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

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07 Aug 2006, 18:31

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32% (01:59) wrong based on 129 sessions

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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?

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

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07 Aug 2006, 19:07

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.

...

Ready?

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:

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.

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

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07 Aug 2006, 19:52

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.

...

Ready?

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:

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.

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

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07 Aug 2006, 20:29

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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, 18:35, edited 1 time in total.

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

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10 Aug 2008, 04:05

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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 < or= x < or = 5 and 6 < or= y < or = 16. How many different triangles with these properties could be constructed?

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

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10 Aug 2008, 05:32

spriya wrote:

ca you explain why x=9*10/2 probability of understanding probability is less

x can have 9 possible with 1 unit length. x can have 8 possible with 2 unit length. x can have 7 possible with 3 unit length. . . . x can have 1 possible with 9 unit length.

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

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10 Aug 2008, 06:47

abhijit_sen wrote:

spriya wrote:

ca you explain why x=9*10/2 probability of understanding probability is less

x can have 9 possible with 1 unit length. x can have 8 possible with 2 unit length. x can have 7 possible with 3 unit length. . . . x can have 1 possible with 9 unit length.

Similarly y will have its own case.

Hope this helps.

I got to work hard in probability
_________________

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

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10 Aug 2008, 08:05

number of ways point P can be selected = 100 ( x can be anything from 4 to 5 (including) and y can be anything from 6 to 16(including)) number of ways point R can be selected = 9 (the y co-orodinate will be same as P and we can choses x co-ordinate in 9 different ways) number of ways point Q can be selected = 11 (the x co-orodinate will be same as P and we can choses y co-ordinate in 11 different ways)

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

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10 Aug 2008, 09:19

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For the rt triange PQR, we will have same Y-corodinate value for point R and P and same X-coordinate value for point Q and P.

So Total possible values for X = ( number of ways X can be selected for point R) * ( number ways X can be selected for point Q and P ) = 10 * 9 ( as Q and R will have same X-coordinate and will be different than that of R ) =90

Similarly, total number of ways for selecting Y = 11* 10 =110

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

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10 Aug 2008, 09:23

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judokan 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 < or= x < or = 5 and 6 < or= y < or = 16. How many different triangles with these properties could be constructed?

(A) 110 (B) 1,100 (C) 9,900 (D) 10,000 (E) 12,100

P can be chosen from 11*10 points and Q can be chosen from 10 points and R from 9 points

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

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24 May 2011, 07:25

I understand the solution given, but doesn't that solution count multiple identical triangles as "different"? That is, we are making a right triangle, so we should only need to worry about the lengths of the two legs. Given the possible x coordinates, there should be 9 possible lengths for PR (1..9). Given the possible y coordinates, there should be 10 possible lengths for PQ (1..10).

The solution provided doesn't give us the number of unique right triangles with integer lengths but instead the total number of unique triangles * the total possible arrangements in the given area for each.

Am I just way off and the wording of the question should have made me understand that we're also looking for unique positions of congruent triangle?

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

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08 Nov 2011, 14:48

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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.

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

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