A certain square is to be drawn on a coordinate plane. One of the vertices must be on the origin, and the square is to have an area of 100. If all coordinates of the vertices must be integers, how many different ways can this square be drawn?
4
6
8
10
12

Each side of the square must have a length of 10. If each side were to be 6, 7, 8, or most other numbers, there could only be four possible squares drawn, because each side, in order to have integer coordinates, would have to be drawn on the x- or y-axis. What makes a length of 10 different is that it could be the hyptoneuse of a pythagorean triple, meaning the vertices could have integer coordinates without lying on the x- or y-axis.

For example, a square could be drawn with the coordinates (0,0), (6,8), (-2, 14) and (-8, 6). (It is tedious and unnecessary to figure out all four coordinates for each square).

If we lable the square abcd, with a at the origin and the letters representing points in a clockwise direction, we can get the number of possible squares by figuring out the number of unique ways ab can be drawn.

a has coordinates (0,0) and b could have coordinates:

(-10,0)
(-8,6)
(6,8)
(0,10)
(6,8)
(8,6)
(10,0)
(8, -6)
(6, -8)
(0, 10)
(-6, -8)
(-8, -6)

There are 12 different ways to draw ab, and so there are 12 ways to draw abcd.

The correct answer is E.

How do we figure out one of the points let alone 12?

Thank you for your help.

You don't have to figure out 12 points separately. You just need to have a couple of insights:

1) First insight:

One way to draw a side AB of length 10, if A is at the origin, is to put B 10 units away from A on one of the axes. This means B could be at (0,10) or (0,-10) or (10,0) or (-10,0).

2) Second insight:

The other way to draw a side AB of length 10, if A is at the origin, is to put B someplace other than on an axis. If the coordinates of B are (x,y), it is possible to draw a right triangle with AB as the hypotenuse and one leg of length x (on or parallel to the x axis) and the other leg of length y (on or parallel to the y axis). (Try drawing AB on a piece of graph paper if you can't picture this triangle).

Since the hypotenuse AB has length =10 , the Pythagorean theorem tells us that x^2 + y^2=10^2. Since we are told that x and y are integers, we know that:

EITHER x^2=6^2 and y^2=8^2
OR x^2=8^2 and y^2=6^2

(This inference is based on the fact that a right triangle whose sides have integer lengths must be a Pythagorean triple, or a multiple of a Pythagorean triple - see p.29 of our Geometry book.)

If x^2=6^2 and y^2=8^2, then B must be (6,8) or (6,-8) or (-6,8) or (-6,-8).
If x^2=8^2 and y^2=6^2, then B must be (8,6) or (8,-6) or (-8,6) or (-8,-6)


Putting all of these possibilities together, we see that there point B could be in any one of the 12 positions we listed.

-Jad

Can someone explain how pythag theorem relates to the area of the square? I'm missing the connection. If one point is 0,0 and the other is 6,8 then the area would be 6*8=48, right? in addition the sides would not be equal...

what does "one of the vertices must be on the origin" mean? does one the square's sides have to lie on one of the axis or that the interesction of 2 sides must be 0,0?

you're making the unwarranted assumption that the sides of the square must be horizontal and vertical. you should know better: the problem statement is simple and straightforward enough that you should suspect some kind of clever angle (heh, angle!) in the solution.

in particular, one side of the square would have endpoints at (0, 0) and (6, 8), and will be oriented diagonally. this is the reason why you must employ the pythagorean theorem: you're finding the length of a diagonal segment in the plane.



vertices means "corners". therefore, this means that one corner of the square must lie at (0, 0). so yes, the intersection of 2 sides.

i also felt that the following comment deserved a response:


this happens to be true here, but there's an additional complication that happens to be rendered irrelevant here by coincidence/symmetry: there are actually two possible squares for each possible location of AB.
simplest illustration: say AB sticks straight up, from (0, 0) to (0, 10). in that case, there are still 2 places you could put the square: (a) to the left of AB, so that the opposite vertex is at (-10, 10), or (b) to the right of AB, so that the opposite vertex is at (10, 10).
the same is true for the diagonal squares: each possible location of AB yields two possible squares, on opposite sides of AB from each other.

so, why aren't there 12 x 2 = 24 squares instead of 12? it's because of the following lucky coincidence: each of the squares is double-counted if we count 2 squares per possible AB. for instance, the square whose vertices are (0, 0) (10, 0) (10, 10) (0, 10) would be counted twice, once under AB(0, 10) and again under AB(10, 0). the same would be true for all the other squares. so the 24 possibilities actually reduce to the 12 that you'd originally have thought just from counting possible AB's.

if you were creating equilateral triangles, for instance, with the same AB's as sides, then there would actually be 24 possible triangles.

just food for thought!

why this does not hold for equilateral triangle why donot we have 12 triangles instead of 24 for equilateral triangle

Think of the 12 possible ab's as spokes of a wheel. (Following the convention used in the discussion above that a is at (0,0) and b is 10 units away at some point (integer, integer)) .

Now imagine a square in which vertices abcd are always labeled counter-clockwise. When we rotate the square around a at (0,0), landing on the 12 possible ab spokes, we prove that there are 12 ways to draw the square. Ron's point is that because each ab spoke is exactly opposite (i.e. 180 degrees from) one of the other spokes, considering squares in which abcd are labeled clock-wise would just produce duplicates of the 12 squares we have already counted.

If it were equilateral triangles we cared about, each of the 12 ab wheel spokes could have 2 triangles based on it: an up-pointing and down-pointing triangle, or a left- and right-pointing triangle, depending on which spoke we are looking at.

Note that we have to ignore the "all coordinates of the vertices must be integers" constraint if thinking of equilateral triangles, though. I think Ron was keeping ab=10 and the coordinates of those points as integers, but allowing non-integer coordinates for c.

_________________
Emily Sledge
Instructor
ManhattanGMAT

Could someone please explain why it's unnecessary to figure out all four coordinates? How can we tell that all coordinates of the vertices are integers if only one of the vertices (6,8) is identified?

For example, a square could be drawn with the coordinates (0,0), (6,8), (-2, 14) and (-8, 6). (It is tedious and unnecessary to figure out all four coordinates for each square).

Funny, I wanted to post, the exact same question. The way I proved it to myself, and I want to stress that it is no way a rigorous proof, was to draw a 14 by 14 square with the following points:
(8,0) (8,14) (-6,14) and (-6,0). Now mark the points (8,6) (2,14) (-6,8) and (0,0).

Connect the freshly drawn points and you will see a 10 by 10 square that has integer points. You know it is a square because all the triangles surrounding it are 3-4-5 triangles. I hope this explanation helps.

Ah, after three years someone asks this question, jenddd.

Students ask about this question every now and then in tutoring or office hours, and I always wonder if they're going to ask this follow-up, but almost nobody does.

Here's one way to think about it: start with one side of the square on the x-axis and one side on the y-axis. The vertices are at (0,0), (10, 0), (10, 10), and (0, 10).

The vertex at (0, 0) will not move, so let's not worry about that. I'm going to focus on the vertices that start at (10,0) and (0, 10), though analogous reasoning would work for the fourth vertex as well. As you rotate the square (let's say counter-clockwise), and the (0, 10) vertex moves to (-6, 8), the (10, 0) vertex moves to (8, 6). The exactly (10, 0) vertex moves the same relative to the x-axis as the (0, 10) moves relative to the y-axis.

And so on....

By the way, cosimogirolamo, your explanation is a fair bit of the way to a rigorous proof, though of course we don't need proofs on the GMAT anyway.