PS--Geometry

[kris610]

On the coordinate plane (6, 2) and (0, 6) are the endpoints of the diagonal of a square. What is the distance between point (0, 0) and the closest vertex of the square?

[parallel_chase]

On the coordinate plane (6, 2) and (0, 6) are the endpoints of the diagonal of a square. What is the distance between point (0, 0) and the closest vertex of the square?

I think end point of the other diagonal is (0,0) since it is a square and all sides are of equal length.

Anyways, whats the OA.

[rs2010]

I feel there is something wrong with the question.

Either it should be rectangle or coordinate points for diagonal should be (0,6) (4,2).

Please provide the answer choices.

[Stuart@KaplanGMAT]

This seems to be crazy-difficult.

We can solve for the diagonal by simple application of the pythagorean theorum. Drawing a right triangle parallel to the axes, we get lengths of 6, 4 and d, where d is the diagonal. So:

6^2 + 4^2 = d^2
52 = d^2
2(sqrt13) = d

Once we know that the diagonal is 2(sqrt13), we can solve for the lengths of the sides of the square, using the 45/45/90 right triangle ratio of:

x : x : x(sqrt2)

Here, we know that x(sqrt2) = 2(sqrt13)

x = 2(sqrt13)/(sqrt2)

let's rationalize the denominator (multiply top and bottom by sqrt2):

x = 2(sqrt13)(sqrt2)/(sqrt2)(sqrt2) = 2(sqrt26)/2 = sqrt26

Believe it or not, that was the easy part of the question!

Now we need to find the bottom left vertex of the square (that one will be closest to the origin). At this point I gave up trying to do this question mathematically (something I probably should have done right from the start) and looked at the diagram I drew.

I noticed that the point (1,1) is equidistant from (0.6) and (6,2) and does, in fact, give a hypotenuse of sqrt26 (5^2 + 1^2 = h^2, or h^2 = 26).

So, we want the distance from (0,0) to (1,1):

1^2 + 1^2 = d^2

2 = d^2

sqrt2 = d

Therefore, the correct answer is sqrt2.

Almost certainly this would be a great question to backsolve (i.e. work backwards from the answers, eyeballing the diagram). In future, please post the choices!

[rs2010]

That is true.

When I got the equations like ( considering point (a,b)

a^2+b^2-12*b=-10
a^2+b^2-12*a-4*b=-14

I thought it is better to start with answer choices.

Last night before sleeping I looked at this and this question bothered me in my dreams.

[Ian Stewart]

Almost certainly this would be a great question to backsolve (i.e. work backwards from the answers, eyeballing the diagram). In future, please post the choices!

I'm curious how it would be helpful here to have answer choices like:

A) sqroot(2)
B) sqroot(3)
C) 4 - 2*sqroot(2)
D) 2
E) sqroot(10)

Backsolving does not seem straightforward, at least not to me.

We have two endpoints of a diagonal of a square. We can use the following:

-the midpoint of one diagonal is the midpoint of the other diagonal;
-the diagonals are perpendicular.

If (0,6) and (6,2) are endpoints, (3,4) is the midpoint.

From (0,6) to (3,4), we go right 3 and down 2; that is, we increase x by 3 and decrease y by 2: the slope is -2/3. Consider the perpendicular diagonal- its slope is the negative reciprocal, i.e. 3/2. From (3,4), on a perpendicular line, to find a point the same distance from (3,4) as (0,6) is, we can decrease x by 2 and decrease y by 3, or we can increase x by 2 and increase y by 3. The endpoints of the other diagonal are (1,1) and (5,7).

The distance from (0,0) to (1,1) is sqrt(2).

[ildude02]

I solved this by drawing the figure in the co-ordinate plance and deduced the co-ordinate points of the square without algebric equations. We know that the diagonal has points, (6, 2) and (0, 6). So the other 2 points in the co-ordinate plane are (0, 2) and (6, 6). The closest distance is between origin(0, 0) and (0, 2) and came up with 2. Not sure where I went wrong with my approaach if it's not 2. Appreciate anyone's response.

[Stuart@KaplanGMAT]

Almost certainly this would be a great question to backsolve (i.e. work backwards from the answers, eyeballing the diagram). In future, please post the choices!

I'm curious how it would be helpful here to have answer choices like:

A) sqroot(2)
B) sqroot(3)
C) 4 - 2*sqroot(2)
D) 2
E) sqroot(10)

Backsolving does not seem straightforward, at least not to me.

Well, we know that the point we want has integer values (otherwise this question would be even more nutty-crazy-hard, way beyond the scope of the GMAT).

So, I'd look at the choices and see to which integer point (if any) that answer gives me, then I'd see if that point does, indeed, form a square. If not, I'd try another point. The scratch paper on the GMAT is cross-lined (i.e. it's graph paper), so it's pretty easy to eyeball and see if things actually work out.

Your solution is a good one, but most test-takers wouldn't see it. The solution I posted certainly works, but again, most test-takers wouldn't see it. Backsolving can be a great alternative when the math doesn't jump out at you.

[Stuart@KaplanGMAT]

I solved this by drawing the figure in the co-ordinate plance and deduced the co-ordinate points of the square without algebric equations. We know that the diagonal has points, (6, 2) and (0, 6). So the other 2 points in the co-ordinate plane are (0, 2) and (6, 6). The closest distance is between origin(0, 0) and (0, 2) and came up with 2. Not sure where I went wrong with my approaach if it's not 2. Appreciate anyone's response.

The problem is that the shape you ended up with is a 4*6 rectangle, not a square.

The actual square in this question doesn't have sides parallel to the axes, it's on a tilt. You'd only get a square with sides parallel to the axes if the slope of the diagonal were 1 or -1.