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Right triangle PQR is to be constructed in the xy-plane so [#permalink]
07 Aug 2006, 17:31
00:00
A
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Difficulty:
(N/A)
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?
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.
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.
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.
Re: Right triangle PQR is to be constructed in the xy-plane so [#permalink]
08 Nov 2011, 13:48
1
This post was BOOKMARKED
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.
gmatclubot
Re: Right triangle PQR is to be constructed in the xy-plane so
[#permalink]
08 Nov 2011, 13:48
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