M15-19

Hi

Is there any shorter way to solve this problem.
I have understood the solution but this seems to be a very tough and complex problem.
I am wondering if this can be solved in some other way.

Or for points (6,2) and (0,6) in this case:
Absolute difference for x is 6-0=6
Absolute difference for y is 6-2=4
Using logic, since this is a square, sides need to be equal and average of 6+4 is 5
Hence, absolute difference of 5 units away is for x is x=1 and for y is y=1, with intersection at point (1, 1) being equidistant to B and D
Then work out the distance as before

The distance between the unknown vertices to the midpoint is half the diagonal:

(x−3)2+(y−4)2=(52√2)2=13(x−3)2+(y−4)2=(522)2=13;


I don't understand what this step is doing?

rachitshah
It seems like the steps from the given explanation accomplish the following:
1) Find the distance of the original diagonal: (6-0)^2 + (2-6)^2 = X^2; X = root 52
2) Find the midpoint of the original diagonal: (6+0)/2, (6+2)/2 = (3,4)
3) Find the equation for the line for the other diagonal, since this will be the line that has the coordinates closest to (0,0)
Key to this step: This line is the perpendicular bisector of the diagonal (the diagonals of squares are perpendicular bisectors = fact in geometry)
You can find the equation for the line the other diagonal by doing the following:
a) y = mx + b (generic equation for line)
b) find the slope of the original diagonal: (6-2)/(0-6) = -2/3 and take the negative reciprocal of this = 3/2 since the other diagonal is perpendicular
c) now you have y = 3x/2 + b; find the value of b by plugging in (3,4) and you get y = 3x/2 - 1/2, which is the same thing as y-4 = 3/2(x-3)
4) Distance from farthest coordinates to midpoint of diagonal = Distance from shortest coordinates to midpoint of diagonal = 1/2 the length of the diagonal (in this case, (root52/2)^2).
5) The distance from the farthest or shortest coordinates of the diagonal to the midpoint can be solved by using pythagorean theorem:
(x-3)^2 +(y-4)^2 = (root52/2)^2. (root52/2)^2 = 13
6) From step 3 you know that y-4 = 3/2(x-3). Plug this into y for step 5 above ==> (x-3)^2 + (y-4)^2 = 13 --> (x-3)^2 + [3/2(x-3)]^2 = 13 -->
(x-3)^2 +9/4(x-3)^2 = 13; solving for x gives you (x-3)^2 = 4, so x can be 1 or 5.

I think this is a high-quality question and I agree with explanation.

This step is finding the distance from vertex A(x,y) to mid point M(3,4) or the length of AM you can say.
since AC is also a diagonal, AM will be half its length (which is\sqrt{52} and diagonals of a square are equal in length.)

I hope it clears your doubt.

this is a very very lengthy problem!
It's great for practice though!

Quicker way to solve this:

The midpoint of the diagonal is (3,4). The vector from the right bottom corner to the midpoint is V1 = <-3, -2>. You need to find a vector perpendicular to this vector and of the same magnitude and add it to the midpoint position vector to find the corners.

Vectors perpendicular meet:

-3x - 2y = 0 or y = (3/2) x
So the form is <x, 3/2x>

The two perpendicular vectors that have the same magnitude as V1 will meet:

x^2 + [(3/2)x]^2 = 13
x = +/- 2

So the two vectors are <2,3> or <-2,-3>

To find the corner closest to the origin add <-2,-3> to the position vector of the midpoint <3,4> to get <1,1>.

The magnitude of this vector sqrt(2) is the answer.

\(m = \frac{(y2 - y1)}{(x2-x1)}\)

\(m = \frac{6-2}{0-6} = \frac{-2}{3}\)

The other diagonal of a square is perpendicular to this one and we know slope of the perpendicular line is always \(\frac{-1}{m} = \frac{-1}{-2/3} = \frac{3}{2}\)

Now mid point of two coordinates or the diagonal of the square is \((3,4)\).

\(Slope = \frac{rise}{run}\)

From mid point of \((3,4)\), we need to back traverse a little. Since line is rising 3 and running 2, we reduce rise by 3 and run by 2 while back tracking to come up with the point \((3-1,4-3)\) i.e. \((1,1)\) which is the vertex of the square nearest to \((0,0)\)

\((0,0)\) and \((1,1)\) be solved using Pythagoras theorem. \(\sqrt{1^2 + 1^2}\) and this comes out to be \(\sqrt{2}\)

Few additional inputs to those which are mentioned above and maybe a graphical presentation will help understand the solution in a better way.

Additional points:- As M is the mid point of BD, in the same way M is the mid point of XY.

Solution ----> we come from M------>Y------->A (with coordinates 1,1)

Here's my 2 cents' worth:

1. Given the coordinates, you can find immediately 2 things:
a) Midpoint = (3,4) - just by doing some quick mental math;
b) Gradient = (-2/3) - again, do a quick calculation using the 2 points given.

2. We know that the unknown vertex MUST fall along the other diagonal passing through the midpoint (3,4) i.e. the diagonal whose gradient is perpendicular to what we just calculated.
------> Gradient of other diagonal = 3/2

3. The unknown vertices, when calculating gradient in relation to the midpoint, MUST yield 3/2.
- We also know that the square is in Quadrant I (top right).
- Observe that:

a) (y-4) / (x-3) = 3/2 --> pick 2 numbers:
- (7-4 = numerator; 5-3 = denominator) --> (5,7) works.
- Alternatively, we also know (1-4)/(1-3) gives us -3/-2 = 3/2 (bingo).

b) (1,1) is what we need since it's the closest to the answer.

Distance is simply the hypotenuse of a right-angled isoceles triangle on the graph paper (1 square grid of 1x1) = Root 2. --> Answer = C

To solve this question we need to identify two equations of line passing through that point, call it (x,y)
We need to get slope and a point to get those lines.

First line: From the diagonal - Ld
Second line : From the side - Ls

Ld:
Slope for the diagonal M1
M2= 6-2/0-6 = -2/3
M1*M2= -1
M1= 3/2
Point- midpoint (3,4)
Eqn 1: 3x-2y=1

Ls:
Slope m3 -
tan 45 = m3-m2/1+m3*m2
m2= -2/3 (from above)
Solve, m3= 1/5
We already have the point (0,6)

Eqn 2:x-6y= -30

Solve eqn 1 & 2
(x,y) = (1,1)

Distance formula = \(\sqrt{2}\)

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.