Triangle ABC will be constructed in a xy-plane according to

Hi All,

While this prompt is a bit wordy (and even looks a bit 'complex), it can be broken down into small 'steps' rather easily.

We're given a number of 'restrictions' in terms of how a right triangle can drawn into an XY-plane:

1) The vertices must be comprised of integer values for the X and Y co-ordinates.
2) Segment AB is parallel to the Y-axis.
3) Angle B is the 90 degree angle.
4) The range of possible X co-ordinates is 0 <= X <= 5 and the range of possible Y-coordinates is -4 <= Y <= 6

We're asked to find the number of possible right triangles.

To start, let's place the first point ('Point A'). Considering the limits on the X and Y co-ordinates, there are 6 possible values for X and 11 possible values for Y. This means that there are (6)(11) = 66 possible places that we can place Point A.

Once we place Point A, we now have limitations on where we can place Point B. Since segment AB must be PARALLEL to the Y-axis, it has to have the SAME X-coordinate as Point A. Since we've placed Point A already, the Y-coordinate of Point B must be DIFFERENT from the one that Point A uses, so there are only 10 possibilities for Point B.

Once we place Point B, we now have limitations on where we can place Point C. Since angle B is a 90 degree angle, segment BC must be PARALLEL to the X-axis and Point C has to have the SAME Y-coordinate as Point B. Since we've placed Point B already, the X-coordinate of Point C must be DIFFERENT from the one that Point B uses, so there are only 5 possibilities for Point C.

In total, that gives us (66)(10)(5) = 3300 possible triangles.

Final Answer:

GMAT assassins aren't born, they're made,
Rich