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How many triangles on the coordinate plane

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How many triangles on the coordinate plane Post Wed Jan 21, 2009 5:14 pm
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  • Lap #[LAPCOUNT] ([LAPTIME])
    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

    Answer: B

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    Last edited by Brent@GMATPrepNow on Tue Nov 04, 2014 7:39 am; edited 3 times in total

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    Post Wed Jan 21, 2009 6:13 pm
    A bit tricky.

    There are 9 points in the restricted plane. There are three points to a triangle.

    9C3 = 9!/3!*6! = 7*8*9/6 = 84

    There are four 3 point sets that don't create triangles (when x is the same for all points, or y is the same for all points).

    So 84-4 = 80

    Post Wed Jan 21, 2009 6:29 pm
    awesomeusername wrote:
    A bit tricky.

    There are 9 points in the restricted plane. There are three points to a triangle.

    9C3 = 9!/3!*6! = 7*8*9/6 = 84

    There are four 3 point sets that don't create triangles (when x is the same for all points, or y is the same for all points).

    So 84-4 = 80
    The first part of your solution looks good (9C3), but we need to subtract more than 4 3-point sets.

    Here's my full solution:

    First recognize that we need to choose 3 of the following 9 points to create a triangle.


    So, for example, if we choose these three points...

    ...we get this triangle.

    Likewise, if we choose these three points...

    ...we get this triangle.

    So, the question really comes down to "In how many ways can we select 3 of the 9 points?"
    Well, notice that the order of the 3 selected points does not matter. In other words, selecting the points (1,2), (1,3) and (3,2) will create the SAME TRIANGLE as selecting the points (3,2), (1,3) and (1,2).
    Since the order of the selected points does not matter, we can use combinations.

    We can select 3 points from 9 points in 9C3 ways ( = 84 ways)

    Aside: If anyone is interested, we have a free video on calculating combinations (like 9C3) in your head: http://www.gmatprepnow.com/module/gmat-counting?id=789

    Now, unfortunately, the correct answer is not 84, because not every selection of 3 points will yield a triangle. For example, if we select these 3 points...

    ...we get a straight line, NOT a triangle.

    So, we must subtract from 84 all of the 3-point selections that DO NOT yield triangles.

    To begin, if the 3 selected points are lined up vertically...

    ...then we don't get a triangle.
    There are 3 different ways to select three points to create a vertical line.

    Also, if the 3 selected points are lined up horizontally...

    ...then we don't get a triangle.
    There are 3 different ways to select three points to create a horizontal line.

    Finally, if the 3 selected points are lined up diagonally...

    ...then we don't get a triangle.
    There are 2 different ways to select three points to create a diagonal line.

    So, the total number of different triangles = 84 - 3 - 3 - 2
    = 76
    = B

    Cheers,
    Brent

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    Last edited by Brent@GMATPrepNow on Sun Oct 27, 2013 8:52 am; edited 1 time in total

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    Post Wed Jan 21, 2009 6:33 pm
    Aha, i missed that! Good question!

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    Post Thu Jan 22, 2009 9:44 am
    I did the same math and made the same mistake. I had drawn out the 9 points, but forgot to think about the straight line combinations.

    -Carrie

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    Post Sun Oct 31, 2010 7:53 pm
    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.?

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    Post Tue Nov 02, 2010 8:42 am
    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.

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    Post Tue Nov 02, 2010 7:42 pm
    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.
    why answer 72 is wrong, answer 86 is right. I do not know, please, help

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    Post Wed Nov 03, 2010 5:18 am
    Answer is 76 .

    9C3 - (3C3 * 4 ) - 4C3
    9C3 => To select all possibles combinations that can form triangles .Then subtract the 4 possibilties where all the 3 points are selected from a line ( with 3 points ) . Similarly subtract the possibility of selecting 3 points from the diagonal line ( with 4 points ). Hope I am correct with the explanation, sounds confusing to myself Smile

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    Post Thu Nov 04, 2010 12:19 pm
    Correct answer is 76 and this is how:

    1. There are total 9 points, lets call them point 1 thru point 9
    2. Point 1 can form a unique pair with rest 8 points
    3. Point 2 can form a unique pair with rest 8 points (but its pair with point 1 is already counted above). So total actual unique pairs for point 2 = 7
    4. Total unique pairs for point 3 = 6
    5. And so on...
    6. total pairs = 8+7+6+5+4+3+2+1 ==> 36
    7. Each unique pair (A,B) can form a unque triangle with any of the rest 7 points (lets represent a point as C) - ABC
    8. So total # of theoretical triangles = 36 * 7
    9. however each triangle A, B, C can be reach via - (I) AB connecting to C, (II) AC connecting to B _or_ (III) BC connecting to A - so each actual triangle ends up being counted 3 times in above calculation
    10. So total triangles = 36 * 7 / 3
    11. There are total of 8 straight line 'pseudo'-triangles (3 horizontal lines, 3 vertical lines and 2 diagonal lines)
    12. So actual triangle count = 36 * 7 / 3 - 8 ==>76

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    Post Thu Nov 11, 2010 4:22 pm
    I do not understand the logic behind these Daily GMAT Questions. The questions are only being answered by other students and never given a confirmed correct answer. If this is really the case, continuing in these discussions would be like the Blind Leading the Blind. Will the instructor giving these questions out ever give the correct answer?

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    Post Thu Nov 11, 2010 6:05 pm
    kmthurman@gmail.com wrote:
    I do not understand the logic behind these Daily GMAT Questions. The questions are only being answered by other students and never given a confirmed correct answer. If this is really the case, continuing in these discussions would be like the Blind Leading the Blind. Will the instructor giving these questions out ever give the correct answer?
    I agree..I subscribed to the daily questions yesterday and got the very first email today....It'll be really helpful if the correct answer is published too, with these daily questions...

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    Post Thu Nov 11, 2010 6:14 pm
    rajs2010 wrote:
    kmthurman@gmail.com wrote:
    I do not understand the logic behind these Daily GMAT Questions. The questions are only being answered by other students and never given a confirmed correct answer. If this is really the case, continuing in these discussions would be like the Blind Leading the Blind. Will the instructor giving these questions out ever give the correct answer?
    I agree..I subscribed to the daily questions yesterday and got the very first email today....It'll be really helpful if the correct answer is published too, with these daily questions...
    If the idea is to promote some healthy discussion first on these questions to explore all the possible ways one could go wrong,it might be a good idea to have the correct answer by the Moderator/Expert in next days email...

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    Post Thu Nov 18, 2010 9:49 am
    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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    Post Thu Dec 09, 2010 4:59 pm
    Hi, I don't understand why each triangle is repeated 6 times, and not 3 ?

    Edit : Finally it's ok, I got it

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