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# Right triangle ABC is to be drawn in the xy-plane so that

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Right triangle ABC is to be drawn in the xy-plane so that [#permalink]

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09 Jan 2010, 03:50
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Right triangle ABC is to be drawn in the xy-plane so that the right angle is at A and AB is parallel to the y-axis. If the x- and y-coordinates of A, B, and C are to be integers that are consistent with the inequalities -6 ≤ x ≤ 2 and 4 ≤ y ≤ 9 , then how many different triangles can be drawn that will meet these conditions?

A. 54
B. 432
C. 2,160
D. 2,916
E. 148,824
[Reveal] Spoiler: OA

Last edited by Bunuel on 15 Jul 2013, 23:17, edited 3 times in total.
Edited the question and added the OA
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09 Jan 2010, 04:34
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hrish88 wrote:
Right triangle ABC is to be drawn in the xy-plane so that the right angle is at A and AB is parallel to the y-axis. If the x- and y-coordinates of A, B, and C are to be integers that are consistent with the inequalities -6 ≤ x ≤ 2 and 4 ≤ y ≤ 9 , then how many different triangles can be drawn that will meet these conditions?

A.54

B.432

C.2160

D.2916

E.148,824

i ve got it right.but this problem is very time consuming.can anyone suggest shorter method

We have the rectangle with dimensions 9*6 (9 horizontal dots and 6 vertical). AB is parallel to y-axis and AC is parallel to x-axis.

Choose the (x,y) coordinates for vertex A: 9C1*6C1;
Choose the x coordinate for vertex C (as y coordinate is fixed by A): 8C1, (9-1=8 as 1 horizontal dot is already occupied by A);
Choose the y coordinate for vertex B (as x coordinate is fixed by A): 5C1, (6-1=5 as 1 vertical dot is already occupied by A).

9C1*6C*8C1*5C1=2160.

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09 Jan 2010, 04:50
Bunuel wrote:
hrish88 wrote:
Right triangle ABC is to be drawn in the xy-plane so that the right angle is at A and AB is parallel to the y-axis. If the x- and y-coordinates of A, B, and C are to be integers that are consistent with the inequalities -6 ≤ x ≤ 2 and 4 ≤ y ≤ 9 , then how many different triangles can be drawn that will meet these conditions?

A.54

B.432

C.2160

D.2916

E.148,824

i ve got it right.but this problem is very time consuming.can anyone suggest shorter method

We have the square with dimensions 9*6(9 horizontal dots and 6 vertical). AB is parallel to y-axis and AC is parallel to x-axis.

Choose the (x,y) coordinates for vertex A: 9C1*6C1;
Choose the x coordinate for vertex C (as y coordinate is fixed by A): 8C1, (9-1=8 as 1 horizontal dot is already occupied by A);
Choose the y coordinate for vertex B (as x coordinate is fixed by A): 5C1, (6-1=5 as 1 vertical dot is already occupied by A).

9C1*6C*8C1*5C1=2160.

OA is C.very nice explanation.you rock man as always.
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14 Oct 2010, 21:06
Bunuel wrote:
hrish88 wrote:
Right triangle ABC is to be drawn in the xy-plane so that the right angle is at A and AB is parallel to the y-axis. If the x- and y-coordinates of A, B, and C are to be integers that are consistent with the inequalities -6 ≤ x ≤ 2 and 4 ≤ y ≤ 9 , then how many different triangles can be drawn that will meet these conditions?

A.54

B.432

C.2160

D.2916

E.148,824

i ve got it right.but this problem is very time consuming.can anyone suggest shorter method

We have the rectangle with dimensions 9*6 (9 horizontal dots and 6 vertical). AB is parallel to y-axis and AC is parallel to x-axis.

Choose the (x,y) coordinates for vertex A: 9C1*6C1;
Choose the x coordinate for vertex C (as y coordinate is fixed by A): 8C1, (9-1=8 as 1 horizontal dot is already occupied by A);
Choose the y coordinate for vertex B (as x coordinate is fixed by A): 5C1, (6-1=5 as 1 vertical dot is already occupied by A).

