GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 26 Jan 2020, 11:31

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# How many triangles with positive area can be drawn on the

Author Message
TAGS:

### Hide Tags

Intern
Joined: 09 Dec 2008
Posts: 22
Location: Vietnam
Schools: Somewhere
How many triangles with positive area can be drawn on the  [#permalink]

### Show Tags

29 Jul 2010, 20:09
9
76
00:00

Difficulty:

95% (hard)

Question Stats:

27% (02:06) correct 73% (02:26) wrong based on 816 sessions

### HideShow timer Statistics

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
Math Expert
Joined: 02 Sep 2009
Posts: 60646

### Show Tags

29 Jul 2010, 20:34
26
40
dungtd wrote:
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)};

So the final answer would be; 9C3-8=84-8=76.

Similar problems with different solutions:
arithmetic-og-question-88380.html
700-question-94644.html
tough-problem-88958.html

Hope it helps.
_________________
##### General Discussion
Intern
Joined: 09 Dec 2008
Posts: 22
Location: Vietnam
Schools: Somewhere

### Show Tags

29 Jul 2010, 22:44
1
Bunuel wrote:
dungtd wrote:
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)};

So the final answer would be; 9C3-8=84-8=76.

Similar problems with different solutions:
arithmetic-og-question-88380.html
700-question-94644.html
tough-problem-88958.html

Hope it helps.

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.
Director
Joined: 23 Apr 2010
Posts: 502

### Show Tags

01 Sep 2010, 00:59
1
Bunuel, I get 72. Here's my logic:

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.
Math Expert
Joined: 02 Sep 2009
Posts: 60646

### Show Tags

01 Sep 2010, 06:54
1
nonameee wrote:
Bunuel, I get 72. Here's my logic:

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.

Hope it's clear.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 60646

### Show Tags

24 Jan 2011, 14:34
1
1
ajit257 wrote:
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.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 60646

### Show Tags

02 Sep 2013, 08:08
1
zachowdhury wrote:
dungtd wrote:
Bunuel wrote:
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)};

So the final answer would be; 9C3-8=84-8=76.

Similar problems with different solutions:
arithmetic-og-question-88380.html
700-question-94644.html
tough-problem-88958.html

Hope it helps.

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.

Hope it's clear.
_________________
Senior Manager
Joined: 13 Jun 2013
Posts: 253
Re: Difficult questions' Easy solutions  [#permalink]

### Show Tags

24 Jun 2014, 02:49
1
2
sameer_kalra wrote:
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:

3 (3 rows) +3 (3 columns) +2 (two diagonals)

therefore total no. of triangles are 84-8=76
Math Expert
Joined: 02 Sep 2009
Posts: 60646
Re: How many triangles with positive area can be drawn on the  [#permalink]

### Show Tags

03 Oct 2017, 01:11
1
siddyj94 wrote:
Bunuel Not able to understand the elimination as well as how did we figure out the collinear points.

If 3 points are collinear (are on the same line) they cannot form a triangle. We have 8 THREE-POINT SETS which are collinear:

3 horizontal: {(1,1), (2,1), (3,1)}; {(1,2) (2,2) (3,2)}; {(1,3) (2,3) (3,3)}.

3 vertical: {(1,1), (1,2), (1,3)}; {(2,1) (2,2) (2,3)}; {(3,1) (3,2) (3,3)}.

2 diagonal {(1,1) (2,2) (3,3)}; {(1,3) (2,2) (3,1)}.
_________________
examPAL Representative
Joined: 07 Dec 2017
Posts: 1155
How many triangles with positive area can be drawn on the  [#permalink]

### Show Tags

31 Dec 2017, 02:36
1
Rw27 wrote:
Can anyone please highlight the difference between the question under discussion and this question https://gmatclub.com/forum/triangle-abc ... 32573.html

How do we distinguish between the two?

The main difference is the constraints.

In this question you are free to choose any 3 points on the plane (and then have to remove colinearities).
This means you choose 3 x coordinates and 3 y coordinates.
So the calculation is 9C3 - colinearities.
Colinearities in this problem are when all three points are on the same line and as explained above there are 8 such points.

In the question you linked to the constraints are much stronger:
Essentially, once you choose points A and C then point B is fixed.
So you're only choosing 2 x coordinates and 2 y coordinates.
You can do this as explained in the link - 6C1 * 11C1 * 5C1 * 10C1 = 3300
You can also calculate 66*(65-colinearities) = 66*50 = 3300
Colinearities in this problem are when points A and C are on the same line and there are 15 such options (10 horizontal, 5 vertical)

If all this is confusing, the bottom line is that the constraints are the main difference - are you choosing 3 points or 2?
Good luck!
_________________
Director
Joined: 23 Apr 2010
Posts: 502

### Show Tags

01 Sep 2010, 07:03
Bunuel, thanks a lot. Yes, I've been trying to get rid of colinear elements. But I haven't considered the case you mentioned.
Manager
Joined: 28 Aug 2010
Posts: 145

### Show Tags

24 Jan 2011, 14:21
Bunuel, is there a quicker way to figure out the collinear points
Manager
Joined: 28 Aug 2010
Posts: 145

### Show Tags

24 Jan 2011, 14:42
yup ...got it ...made a smal mistake ...Thanks a lot
Manager
Joined: 31 Mar 2013
Posts: 60
Location: India
GPA: 3.02

### Show Tags

07 Jul 2013, 07:10
Bunuel wrote:
ajit257 wrote:
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.
Math Expert
Joined: 02 Sep 2009
Posts: 60646

### Show Tags

07 Jul 2013, 08:12
emailmkarthik wrote:
Bunuel wrote:
ajit257 wrote:
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.
_________________
Manager
Joined: 31 Mar 2013
Posts: 60
Location: India
GPA: 3.02

### Show Tags

07 Jul 2013, 09:30
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?
Math Expert
Joined: 02 Sep 2009
Posts: 60646

### Show Tags

07 Jul 2013, 09:37
Intern
Joined: 09 Jun 2012
Posts: 27

### Show Tags

25 Jul 2013, 03:36
Bunuel wrote:
ajit257 wrote:
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.
Intern
Joined: 28 Mar 2011
Posts: 7

### Show Tags

02 Sep 2013, 08:02
dungtd wrote:
Bunuel wrote:
dungtd wrote:
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)};

So the final answer would be; 9C3-8=84-8=76.

Similar problems with different solutions:
arithmetic-og-question-88380.html
700-question-94644.html
tough-problem-88958.html

Hope it helps.

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?
Intern
Joined: 07 Jan 2013
Posts: 15
Location: Poland
GRE 1: Q161 V153
GPA: 3.8

### Show Tags

23 Oct 2013, 12:50
Bunuel wrote:
dungtd wrote:
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)};

So the final answer would be; 9C3-8=84-8=76.

Similar problems with different solutions:
arithmetic-og-question-88380.html
700-question-94644.html
tough-problem-88958.html

Hope it helps.

OMG. This question was so tough to understand. Thank you Bunuel.
Re: Triangles and letters!   [#permalink] 23 Oct 2013, 12:50

Go to page    1   2    Next  [ 24 posts ]

Display posts from previous: Sort by