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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] New post 09 Jan 2010, 04:50
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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 02 Apr 2012, 18:29, edited 2 times in total.
Edited the question and added the OA
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Re: tough problem [#permalink] New post 09 Jan 2010, 05:34
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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.

Answer: C.

Good one. +1 for it. Hope I didn't mess it up.
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Re: tough problem [#permalink] New post 09 Jan 2010, 05:50
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.

Answer: C.

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


OA is C.very nice explanation.you rock man as always.
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Re: tough problem [#permalink] New post 14 Oct 2010, 22:06
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.

Answer: C.

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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Re: tough problem [#permalink] New post 15 Oct 2010, 02:58
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.

Answer: C.

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....


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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Re: tough problem [#permalink] New post 17 Oct 2010, 20:24
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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] New post 25 Jan 2013, 07:08
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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

Answer: C
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Re: Right triangle ABC is to be drawn in the xy-plane so that [#permalink] New post 25 Jan 2013, 13:03
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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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Re: tough problem [#permalink] New post 18 May 2013, 10:49
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.

Answer: C.

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


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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Re: tough problem [#permalink] New post 19 May 2013, 04:09
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.

Answer: C.

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


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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PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!


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Re: tough problem   [#permalink] 19 May 2013, 04:09
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