Permutations: Triangle and rectangle vertices.

[vineeshp]

Source: GROCKIT.

Hi,

I need help with the question Q1 below.

Q2) 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?

To solve this, I went with the following.
that A can be chosen in 9 * 6 = 54 ways.
B will have the same Y coordinate only x varies and should not be identical to A. So B can be chosen in 9-1 = 8 ways.
C can similarly be chosen in 6 -1 = 5 ways.
So total no. of ways of choosing the triangle is 54 * 8 * 5 = 2160 ways. This came out correct!

Let us go to Q1:
Q1) Rectangle ABCD is constructed in the coordinate plane parallel to the x- and y-axes. If the x- and y-coordinates of each of the points are integers which satisfy 3 ≤ x ≤ 11 and -5 ≤ y ≤ 5, how many possible ways are there to construct rectangle ABCD?

(Note that two rectangles that have the same four vertices that are labeled differently are considered to be the same rectangle.)

Answer is 1980.

Now in Q1 above, I must be able to employ the same technique I used for the triangle in Q1 because a rectangle's 4th coordinate must use the x and y coordinates of the adjacent vertices. When I apply that method I get to 7920. But the answer seems to be 1980.

Where am I going wrong?

[Ashley@VeritasPrep]

YOUR QUESTION IS AWESOME. "Why does my seemingly totally logical method not actually work" questions are my favorite things to think about GMAT-wise. And this one was a challenge.

So, notice that the answer you came up with is exactly four times the correct answer. That's no coincidence. Your method is perfect for the Right Triangle problem (and, I should add, MUCH more efficient than the method I used -- thanks!), but as you use it, it relies on the fact that we KNOW, in the triangle problem, that A is the vertex at the right angle (and of course, in a triangle, this will be true of only one of the vertices). In the rectangle problem, you can still start with that, but then -- and here's the kicker -- you must account for the fact that any given Rectangle R that you form starting from a fixed point A is going to wind up identical to three other rectangles you'll form (and have counted into your total) -- specifically, the ones in which you place A in the locations that B, C, and D (the other three vertices) are occupying in this Rectangle R. So for any given rectangle, your method will count it four times, not once. The very tiny hint to this is the problem's parenthetical:

(Note that two rectangles that have the same four vertices that are labeled differently are considered to be the same rectangle.)

So all that means is that you need to take the answer you initially come out with and divide it by 4, to account for the fact that there are four possible rotations of the vertices that will actually create a rectangle in that exact same place. 7920/4 = 1980, so there you go.

Again, great question! Hope this helps!

[vineeshp]

Hi Ashley,

Can we exchange our GMAT scores? I will return it after my retake.

Hey the question is a Grockit question. Thanks for giving my method some credit.

I have quickly gone through your answer and I think I understand. I have to read again (but not right now, as I am sneaking in this reply from office during office hours.)

Thanks a zillion.