knewton, error in question?

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knewton, error in question?

by qwe12 » Mon Aug 02, 2010 10:52 pm
Question #23 (correct)
Right triangle LMN is to be constructed in the xy-plane so that the right angle is at point L and LM is parallel to the x-axis. The x- and y- coordinates of L, M, and N are to be integers that satisfy the inequalities -3 < x < 4 and 3 < y < 11. How many different triangles with these properties could be constructed?


(A) 72

(B) 576

(C) 4032 your answer correct

(D) 4608

(E) 6336
Explanation
We are asked to find out how many possible triangles could exist within an area on the xy-plane horizontally bounded by x = -3 and x = 4 (8 possible integer values of x, including 0) and by y = 3 and y = 11 vertically (9 possible integer values). Point L lies on one of the points within this rectangular area, which means that L can lie in any one of 72 possible points (8 x 9).

Because we know that LM is parallel to the x-axis (a horizontal line), we know that point M always has the same y-value as L. We do not know the orientation of the triangle, so point M could be either to the left or right of point L on the plane, meaning that it could lie on any point on the same horizontal line as L, within the boundaries of the problem, except the point on which L already lies. This means that there are 7 possible values of point M for every point that L lies on. We can thus multiply the 72 possible values of L with the 7 possible values of M to get all the possible placements of L and M: 504.

********** POSSIBLE ERROR
********** So shouldn't this be 8 possible values for point M? So 72 * 8 instead of 72* 7?
********** POSSIBLE ERROR

Every one of these possible triangles has a right angle at L, which means that point N has the same x-coordinate as L. However, it could be above or below L, and can thus be in any of 8 different points within the boundaries (the 9 possible y-values minus the one on which point L lies). This means that for every combination of L and M, there are 8 different triangles that can be made. We multiply 504 by 8 to get 4032.

Answer choice C is correct.

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by Ian Stewart » Mon Aug 02, 2010 11:11 pm
qwe12 wrote:Question #23 (correct)
Right triangle LMN is to be constructed in the xy-plane so that the right angle is at point L and LM is parallel to the x-axis. The x- and y- coordinates of L, M, and N are to be integers that satisfy the inequalities -3 < x < 4 and 3 < y < 11. How many different triangles with these properties could be constructed?


(A) 72

(B) 576

(C) 4032 your answer correct

(D) 4608

(E) 6336
Explanation
We are asked to find out how many possible triangles could exist within an area on the xy-plane horizontally bounded by x = -3 and x = 4 (8 possible integer values of x, including 0) and by y = 3 and y = 11 vertically (9 possible integer values). Point L lies on one of the points within this rectangular area, which means that L can lie in any one of 72 possible points (8 x 9).

Because we know that LM is parallel to the x-axis (a horizontal line), we know that point M always has the same y-value as L. We do not know the orientation of the triangle, so point M could be either to the left or right of point L on the plane, meaning that it could lie on any point on the same horizontal line as L, within the boundaries of the problem, except the point on which L already lies. This means that there are 7 possible values of point M for every point that L lies on. We can thus multiply the 72 possible values of L with the 7 possible values of M to get all the possible placements of L and M: 504.

********** POSSIBLE ERROR
********** So shouldn't this be 8 possible values for point M? So 72 * 8 instead of 72* 7?
********** POSSIBLE ERROR

Every one of these possible triangles has a right angle at L, which means that point N has the same x-coordinate as L. However, it could be above or below L, and can thus be in any of 8 different points within the boundaries (the 9 possible y-values minus the one on which point L lies). This means that for every combination of L and M, there are 8 different triangles that can be made. We multiply 504 by 8 to get 4032.

Answer choice C is correct.
That looks okay to me. While there are 8 different possible integer values of x in the range -3 < x < 4, once we've chosen a location for point L, we can't put M in the same place as L. That leaves only seven remaining values for the x coordinate of M.
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by sandysai » Wed Aug 04, 2010 7:15 pm
Hi

One basic question;

Given question "The satisfy the inequalities -3 < x < 4 and 3 < y < 11. How many different triangles with these properties could be constructed? "

The possible values of X = (-2,-1,0,1,2,3) --> That is 6 possible values.. isn't it ? How was 8 derived at.

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by Ian Stewart » Wed Aug 04, 2010 9:32 pm
sandysai wrote:Hi

One basic question;

Given question "The satisfy the inequalities -3 < x < 4 and 3 < y < 11. How many different triangles with these properties could be constructed? "

The possible values of X = (-2,-1,0,1,2,3) --> That is 6 possible values.. isn't it ? How was 8 derived at.
I'm guessing, reading the solution, that there was a typo in the original post; those '<' symbols are surely supposed to be '<' symbols.
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by qwe12 » Wed Aug 04, 2010 10:15 pm
Ian Stewart wrote:
sandysai wrote:Hi

One basic question;

Given question "The satisfy the inequalities -3 < x < 4 and 3 < y < 11. How many different triangles with these properties could be constructed? "

The possible values of X = (-2,-1,0,1,2,3) --> That is 6 possible values.. isn't it ? How was 8 derived at.
I'm guessing, reading the solution, that there was a typo in the original post; those '<' symbols are surely supposed to be '<' symbols.
yes, ian there was a typo. a copy paste error. it must be <=. thx