9C1*6C*8C1*5C1=2160.

Good one. +1 for it. Hope I didn't mess it up.

so what about the triangles that look like the mirror images of the ones above? - think, switching the co-ords of A and C along x axis and switching A and B along y axis....
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15 Oct 2010, 01:58
BlitzHN wrote:
Bunuel wrote:
hrish88 wrote:
Right triangle ABC is to be drawn in the xy-plane so that the right angle is at A and AB is parallel to the y-axis. If the x- and y-coordinates of A, B, and C are to be integers that are consistent with the inequalities -6 ≤ x ≤ 2 and 4 ≤ y ≤ 9 , then how many different triangles can be drawn that will meet these conditions?

A.54

B.432

C.2160

D.2916

E.148,824

i ve got it right.but this problem is very time consuming.can anyone suggest shorter method

We have the rectangle with dimensions 9*6 (9 horizontal dots and 6 vertical). AB is parallel to y-axis and AC is parallel to x-axis.

Choose the (x,y) coordinates for vertex A: 9C1*6C1;
Choose the x coordinate for vertex C (as y coordinate is fixed by A): 8C1, (9-1=8 as 1 horizontal dot is already occupied by A);
Choose the y coordinate for vertex B (as x coordinate is fixed by A): 5C1, (6-1=5 as 1 vertical dot is already occupied by A).

9C1*6C*8C1*5C1=2160.

so what about the triangles that look like the mirror images of the ones above? - think, switching the co-ords of A and C along x axis and switching A and B along y axis....

Above solution counts all position:

AC and CA;

A
B
and
B
A;

For example point C with 8C1 can be placed to the right as well to the left of A and point B with 5C1 can be placed below as well as above of A. So all cases are covered.

More here: arithmetic-og-question-88380.html?hilit=dimensions

Hope it helps.
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17 Oct 2010, 19:24
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C.

I am not sure if this approach is correct. I used Elimination. There can be only 5 possible values of C if we fix A. So the number of triangles possible has to be multiple of 5. The only answer that satisfies the criterion is C.
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Re: Right triangle ABC is to be drawn in the xy-plane so that [#permalink]

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25 Jan 2013, 06:08
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hrish88 wrote:
Right triangle ABC is to be drawn in the xy-plane so that the right angle is at A and AB is parallel to the y-axis. If the x- and y-coordinates of A, B, and C are to be integers that are consistent with the inequalities -6 ≤ x ≤ 2 and 4 ≤ y ≤ 9 , then how many different triangles can be drawn that will meet these conditions?

A. 54
B. 432
C. 2,160
D. 2,916
E. 148,824

First, get the integer points available for x-axis: 2 - (-6) + 1 = 9
Second, get the interger points available for y-axis: 9-4+1 = 6

How many ways to select the location of line AB in the x-axis? 9
How many ways to select the location of point C in the x-axis? 8 (Note: we cannot select the location of line AB)
How many ways to select the location of the base? 2 (Is it BC or AB?)
How many ways to position line AB parallel to y axis? 6!/2!4! = 15

Multiple all that:$$9*8*2*15 = 2,160$$

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Re: Right triangle ABC is to be drawn in the xy-plane so that [#permalink]

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25 Jan 2013, 12:03
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Slightly different way of thinking:

On the 9x6 grid of possibilities, I can imagine a bunch of rectangles (with sides parallel to x and y axes). Each of these rectangles contains 4 triangles that fit the description of the question stem.

therefore:

Answer = ( # of Rectangles I can make on the grid) x 4

To create the rectangle, I need to pick 2 points on the x direction, and 2 points on the y direction. Therefore:

Answer = C(9,2) * C(6,2) * 4 = 36 * 15 * 4 = 2160 (OPTION C)
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18 May 2013, 09:49
Bunuel wrote:
hrish88 wrote:
Right triangle ABC is to be drawn in the xy-plane so that the right angle is at A and AB is parallel to the y-axis. If the x- and y-coordinates of A, B, and C are to be integers that are consistent with the inequalities -6 ≤ x ≤ 2 and 4 ≤ y ≤ 9 , then how many different triangles can be drawn that will meet these conditions?

