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Right triangle PQR is to be constructed in the xy-plane so that the ri
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23 Dec 2009, 19:52

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Right triangle PQR is to be constructed 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 and 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?

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12 Apr 2010, 05:23

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gmatJP wrote:

Can anyone please tell me how to figure this out...

Right triangle PQR is to be constructed 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 and R are to be integers that satisfy the inequalitites -4≤ X≤ 5 and 6≤ y≤ 16. How many different triangles with these properties could be constructed?

A)110 B)1100 C)9900 D)10000 E)12100

thanks..

We have the rectangle with dimensions 10*11 (10 horizontal dots and 11 vertical). PQ is parallel to y-axis and PR is parallel to x-axis.

Choose the (x,y) coordinates for vertex P (right angle): 10C1*11C1; Choose the x coordinate for vertex R (as y coordinate is fixed by A): 9C1, (10-1=9 as 1 horizontal dot is already occupied by A); Choose the y coordinate for vertex Q (as x coordinate is fixed by A): 10C1, (11-1=10 as 1 vertical dot is already occupied by A).

Re: Right triangle PQR is to be constructed in the xy-plane so that the ri
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23 Dec 2009, 20:50

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10

imo C total values for x=10;y=11

x1,y1=10*11.......................coordinates of 1st pnt x2,y2=9*1(y2=y1)............... coordinates of 2nd pnt y coordinates will be same as that of 1st pnt bcoz it is parallel to x axis x3,y3=1*10(x2=x3)..........coordinates of 3rd pt. x coordinates will be same as that of 2nd point bcoz to make a right angle it has to be parallel to y axis

Status: Waiting to hear from University of Texas at Austin

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Location: Changchun, China

Schools: University of Texas at Austin, Michigan State

Re: Right triangle PQR is to be constructed in the xy-plane so that the ri
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12 Jul 2010, 05:49

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2 more hours and I have it figured out.

Thinking about this on the coordinate plane confuses me more than just thinking about it as a combination problem.

P (A,B) R (C,B) Q (A,D)

So you need four values to solve the problem.

2 x-values {A and C}

2 y-values {B and D}

When you approach the question this way the orientation of the triangle is irrelevant. Because by using the same x-value for point P and point Q along with the same y-value for point point P and point R. The right angle is at P and PR is parallel to the x-axis.

So I need 2 x values from -4≤ X≤ 5. Pick the first one from 10 choices Pick the second one from 9 choices

10*9 = 90

and I need 2 y values from 6≤ y≤ 16 Pick the first one from 11 choices Pick the second one from 10 choices

11*10 = 110

So combine the 2 and we have 9900 (110*90) possible triangles.

I realized my error when I started to make a table of the possible coordinates for point P using the triangle in the upper right of the four choices

With point P and (4,6) I realized that I only have one possible length for PR *If P is (4,6) and the triangle has the orientation in the upper right of my image. *Then R must be (5,6)

That got me back to the starting line and my wonderful Chinese girlfriend helped me going in the correct direction.

She is junior in college and an English major by the way, with no math since high school.

Re: Right triangle PQR is to be constructed in the xy-plane so that the ri
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28 Jan 2011, 06:16

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Official answer is "C" - 9900

This was a hard question for me.

Anyway, I took the following approach to solve this.

****************************************

When y=6 We know -4<=x<=+5, there are 10 integral points. The points are: {-4,6},{-3,6},{-2,6},{-1,6},{0,6},{1,6},{2,6},{3,6},{4,6},{5,6} So, number of distinct line segments that can be formed are 10C2=45. And these line segments PR as it should be parallel to x-axis. With every line segment there can be 20 triangles:

Let's see how:

consider line segment with points: P={-4,6} and R={1,6} We know there are 10 points above P that can serve us as Q. Q should always be vertically above or below point P because P is the right angle. So Q can be: {-4,7},{-4,8},{-4,9},{-4,10},{-4,11},{-4,12},{-4,13},{-4,14},{-4,15},{-4,16} See for yourself that now: for just one segment PQ we have 10 triangles. PQR. However, we can flip the points on the same line segment PR, so that P={1,6} and R={-4,6} We know there are 10 points above P that can serve us as Q. Q should always be vertically above or below point P because P is the right angle. So in this case Q can be: {1,7},{1,8},{1,9},{1,10},{1,11},{1,12},{1,13},{1,14},{1,15},{1,16} See for yourself that for just one segment PQ we have another 10 triangles. So, a total of 20 distinct triangles for just one line segment.

thus: a total of 20 * 10C2 = 20 * 45 = 900 triangles for all line segments where PR is on y=6.

Remember, we were talking only about PR that lies on y=6 line.

There are 11 such lines between y=6 and y=16 i.e. 16-6+1.

Thus; a total of 900 * 11 = 9900 triangles can be formed with given conditions. *******************************************************************************

Re: Right triangle PQR is to be constructed in the xy-plane so that the ri
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18 Oct 2013, 06:51

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1

In the figure shown

When P is fixed , q can take 10 positions( remaining 10 co-ordinates up the y axis) and R can take 9 positions ( x-axis) . Hence 10*9 = 90 traingles and p itself can take - 10 *11 positions =110 hence 90*110 = 9900

Re: Right triangle PQR is to be constructed in the xy-plane so that the ri
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12 Apr 2010, 03:56

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Hello, Pls explain.

x1,y1=10*11.......................coordinates of 1st pnt x2,y2=9*1(y2=y1)............... coordinates of 2nd pnt y coordinates will be same as that of 1st pnt bcoz it is parallel to x axis x3,y3=1*10(x2=x3)..........coordinates of 3rd pt. x coordinates will be same as that of 2nd point bcoz to make a right angle it has to be parallel to y axis

you mentioned, y2=y1. but calculation value is different. same for x2 and x3.

Status: Waiting to hear from University of Texas at Austin

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Posts: 58

Location: Changchun, China

Schools: University of Texas at Austin, Michigan State

Re: Right triangle PQR is to be constructed in the xy-plane so that the ri
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11 Jul 2010, 21:38

This question as been bothering me for approaching 24 hours. Please help!

The question reads:

Quote:

Right triangle PQR is to be constructed 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 and R are to be integers that satisfy the inequalitites -4≤ X≤ 5 and 6≤ y≤ 16. How many different triangles with these properties could be constructed?

A)110 B)1100 C)9900 D)10000 E)12100

First: I see these four basic triangle orientations. I do not see why the question does not consider them.

Attachment:

4triangles.JPG [ 14.25 KiB | Viewed 22689 times ]

Each triangle shows PR parallel to the x-axis Each triangle shows the right angle at P

Second: If we exclude all the triangles except the one in the upper right. I keep coming up with only 8100 possible triangles

Because Point P and Point R share a y value, they must both be integers, and they must be different (the triangles must have 3 sides) Because Point P and Point Q share an x value, they must both be integers, and they must be different (the triangles must have 3 sides) We know Point P coordinates are ( -4≤ X≤ 4 , 6≤ y≤ 15)

Because we know that each x and each y must be integers, the two legs of the triangle have limited possible lengths

1 ≤ length PR ≤ 9

1 ≤ length PQ ≤ 10

With this in mind I find only 8100 possible triangles resulting from ( 90 coordinate pairs possible for P times 9 lengths for PR times 10 lengths for PQ = 90 * 9 * 10)

Re: Right triangle PQR is to be constructed in the xy-plane so that the ri
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07 Jan 2017, 14:19

This was a very hard question for me.

I know that the set of vertices that should satisfy the inequalities apply to all P,Q and R. I took a look at the options and noticed only C that could be divisible by 3.. since set of all the vertices/3. Hence picked C.

Re: Right triangle PQR is to be constructed in the xy-plane so that the ri
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01 Feb 2019, 21:59

lets say the coordinates of P, Q, R are (x1,y1), (x1,y2), (x2,y1).

Now we want (x1,x2) combinations * (y1,y2) combinations x1,x2 & y1,y2 can never be equal to form a triangle and they both satisfy -4≤ X≤ 5 and 6≤ y≤ 16.

x1,x2 combinations = 10 * 10 (total combinations) - 10(both are same) = 90 y1,y2 combinations = 11 * 11 (total combinations) - 11(when both are same) = 110

Hence total number of combinations = 90 * 110 = 9900

A perfect example of permutations and combinations with Co-ordinate geometry.

Re: Right triangle PQR is to be constructed in the xy-plane so that the ri
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01 Feb 2019, 23:39

gmatJP wrote:

Right triangle PQR is to be constructed 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 and 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?

A. 110 B. 1100 C. 9900 D. 10000 E. 12100

total x coordinates= 10 y coordinates= 11 so for P we can have = 10c1*11c1 since its a right angled triangle so point Q would be fixed and value of x =1 and y can be choosen in 10c1 ways for point R , value of x would be 1 less than 10 ; 9c1 and y would be fixed so 1 10c1*11c1*10c1*9c1 = 9900 IMO C

Re: Right triangle PQR is to be constructed in the xy-plane so that the ri
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25 May 2020, 07:22

Top Contributor

gmatJP wrote:

Right triangle PQR is to be constructed 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 and 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?

A. 110 B. 1100 C. 9900 D. 10000 E. 12100

Take the task of building triangles and break it into stages.

Stage 1: Select any point where the right angle will be (point P). The point can be selected from a 10x11 grid. So, there 110 points to choose from. This means that stage 1 can be completed in 110 ways.

Stage 2: Select a point that is on the same horizontal line as the first point. This point will be point R. The 2 legs of the right triangle are parallel to the x- and y-axes. The first point we select (in stage 1) dictates the y-coordinate of point R. In how many ways can we select the x-coordinate of point R? Well, we can choose any of the 10 coordinates from -4 to 5 inclusive EXCEPT for the x-coordinate we chose for point P (in stage 1). So, there are 9 coordinates to choose from. This means that stage 2 can be completed in 9 ways.

Stage 3: Select a point that is on the same vertical line as the first point. This point will be point Q. The 2 legs of the right triangle are parallel to the x- and y-axes. The first point we select (in stage 1) dictates the x-coordinate of point Q. In how many ways can we select the y-coordinate of point Q? Well, we can choose any of the 11 coordinates from 6 to 16 inclusive EXCEPT for the y-coordinate we chose for point P (in stage 1). So, there are 10 coordinates to choose from. This means that stage 3 can be completed in 10 ways.

So, by the Fundamental Counting Principle (FCP), the total number of triangles = (110)(9)(10) = 9900

Answer: C

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.