In a plane if there are n points out of which m points are collinear, then the number of triangles that can be formed by joining them is nC3 - mC3
Here, n=9 and m=8(as we have 8 sets of points that are collinear(3 horizontal, 3 vertical, and 2 diagonal))
Therefore, the number of triangles is:
\(9C3 - 8C3 ----> 84 - 56 -----> 28\)
Where am I going wrong? Can someone help please! Thanks.
Bunuel, is there a quicker way to figure out the collinear points
Points are collinear if they lie on a straight line. So I think it should be quite obvious that 3 vertical points of three, 3 horizontal points of three and 2 diagonal points of three are collinear. Total of 3+3+2=8 collinear points of three.
In a plane if there are n points out of which m points are collinear, then the number of triangles that can be formed by joining them is nC3 - mC3
Here, n=9 and m=8(as we have 8 sets of points that are collinear(3 horizontal, 3 vertical, and 2 diagonal))
Therefore, the number of triangles is:
\(9C3 - 8C3 ----> 84 - 56 -----> 28\)
Where am I going wrong? Can someone help please! Thanks.
We don't have 8 points which are on one line (collinear). We have 8 3-POINT SETS which are collinear.
_________________
Thanks Bunuel. So what will the updated formula be in this case? Is it possible to use this formula for such type problems. Is there a generalized rule?
Thanks Bunuel. So what will the updated formula be in this case? Is it possible to use this formula for such type problems. Is there a generalized rule?
I don't think you need any special formulas for such kind of questions.
Bunuel, is there a quicker way to figure out the collinear points
Points are collinear if they lie on a straight line. So I think it should be quite obvious that 3 vertical points of three, 3 horizontal points of three and 2 diagonal points of three are collinear. Total of 3+3+2=8 collinear points of three.
Totally missed out on the collinearity of points and solved for the answer to be 9C3=84. Thanks Bunuel for throwing lights on excluding collinear points! And thanks again for the additional questions posted.
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
Thanks for your kind help!
It would be better if you draw it while reading this explanation. With the restriction given (1≤x≤3 and 1≤y≤3) we get 9 points, from which we can form the triangle: (1,1), (1,2), (1,3), (2,1)...
From this 9 points any three (9C3) will form the triangle BUT THE SETS of three points which are collinear.
We'll have 8 sets of collinear points of three: 3 horizontal {(1,1),(2,1),(3,1)}; {(1,2)(2,2)(3,2)}; ... 3 vertical 2 diagonal {(1,1)(2,2)(3,3)}; {(1,3)(2,2)(3,1)};
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
It would be better if you draw it while reading this explanation. With the restriction given (1≤x≤3 and 1≤y≤3) we get 9 points, from which we can form the triangle: (1,1), (1,2), (1,3), (2,1)...
From this 9 points any three (9C3) will form the triangle BUT THE SETS of three points which are collinear.
We'll have 8 sets of collinear points of three: 3 horizontal {(1,1),(2,1),(3,1)}; {(1,2)(2,2)(3,2)}; ... 3 vertical 2 diagonal {(1,1)(2,2)(3,3)}; {(1,3)(2,2)(3,1)};
Thanks a lot! I got stuck when trying to understand the set of three points which are collinear to make a triangle.
By the way, I like your avatar. It reminds me about my ancestors in my country.
I don't get the collinear thing. why you deduct 8 from 84?
If 3 points are collinear (are on the same line) they cannot form a triangle. We have 8 THREE-POINT SETS which are collinear, so 8 triangles cannot be formed.
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
Thanks for your kind help!
It would be better if you draw it while reading this explanation. With the restriction given (1≤x≤3 and 1≤y≤3) we get 9 points, from which we can form the triangle: (1,1), (1,2), (1,3), (2,1)...
From this 9 points any three (9C3) will form the triangle BUT THE SETS of three points which are collinear.
We'll have 8 sets of collinear points of three: 3 horizontal {(1,1),(2,1),(3,1)}; {(1,2)(2,2)(3,2)}; ... 3 vertical 2 diagonal {(1,1)(2,2)(3,3)}; {(1,3)(2,2)(3,1)};
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
inequalities 1≤x≤3 and 1≤y≤3 ; represents 9 points on the xy plane. now out of these 9 points total no. of triangles that can be formed =9C3=84
if we put all these points in the xy plane we will see that it will form a square of 3x3. also, point (1,1), (1,2),(1,3) forms the straight line, thus if three points selected lies on this line, then it will not result into a triangle. therefore we will have to subtract all such cases which are:
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27% (02:06) correct 