A.54

B.432

C.2160

D.2916

E.148,824

i ve got it right.but this problem is very time consuming.can anyone suggest shorter method

We have the rectangle with dimensions 9*6 (9 horizontal dots and 6 vertical). AB is parallel to y-axis and AC is parallel to x-axis.

Choose the (x,y) coordinates for vertex A: 9C1*6C1;
Choose the x coordinate for vertex C (as y coordinate is fixed by A): 8C1, (9-1=8 as 1 horizontal dot is already occupied by A);
Choose the y coordinate for vertex B (as x coordinate is fixed by A): 5C1, (6-1=5 as 1 vertical dot is already occupied by A).

9C1*6C*8C1*5C1=2160.

Hi Bunuel ,

That was a fantastic solution , but i have a small doubt . How do we ensure that by selecting points in this way the properties of a triangle are satisfied always . Could there be some points through which we cannot even form a triangle leave alone right angled triangle. I hope i am clear in my question .
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19 May 2013, 03:09
venkat18290 wrote:
Bunuel wrote:
hrish88 wrote:
Right triangle ABC is to be drawn in the xy-plane so that the right angle is at A and AB is parallel to the y-axis. If the x- and y-coordinates of A, B, and C are to be integers that are consistent with the inequalities -6 ≤ x ≤ 2 and 4 ≤ y ≤ 9 , then how many different triangles can be drawn that will meet these conditions?

A.54

B.432

C.2160

D.2916

E.148,824

i ve got it right.but this problem is very time consuming.can anyone suggest shorter method

We have the rectangle with dimensions 9*6 (9 horizontal dots and 6 vertical). AB is parallel to y-axis and AC is parallel to x-axis.

Choose the (x,y) coordinates for vertex A: 9C1*6C1;
Choose the x coordinate for vertex C (as y coordinate is fixed by A): 8C1, (9-1=8 as 1 horizontal dot is already occupied by A);
Choose the y coordinate for vertex B (as x coordinate is fixed by A): 5C1, (6-1=5 as 1 vertical dot is already occupied by A).

9C1*6C*8C1*5C1=2160.

Hi Bunuel ,

That was a fantastic solution , but i have a small doubt . How do we ensure that by selecting points in this way the properties of a triangle are satisfied always . Could there be some points through which we cannot even form a triangle leave alone right angled triangle. I hope i am clear in my question .

ANY 3 non-collinear points on a plane form a triangle.
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Re: Right triangle ABC is to be drawn in the xy-plane so that [#permalink]

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30 May 2013, 20:14
Bunuel... you're a freaking genius. Get a job with NASA already.
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Re: Right triangle ABC is to be drawn in the xy-plane so that [#permalink]

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13 Nov 2013, 07:38
Another way of looking at the problem.
According to the given constraints, the co-ordinates have to be chosen this way :-
A(a,b) B(a,c) C(d,b) where a,b,c and d are arbitrary integers. If you check this satisfies the constraint that AB must be parallel to the Y-axis.
Drawing the triangle and rotating it will give you a rectangle whose sides will measure length= |b-c| and breadth= |a-d|.
This rectangle's area will be = |b-c| X |a-d|
Now after having realized this, you just have to choose values from the given ranges such that the area is always non-zero,
and this can be done in the following way,
!.
selecting a and d from the range [-6,2] which has 9 elements, derived as --> 2 - (-6) +1 = 9.

9C2 X 2 (2 because both a>d and d>a are permissible).

2. selecting b and c similarly
6C2 X 2.

