PS--Geometry

kris610 · Posted: Fri Jul 04, 2008 7:22 am Post subject: PS--Geometry

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 · Posted: Fri Jul 04, 2008 8:45 am Post subject: Re: PS--Geometry

hemantsood · Posted: Fri Jul 04, 2008 8:18 pm Post subject:

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 Kovinsky · Posted: Fri Jul 04, 2008 10:02 pm Post subject:

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!

hemantsood · Posted: Sat Jul 05, 2008 5:53 am Post subject:

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. Sad

Ian Stewart · Posted: Sat Jul 05, 2008 11:07 am Post subject:

ildude02 · Posted: Sat Jul 05, 2008 4:25 pm Post subject:

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 Kovinsky · Posted: Sat Jul 05, 2008 5:18 pm Post subject:

Stuart Kovinsky · Posted: Sat Jul 05, 2008 5:20 pm Post subject: