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# Right triangle PQR is to be constructed in the xy-plane so

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Right triangle PQR is to be constructed in the xy-plane so [#permalink]

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23 Dec 2009, 20:52
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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 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
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Re: Arithmetic/OG question.. [#permalink]

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23 Dec 2009, 21:50
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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

tot ways=10*11*9*1*1*10=9900

oa pls
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Re: Arithmetic/OG question.. [#permalink]

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12 Apr 2010, 04:56
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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.
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Re: Arithmetic/OG question.. [#permalink]

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12 Apr 2010, 06:23
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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).

10C1*11C1*9C1*10C1=9900.

Answer: C.
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Re: Arithmetic/OG question.. [#permalink]

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12 Apr 2010, 13:50
10P2 *11C1*10C1 gives the same

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Re: Arithmetic/OG question.. [#permalink]

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16 Apr 2010, 08:23
[quote="xcusemeplz2009"]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

tot ways=10*11*9*1*1*10=9900

Gr8...how do you get such insight...lots of practice ?

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Re: Arithmetic/OG question.. [#permalink]

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17 Apr 2010, 23:57
target2011 wrote:
Hello,
Pls explain.
you mentioned, y2=y1. but calculation value is different. same for x2 and x3.

y2=y1 means the value of y2 will be same as that of y1 , therefore the # of ways in which we can select y2 is only 1
and that for x2 will be 9 as one value is taken by x1
hth
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Re: Arithmetic/OG question...Too many triangles! [#permalink]

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11 Jul 2010, 07:23
First a picture to summarize the discussion so far.
Attachment:

Triangles.JPG [ 8.26 KiB | Viewed 14901 times ]

I disagree with 10 possible x values for P, because the length of PR must be an integer

P(X1,Y1)

X1 can be {-4,4} but not 5
Y1 can be {6,15} but not 15

I actually come up with 8100 triangles the regular way

And then a question

What in the question excludes a triangle like this, and why is this triangle not considered different?

Attachment:

Whynot.JPG [ 3.51 KiB | Viewed 14892 times ]

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Right Triangle PQR, OG11 Q248 (Why not 4 Triangles?) [#permalink]

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11 Jul 2010, 22:38
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 14958 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)

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Re: Right Triangle PQR, OG11 Q248 (Why not 4 Triangles?) [#permalink]

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12 Jul 2010, 06:49
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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.

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Re: Applied Geometry PS, OG12 #229 - Better Explanation Wanted! [#permalink]

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18 Sep 2010, 03:45
X can have 10 values & Y can 11 values.
Just choose a random Point P. Now, P has both x & y values. So, P can have 11*10=110 values.
PR is vertical. So, the y coordinates of P & R would be equal. Since PQR is a right triangle, QR would be perpendicular to the X axis & eventually parallel to the Y axis. So, the X coordinates of P & Q would be the same.

Since, we've fixed P. Q has the same value for x as that of P. So, there will be 10 (since 1 value of y is already taken by P) possible options for y coordinates for Q. Similarly, as P & R have the same value for y coordinates, R will have 9 possible values for x.
So, the number of triangles would be 110*9*10=9,900.
We're just calculating the number of variations after fixing a point.. P in this case. You'd get the solution if you fixed either Q or R.

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Re: Tough Coordinate Geo Problem. OG PS #229 [#permalink]

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30 Dec 2010, 01:25
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28 Jan 2011, 07:16
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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.
*******************************************************************************

~fluke
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Re: Arithmetic/OG question.. [#permalink]

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31 Jan 2011, 11:40
I could not solve this problem in the allotted time, but here is what my tutor says - actually same as aneeshdatar suggested, but in other words:

It is an area bounded by x = -4 and x = 5 (10 options for x) and by y = 6 and y = 16 (11 options for y). Point P can have 110 possible points (10*11).

Since PR is parallel to the x-axis, we know that point R must have the same y-coordinate as P. Point R could be either to the left or right of point P on the plane, i.e. it could lie on any point on the same horizontal line as P, save for the point on which P already lies. This means that there are 9 possible values of point R for every point P. By multiplication of 110 possible values of P with the 9 possible values of R, we get 990 possible placements.

Whereas all triangles have a right angle at P, point Q has the same x-coordinate as P. Following the same principle as above, we can conclude that point Q can be in any of 10 different points (11-1). We multiply 990 by 10 and get 9900.

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Re: Right triangle PQR is to be constructed in the xy-plane so [#permalink]

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Re: Right triangle PQR is to be constructed in the xy-plane so [#permalink]

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11 Oct 2013, 22:09
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

CHECK attached file for clear explanation.
Attachments

Solution 102.docx [21.72 KiB]
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Re: Right triangle PQR is to be constructed in the xy-plane so [#permalink]

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18 Oct 2013, 07:51
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
Attachments

traingle.png [ 4.86 KiB | Viewed 11114 times ]

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Re: Right triangle PQR is to be constructed in the xy-plane so [#permalink]

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08 Mar 2014, 05:50
hi, can you please help me to find error with below approach.

here we need to select 3 points such that
A = (x,y)
B = (x1,y)
C = (x,y1)

So effectively we need to select 2 values of X : x and x1 and 2 values of Y : y and y1

2 values of X can be selected in 10C2 ways = 45
2 values of Y can be selected in 11C2 ways = 55

so number of triangles will be 45x55 = 2475
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Re: Right triangle PQR is to be constructed in the xy-plane so [#permalink]

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15 Apr 2014, 23:56
Hi ankur1901,

We are not doing any selection there. We are just finding how many different triangles with the given inequality -4≤ X≤ 5 and 6≤ y≤ 16 could be constructed

If we assume point P to be fixed, coordinates for vertex P (right angle): 10C1*11C1;
PR is parallel to the x-axis , so no need to choose the Y value for Co-ordinate R. x coordinate for vertex R: 9C1;
Y coordinate for vertex Q :10C1; (X coordinate already fixed by point P).
10C1*11C1*9C1*10C1=9900.

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Re: Right triangle PQR is to be constructed in the xy-plane so [#permalink]

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14 Jul 2014, 02:09
I still can't get the solution. I do understand the chosen approach but still think there ist something wrong.

The vertices of a triangle are named consecutive and counterclockwise.

Given that PR has to be horizontal and PQ has to be vertical there are for example no possible solutions if P= (-4/6). Every triangle fulfilling the mentioned assumptions would be named PRQ and not PQR.

Is anybody able to help me ?

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Re: Right triangle PQR is to be constructed in the xy-plane so   [#permalink] 14 Jul 2014, 02:09

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