3. Multiplying the two terms :-
9C2 X 6C2 X 2 X 2 = 2160.
Kudos if you liked it.
Do have a look at this approach Bunuel
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13 Nov 2013, 08:38
Bunuel wrote:
hrish88 wrote:
Right triangle ABC is to be drawn in the xy-plane so that the right angle is at A and AB is parallel to the y-axis. If the x- and y-coordinates of A, B, and C are to be integers that are consistent with the inequalities -6 ≤ x ≤ 2 and 4 ≤ y ≤ 9 , then how many different triangles can be drawn that will meet these conditions?

A.54

B.432

C.2160

D.2916

E.148,824

i ve got it right.but this problem is very time consuming.can anyone suggest shorter method

We have the rectangle with dimensions 9*6 (9 horizontal dots and 6 vertical). AB is parallel to y-axis and AC is parallel to x-axis.

Choose the (x,y) coordinates for vertex A: 9C1*6C1;
Choose the x coordinate for vertex C (as y coordinate is fixed by A): 8C1, (9-1=8 as 1 horizontal dot is already occupied by A);
Choose the y coordinate for vertex B (as x coordinate is fixed by A): 5C1, (6-1=5 as 1 vertical dot is already occupied by A).

9C1*6C*8C1*5C1=2160.

Kudos Bunuel. Nice explanation.
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Re: Right triangle ABC is to be drawn in the xy-plane so that [#permalink]

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31 Jan 2015, 06:24
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Re: Right triangle ABC is to be drawn in the xy-plane so that [#permalink]

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15 Jul 2015, 00:29
Bunuel wrote:
hrish88 wrote:
Right triangle ABC is to be drawn in the xy-plane so that the right angle is at A and AB is parallel to the y-axis. If the x- and y-coordinates of A, B, and C are to be integers that are consistent with the inequalities -6 ≤ x ≤ 2 and 4 ≤ y ≤ 9 , then how many different triangles can be drawn that will meet these conditions?

A.54

B.432

C.2160

D.2916

E.148,824

i ve got it right.but this problem is very time consuming.can anyone suggest shorter method

We have the rectangle with dimensions 9*6 (9 horizontal dots and 6 vertical). AB is parallel to y-axis and AC is parallel to x-axis.

Choose the (x,y) coordinates for vertex A: 9C1*6C1;
Choose the x coordinate for vertex C (as y coordinate is fixed by A): 8C1, (9-1=8 as 1 horizontal dot is already occupied by A);
Choose the y coordinate for vertex B (as x coordinate is fixed by A): 5C1, (6-1=5 as 1 vertical dot is already occupied by A).

9C1*6C*8C1*5C1=2160.

Bunuel, Can the x coordinate for A be 0? Wouldn't that make A lie on the y-axis and as a result not let AB be parallel to the y-axis?
I calculated the options for A as : 8C1*6C1

What am I missing?
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31 Jul 2016, 08:06
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Re: Right triangle ABC is to be drawn in the xy-plane so that [#permalink]

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28 Aug 2016, 19:05
Bunuel wrote:

A.54

B.432

C.2160

D.2916

E.148,824

i ve got it right.but this problem is very time consuming.can anyone suggest shorter method

We have the rectangle with dimensions 9*6 (9 horizontal dots and 6 vertical). AB is parallel to y-axis and AC is parallel to x-axis.

Choose the (x,y) coordinates for vertex A: 9C1*6C1;
Choose the x coordinate for vertex C (as y coordinate is fixed by A): 8C1, (9-1=8 as 1 horizontal dot is already occupied by A);
Choose the y coordinate for vertex B (as x coordinate is fixed by A): 5C1, (6-1=5 as 1 vertical dot is already occupied by A).

9C1*6C*8C1*5C1=2160.

Will all these be right angled triangles? Will all these obey Pythagorean theory? I was thinking about right angle triads (like 3,4,5 etc) and messed it up
Re: Right triangle ABC is to be drawn in the xy-plane so that   [#permalink] 28 Aug 2016, 19:05
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