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Right triangle PQR is to be constructed in the xy-plane so [#permalink]
24 Jun 2007, 20:29

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?

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

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

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

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

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.

[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

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.

the right angle is at P and PR parallel to the x-axis.

I don't think hypotenuse can be parallel to x axis.

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).

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)

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.

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.

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

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

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...