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14 Nov 2004, 14:09
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Right traingle PQR is to be constrcuted in the xy- plane so that the right angle is at P and PR is 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<=y<=16. How many different triangles with these properties could be constructed?
6<=y<=16. How many different triangles with these properties could be constructed?
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14 Nov 2004, 19:08
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This problem OG seems to be causing much trouble but it is really not that complicated. I already solved this in the past but will try to be more precise in my explanation this time. In the below diagram, you will see that there are 4 variables you need to find {x1,x2,y1,y2}. Remember that right angle has to be at P. P will then have the same x-value as Q (x1 it is) and R will have the same y-value as P (y1 it is). Now, this is where this will become a permutation problem.
Let's start with the x's:
You have 10 ways of picking x1 for P(from -4 to 5 inclusive)
You have 9 remaining ways of picking x2 for R
Total permutations for x's: 10*9 ways of picking 2 values where order counts
For y's:
You have 11 ways of picking y1 for P(from 6 to 16 inclusive)
You have 10 remaining ways of picking y2 for Q
Total permutations for y's: 11*10
Total permutations, number of ways of picking 4 values, x1, x2, y1, y2, to form a right triangle: 11*10*10*9 = 9900
Let's start with the x's:
You have 10 ways of picking x1 for P(from -4 to 5 inclusive)
You have 9 remaining ways of picking x2 for R
Total permutations for x's: 10*9 ways of picking 2 values where order counts
For y's:
You have 11 ways of picking y1 for P(from 6 to 16 inclusive)
You have 10 remaining ways of picking y2 for Q
Total permutations for y's: 11*10
Total permutations, number of ways of picking 4 values, x1, x2, y1, y2, to form a right triangle: 11*10*10*9 = 9900
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15 Nov 2004, 00:54
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Thanks a lot Paul. That was really really helpful. Couldnt understand a thing from the explanation provided by the OG. :-D
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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
