Counting....Right Triangles

This topic has expert replies
User avatar
Legendary Member
Posts: 504
Joined: Tue Apr 19, 2011 1:40 pm
Thanked: 114 times
Followed by:11 members

Counting....Right Triangles

by knight247 » Sat Oct 01, 2011 11:42 pm
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

OA C. Detailed explanations would be appreciated.
Source: — Problem Solving |

User avatar
Master | Next Rank: 500 Posts
Posts: 496
Joined: Tue Jun 07, 2011 5:34 am
Thanked: 38 times
Followed by:1 members

by sl750 » Sun Oct 02, 2011 3:10 am
Total number of points for x coordinate = 9
Total number of points for y coordinate = 6

For point A, we have 9*6 ways to chose the x and y coordinate
For point B, only the y coordinate is different and we cannot use the same y coordinate as used by point A, so we have 9*5 ways to chose the x and y coordinate
For point C, only the x coordinate is different and we cannot use the same x coordinate as used by point A, so we have 8*6 ways to chose the x and y coordinate

Therefore, number of ways to create different right triangles is 9*6*5*8 =2160

User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Sun Oct 02, 2011 3:17 am
knight247 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

OA C. Detailed explanations would be appreciated.
When a question asks for the number of triangles that can be constructed, it's not a geometry question but a combinations question. Why? Because a triangle is a combination of 3 points.

We need to determine how many ways we can combine A, B and C to form a triangle that satisfies all the conditions given.
For each point, we need to choose an x value and a y value.

Point A:
x value: -6≤x≤2, giving us 9 choices.
y value: 4≤y≤9, giving us 6 choices.

To combine our choices for x with our choices for y, we multiply:
9*6 = 54.

Point C:
x value: In order to construct a right triangle, C has to have the same x coordinate as A (so that C is directly above A and we get a right angle). So we have only 1 choice for x: it must be the same integer that we chose for A's x value.

y value: If A and C share the same x value, they can't have the same y value, or they will be the same point. We used 1 of our 6 choices for y when we chose A, so we have 6-1 = 5 choices for C's y value.

To combine our choices for x with our choices for y, we multiply:
1*5 = 5.

Point B:
y value: For AB to be parallel to the x axis, A and B have to share the same y value. So the number of choices for y is 1; it must be the same integer that we chose for A's y value.

x value: If A and B share the same y value, they can't have the same x value, or they will be the same point. We used 1 of our 9 choices for x when we chose A, so we have 9-1 = 8 choices for B's x value.

To combine our choices for x with our choices for y, we multiply:
8*1 = 8.

Now we need to count how many ways A, B and C can be combined to form a triangle.
To combine our choices for A with our choices for B and C, we multiply:
54*5*8 = 2160.

The correct answer is C.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3

Junior | Next Rank: 30 Posts
Posts: 11
Joined: Tue Sep 28, 2010 1:17 am

by cmal » Mon Oct 17, 2011 1:18 am
Hey wouldnt this be something like this

15 points in total =9+6
15c3=455
if they are straight lines on the y axis or x-axis

hence

15c3-(9-6)?
=440
so i would pick B