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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? (A) 72 (B) 76 (C) 78 (D) 80 (E) 84
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Re: Triangles and letters!
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29 Jul 2010, 20:34
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; 9C38=848=76. Answer: B. Similar problems with different solutions: arithmeticogquestion88380.html700question94644.htmltoughproblem88958.htmlHope it helps.
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Re: Triangles and letters!
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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; 9C38=848=76. Answer: B. Similar problems with different solutions: arithmeticogquestion88380.html700question94644.htmltoughproblem88958.htmlHope 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!
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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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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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Re: Triangles and letters!
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24 Jan 2011, 14:34
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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Re: Triangles and letters!
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02 Sep 2013, 08:08
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; 9C38=848=76. Answer: B. Similar problems with different solutions: arithmeticogquestion88380.html700question94644.htmltoughproblem88958.htmlHope 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 THREEPOINT SETS which are collinear, so 8 triangles cannot be formed. Hope it's clear.
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24 Jun 2014, 02:49
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 848=76



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Re: How many triangles with positive area can be drawn on the
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03 Oct 2017, 01:11
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 THREEPOINT 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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How many triangles with positive area can be drawn on the
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31 Dec 2017, 02:36
Rw27 wrote: Can anyone please highlight the difference between the question under discussion and this question https://gmatclub.com/forum/triangleabc ... 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*(65colinearities) = 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!
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Re: Triangles and letters!
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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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Re: Triangles and letters!
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24 Jan 2011, 14:21
Bunuel, is there a quicker way to figure out the collinear points



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24 Jan 2011, 14:42
yup ...got it ...made a smal mistake ...Thanks a lot



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Re: Triangles and letters!
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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  mC3Here, 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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Re: Triangles and letters!
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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  mC3Here, 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 3POINT SETS which are collinear.
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Re: Triangles and letters!
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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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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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Re: Triangles and letters!
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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; 9C38=848=76. Answer: B. Similar problems with different solutions: arithmeticogquestion88380.html700question94644.htmltoughproblem88958.htmlHope 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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Re: Triangles and letters!
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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; 9C38=848=76. Answer: B. Similar problems with different solutions: arithmeticogquestion88380.html700question94644.htmltoughproblem88958.htmlHope it helps. OMG. This question was so tough to understand. Thank you Bunuel.




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