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

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VP
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Right triangle PQR is to be constructed in the xy-plane so [#permalink]  07 Aug 2006, 17:31
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Question Stats:

33% (00:00) correct 67% (01:02) wrong based on 6 sessions
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

Note <= is less than or equal to
Intern
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Re: PS: Triangles [#permalink]  07 Aug 2006, 18:07
Quick note: "[ u]<[/u] yields <," so this looks a little cleaner:

haas_mba07 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

This is a tricky question posed by the OP, so this may be a tricky response. ^^ What I initially did was get a blank sheet of paper and make a 11 (tall) x 10 (wide) grid of dots.

Go on, you do it too.

...

The lower left point is (-4, 6) and the upper right is (5, 16).

Start by placing P at (-4, 6). You'll notice that if you fix R at (-3, 6) you can create 10 unique triangles by placing Q's at (-4, 7) to (-4, 16).

Now place P at (-4, 6) again. Fix Q at (-4, 7) and you can get 9 unique triangles by placing R's at (-3, 6) through (5, 6).

If you place P at (-3, 6), you can get 10 unique Q's and 8 unique R's. Go through (4, 6) and you'd have 10 Q's and 1 R.

If you place P at (-4, 7), you can get 9 unique Q's and 9 unique R's. Go through (-4, 15) and you'd have 1 Q and 9 R.

The point is, the pattern of this methodology yields (10 + 9 + ... + 2 + 1) * (9 + 8 + ... + 2 + 1) triangles. However, those only extend up and right. because triangles can exist in four different ways (up/right, up/left, down/left, down/right), your final calculation should be:

(10 + 9 + ... + 2 + 1) * (9 + 8 + ... + 2 + 1) * 4
55 * 45 * 4
2,475 * 4
9,900

So I'm sticking with C.

Note that 9,900 = 11 * 10 * 9, so there's probably an easier way to do this, and by probably, I mean I hope so because this solution was thorough but sloooooooooow.
VP
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Re: PS: Triangles [#permalink]  07 Aug 2006, 18:52
Innovative way of solving..

Thanks for the [ u]<[/u] tip...

As for time, this one took quite some time for me... but once you get the trick it is easier.

I'll wait for some more answers before posting OA...

Zoelef wrote:
Quick note: "[ u]<[/u] yields <," so this looks a little cleaner:

haas_mba07 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

This is a tricky question posed by the OP, so this may be a tricky response. ^^ What I initially did was get a blank sheet of paper and make a 11 (tall) x 10 (wide) grid of dots.

Go on, you do it too.

...

The lower left point is (-4, 6) and the upper right is (5, 16).

Start by placing P at (-4, 6). You'll notice that if you fix R at (-3, 6) you can create 10 unique triangles by placing Q's at (-4, 7) to (-4, 16).

Now place P at (-4, 6) again. Fix Q at (-4, 7) and you can get 9 unique triangles by placing R's at (-3, 6) through (5, 6).

If you place P at (-3, 6), you can get 10 unique Q's and 8 unique R's. Go through (4, 6) and you'd have 10 Q's and 1 R.

If you place P at (-4, 7), you can get 9 unique Q's and 9 unique R's. Go through (-4, 15) and you'd have 1 Q and 9 R.

The point is, the pattern of this methodology yields (10 + 9 + ... + 2 + 1) * (9 + 8 + ... + 2 + 1) triangles. However, those only extend up and right. because triangles can exist in four different ways (up/right, up/left, down/left, down/right), your final calculation should be:

(10 + 9 + ... + 2 + 1) * (9 + 8 + ... + 2 + 1) * 4
55 * 45 * 4
2,475 * 4
9,900

So I'm sticking with C.

Note that 9,900 = 11 * 10 * 9, so there's probably an easier way to do this, and by probably, I mean I hope so because this solution was thorough but sloooooooooow.
Manager
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I also got 9900, but I approached like a combinatorics problem.

We know that P,Q, R can only lie on a grid 10x11

Let's start by fixing P' x-coordinate: we have 10 possible places for it

Then we choose x-coordinate for R: since PR || x-axis, R's x-coordinate can not coincide with P's x-coord, so we have only 9 possible choices left.

Then, we fix P's (and R's too) y-coordinate: 11 possible choices.

Finally, after we fixed P and R, Q's x-coordinate must be the same as P's (because P is the right angle and PR || x-axis), so the only choice is for y-coordinate, and we have 10 possible choices left.

So, total number of different triangles: 10*9*11*10=9,900

(C)

Last edited by v1rok on 08 Aug 2006, 17:35, edited 1 time in total.
Intern
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I realized a quicker way driving to work today.

11 x 10 grid = 110 points

each point has 10 possible y's and 9 possible x's (going both ways; up/down or left/right)

11 * 10 * 10 * 9 = 9,900

Still (C), but with 400% more conciseness.
Senior Manager
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Re: [#permalink]  08 Nov 2011, 02:50
Zoelef wrote:
I realized a quicker way driving to work today.

11 x 10 grid = 110 points

each point has 10 possible y's and 9 possible x's (going both ways; up/down or left/right)

11 * 10 * 10 * 9 = 9,900

Still (C), but with 400% more conciseness.

How do you get a 11*10 grid ?
Intern
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Location: Virginia, USA
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Re: Right triangle PQR is to be constructed in the xy-plane so [#permalink]  08 Nov 2011, 13:48
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Trying to explain the concept

Consider a grid with 3 points on X axis and 3 points on Y axis.
Now right angle triangel is to be drawn such that one line is parallel to X axis.
Just consider botton row of a grid with 3 points.
Two points A and B can be selected from 3 points by 3C2 = 3 ways
Now we have to select 3rd point from verticle line. We have 2 points available in line with point A and 2 points available in line with point B.
So 3rd point C can be selected by (2 + 2) ways.
So with bottom row of a grid we can draw 3 * (2 + 2) right angle triangles.
But we have total 3 horizontal rows so total number of right angle triangel = (3 * (2 + 2)) * 3

Now consider given example where X point can be from -4 to 5 (total 10 points)
and Y point can be from 6 to 16 (total 11 points)

Again consider just bottom row.
Two points A and B can be selected from 3 points by 10C2 = 45 ways
To select point C we have 10 points in line with A and 10 points in line with B.
So total number of right angle triangles can be drawn from bottom row = 45 * (10 +10)
but we have total 11 rows so total number of triangle = (45 * (10 + 10)*11 = 9900.

Hope this helps.
Re: Right triangle PQR is to be constructed in the xy-plane so   [#permalink] 08 Nov 2011, 13:48
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