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Right triangle PQR is to be constructed in the xyplane so [#permalink]
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24 Jun 2007, 21:29
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Right triangle PQR is to be constructed in the xyplane so that the right angle is at P and PR parallel to the xaxis. 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?



Intern
Joined: 27 Jun 2007
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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 xaxis, analogously 55 combinations parallel to yaxis) rectangle combinations in that region. Thus, the number of right triangles is 4*45*55=9900. BINGO



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Joined: 04 Jun 2007
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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 xaxis, analogously 55 combinations parallel to yaxis) rectangle combinations in that region. Thus, the number of right triangles is 4*45*55=9900. BINGO
Great solution !!



Manager
Joined: 23 Apr 2007
Posts: 132

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 xaxis, analogously 55 combinations parallel to yaxis) rectangle combinations in that region. Thus, the number of right triangles is 4*45*55=9900. BINGO



Senior Manager
Joined: 21 Jun 2006
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Yeah.. still not clear on the 55*45.. please explain.



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Joined: 03 Jun 2007
Posts: 376

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 xaxis, analogously 55 combinations parallel to yaxis) 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



Intern
Joined: 27 Jun 2007
Posts: 6

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 xaxis, analogously 55 combinations parallel to yaxis) 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 xaxis. 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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Joined: 27 Jun 2007
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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



Senior Manager
Joined: 19 Feb 2007
Posts: 325

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 xaxis, analogously 55 combinations parallel to yaxis) 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 xaxis. 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.
the right angle is at P and PR parallel to the xaxis.
I don't think hypotenuse can be parallel to x axis.



Senior Manager
Joined: 19 Feb 2007
Posts: 325

Can u pls explain how u got 55*45.
Is the ans 4 *9 * 10 = 360??



Senior Manager
Joined: 04 Jun 2007
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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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Joined: 19 Feb 2007
Posts: 325

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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Joined: 04 Jun 2007
Posts: 345

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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Joined: 19 Feb 2007
Posts: 325

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.



Director
Joined: 09 Aug 2006
Posts: 755

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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Joined: 23 Apr 2007
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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!!!!



Intern
Joined: 27 Jun 2007
Posts: 6

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










