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Right triangle PQR is to be constructed in the xy-plane so [#permalink]
 24 Jun 2007, 21:29
Right triangle PQR is to be constructed in the xy-plane so that the right angle is at P and PR parallel to the x-axis. The x and y coordinates of P,Q,R are to be integers that satisfy the inequalities -4<=x<=5 and 6<=5<=16. How many different triangles with these properties could be constructed?
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Solution:
1)As -4<=x<=5 and 6<=y<=16, all triangles must be inside the rectangle with the dimension of 9x10.
2)We claim that from a rectangle we can make 4 distinct right triangles.
There are 45*55 (why?—there are 45 combinations for the side of a rectangle which is parallel to x-axis, analogously 55 combinations parallel to y-axis) rectangle combinations in that region. Thus, the number of right triangles is 4*45*55=9900. BINGO
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checkmate wrote: Solution:
1)As -4<=x<=5 and 6<=y<=16, all triangles must be inside the rectangle with the dimension of 9x10.
2)We claim that from a rectangle we can make 4 distinct right triangles.
There are 45*55 (why?—there are 45 combinations for the side of a rectangle which is parallel to x-axis, analogously 55 combinations parallel to y-axis) rectangle combinations in that region. Thus, the number of right triangles is 4*45*55=9900. BINGO
Great solution !! 
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We claim that from a rectangle we can make 4 distinct right triangles.
Is this a rule/limitation?
Could you please explain how you got 45 and 55??
Thanks!
checkmate wrote: Solution:
1)As -4<=x<=5 and 6<=y<=16, all triangles must be inside the rectangle with the dimension of 9x10.
2)We claim that from a rectangle we can make 4 distinct right triangles.
There are 45*55 (why?—there are 45 combinations for the side of a rectangle which is parallel to x-axis, analogously 55 combinations parallel to y-axis) rectangle combinations in that region. Thus, the number of right triangles is 4*45*55=9900. BINGO
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Yeah.. still not clear on the 55*45.. please explain.
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checkmate wrote: Solution:
1)As -4<=x<=5 and 6<=y<=16, all triangles must be inside the rectangle with the dimension of 9x10.
2)We claim that from a rectangle we can make 4 distinct right triangles.
There are 45*55 (why?—there are 45 combinations for the side of a rectangle which is parallel to x-axis, analogously 55 combinations parallel to y-axis) rectangle combinations in that region. Thus, the number of right triangles is 4*45*55=9900. BINGO
Simply amazing. This one stumped me. 45*55 aspect was truly good
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checkmate wrote: Solution:
1)As -4<=x<=5 and 6<=y<=16, all triangles must be inside the rectangle with the dimension of 9x10.
2)We claim that from a rectangle we can make 4 distinct right triangles.
There are 45*55 (why?—there are 45 combinations for the side of a rectangle which is parallel to x-axis, analogously 55 combinations parallel to y-axis) rectangle combinations in that region. Thus, the number of right triangles is 4*45*55=9900. BINGO
I feel sorry for the insufficient answer, I was browsing the net for another purpose. Here is the point I want to add: In my solution, I supposed that one of the legs of the triangle is parallel to x-axis. By the way, hypotenuse can also be parallel, can't it? We don't know which side is the PR line segment. I guess, there are still some triangles which sum up to 10065.
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[quote="AugiTh"]We claim that from a rectangle we can make 4 distinct right triangles.
Is this a rule/limitation?
Could you please explain how you got 45 and 55??
Thanks!
It's neigher rule nor limitation! each diagonal separate a rectangle into 2 right triangles, and there are 2 diagonals.
45 and 55 are combinations of 2 points out of 9 and 10, respectively
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checkmate wrote: checkmate wrote: Solution:
1)As -4<=x<=5 and 6<=y<=16, all triangles must be inside the rectangle with the dimension of 9x10.
2)We claim that from a rectangle we can make 4 distinct right triangles.
There are 45*55 (why?—there are 45 combinations for the side of a rectangle which is parallel to x-axis, analogously 55 combinations parallel to y-axis) rectangle combinations in that region. Thus, the number of right triangles is 4*45*55=9900. BINGO I feel sorry for the insufficient answer, I was browsing the net for another purpose. Here is the point I want to add: In my solution, I supposed that one of the legs of the triangle is parallel to x-axis. By the way, hypotenuse can also be parallel, can't it? We don't know which side is the PR line segment. I guess, there are still some triangles which sum up to 10065. I don't think hypotenuse can be parallel to x axis.
