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How many triangles with positive area can be drawn on the

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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?

(A) 72
(B) 76
(C) 78
(D) 80
(E) 84
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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

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)};

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

Answer: B.

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

Hope it helps.
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Re: Triangles and letters! [#permalink]

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New post 29 Jul 2010, 22:44
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

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)};

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

Answer: B.

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.

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Re: Triangles and letters! [#permalink]

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New post 01 Sep 2010, 00:59
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.

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New post 01 Sep 2010, 06:54
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.
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New post 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.

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New post 24 Jan 2011, 14:21
Bunuel, is there a quicker way to figure out the collinear points
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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.
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New post 24 Jan 2011, 14:42
yup ...got it ...made a smal mistake ...Thanks a lot
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Re: Triangles and letters! [#permalink]

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New post 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.

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New post 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.
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Re: Triangles and letters! [#permalink]

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New post 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?

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emailmkarthik wrote:
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.

Check similar questions to practice:
if-4-points-are-indicated-on-a-line-and-5-points-are-132677.html
abcde-is-a-regular-pentagon-with-f-at-its-center-how-many-d-86284.html
abcde-is-a-regular-pentagon-with-f-at-its-center-how-many-133328.html
the-sides-bc-ca-ab-of-triangle-abc-have-3-4-5-interior-109690.html
gmat-diagnostic-test-question-79373.html
m03-71107.html
how-many-triangles-with-positive-area-can-be-drawn-on-the-98236.html
right-triangle-abc-is-to-be-drawn-in-the-xy-plane-so-that-88958.html
how-many-circles-can-be-drawn-on-the-line-segment-128149.html
how-many-triangles-can-be-inscribed-in-the-heptagon-pictured-154277.html

Hope it helps.
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New post 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.

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New post 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

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)};

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

Answer: B.

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?

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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.

Answer: B.

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.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Triangles and letters! [#permalink]

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New post 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

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)};

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

Answer: B.

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.

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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

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Re: How many triangles with positive area can be drawn on the [#permalink]

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New post 03 Oct 2017, 00:57
Bunuel Not able to understand the elimination as well as how did we figure out the collinear points.

Can you please help!

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Re: How many triangles with positive area can be drawn on the [#permalink]

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siddyj94 wrote:
Bunuel Not able to understand the elimination as well as how did we figure out the collinear points.

Can you please help!


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)}.
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Re: How many triangles with positive area can be drawn on the   [#permalink] 03 Oct 2017, 01:11

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