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Can u pls explain how u got 55*45.
Is the ans 4 *9 * 10 = 360??
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sidbidus wrote: Can u pls explain how u got 55*45.
Is the ans 4 *9 * 10 = 360??
I do not know how checkmate actually came up with 45 * 55, but I will take a shot.
The base of the triangle has two points P and Q. They cannot be the same point. So, for example, if we have P at -4, there are 9 possible values of Q (-3 to 5). Similarly, for P at -3, there are 8 possible points for Q. P cannot be more than 4. So total number of possible bases = 9+8+7+6+5+4+3+2+=45.
Similarly, we can count 55 points for the points P and R (perpendicular side).
Is this right checkmate ?
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sumande wrote: sidbidus wrote: Can u pls explain how u got 55*45.
Is the ans 4 *9 * 10 = 360?? I do not know how checkmate actually came up with 45 * 55, but I will take a shot. The base of the triangle has two points P and Q. They cannot be the same point. So, for example, if we have P at -4, there are 9 possible values of Q (-3 to 5). Similarly, for P at -3, there are 8 possible points for Q. P cannot be more than 4. So total number of possible bases = 9+8+7+6+5+4+3+2+=45. Similarly, we can count 55 points for the points P and R (perpendicular side). Is this right checkmate ? when p is at -3, Q can also be at -4 and (-2 to 5)
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sidbidus wrote: sumande wrote: sidbidus wrote: Can u pls explain how u got 55*45.
Is the ans 4 *9 * 10 = 360?? I do not know how checkmate actually came up with 45 * 55, but I will take a shot. The base of the triangle has two points P and Q. They cannot be the same point. So, for example, if we have P at -4, there are 9 possible values of Q (-3 to 5). Similarly, for P at -3, there are 8 possible points for Q. P cannot be more than 4. So total number of possible bases = 9+8+7+6+5+4+3+2+=45. Similarly, we can count 55 points for the points P and R (perpendicular side). Is this right checkmate ? when p is at -3, Q can also be at -4 and (-2 to 5) Agree with you. But that is why we are multiplying by 4.
Take a look at the file attached. There are four possible right angle traingles in a rectangle. We are counting for one and then multiplying by 4.
This takes care of your contention.
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sumande wrote: sidbidus wrote: sumande wrote: sidbidus wrote: Can u pls explain how u got 55*45.
Is the ans 4 *9 * 10 = 360?? I do not know how checkmate actually came up with 45 * 55, but I will take a shot. The base of the triangle has two points P and Q. They cannot be the same point. So, for example, if we have P at -4, there are 9 possible values of Q (-3 to 5). Similarly, for P at -3, there are 8 possible points for Q. P cannot be more than 4. So total number of possible bases = 9+8+7+6+5+4+3+2+=45. Similarly, we can count 55 points for the points P and R (perpendicular side). Is this right checkmate ? when p is at -3, Q can also be at -4 and (-2 to 5) Agree with you. But that is why we are multiplying by 4. Take a look at the file attached. There are four possible right angle traingles in a rectangle. We are counting for one and then multiplying by 4. This takes care of your contention. ya that's correct.
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Since -4<=x<=5 and 6<=y<=16,
For point P: There are 10 possible values of x and 11 possible values of y giving a total of 10*11=110 possibilities
For point Q: Since this point is vertically above point P, it has the same x value as point P and 10 possible y values giving a total of 1*10=10 possibilities
For Point R: Since PR is || to x axis, point R has the same y value as point P and 9 possible values for point x giving a total of 1*9=9 possibilities
Total possibilities for the 3 points = 110*10*9 =9900
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GK_Gmat wrote: Since -4<=x<=5 and 6<=y<=16,
For point P: There are 10 possible values of x and 11 possible values of y giving a total of 10*11=110 possibilities
For point Q: Since this point is vertically above point P, it has the same x value as point P and 10 possible y values giving a total of 1*10=10 possibilities
For Point R: Since PR is || to x axis, point R has the same y value as point P and 9 possible values for point x giving a total of 1*9=9 possibilities
Total possibilities for the 3 points = 110*10*9 =9900
Excelletn.....thanks a lot GK_Gmat!!!!
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sidbidus wrote: Can u pls explain how u got 55*45.
Is the ans 4 *9 * 10 = 360??
From yours you will only get right triangles with unit sides
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