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m15q19: (6, 2) and (0, 6) are the endpoints of the diagonal

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Re: Coordinate [#permalink]

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29 Jul 2008, 04:46
Its a diagonal of a square.

I really doubt it is a gmat level prob. The only way i can think of solving it ...

1). Slope of given diagonal (call it d1) is -2/3. Hence slope of other diagonal (call it d2) is 3/2.
2). Use 3/2 to find the equation of d2. (it comes out to be 3x-2y-1=0).
2). The midpoint of the d1 is (3,4). This should also be the midpoint of d2.
3). Length of the d1 = length of d2 = 2 * (13^0.5) call it D.

Therefore you need two points at distance D from (3,4) on the line d2. This involves use of sin and cos functions on argument z, where tan(z)=slope of d2=3/2.
Had the angle z been 30 or 60 or 0 degrees, it would have been fairly simple, but with tan(z)=3/2.

The points would be ( 3+D cos(z), 4+D sin(z) ) and ( 3-D cos(z), 4-D sin(z) ). Find the distance of these from the origin to get your answer.

Hope I am not missing something very simple...

Last edited by bhushangiri on 29 Jul 2008, 05:03, edited 1 time in total.
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Re: Coordinate [#permalink]

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29 Jul 2008, 05:11
Yup.. Check out the solution by ian stewart in the thread posted by Durgesh. Neat solution.

I quote IanStewart here..
IanStewart wrote:
".......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). "
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Re: solve in 2 min... [#permalink]

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27 Aug 2008, 21:26
Guessing C, took too long to solve until recall the formula for finding the distance between 2 points.

$$sqrt{(x1-x2)^2 + (y1-y2)^2}$$

so the length of the diag is $$sqrt{52}$$

the diag is also $$X sqrt{52}$$, so one side is X, and X = $$sqrt{26}$$, which is about 5

guessing the closest vertex to the origin is at point (1,1), the diagonal of that square is $$1sqrt{2}$$
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Re: solve in 2 min... [#permalink]

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28 Aug 2008, 15:51
x2suresh wrote:
arjtryarjtry wrote:
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?

* {1}/sqrt{2}
* 1
* sqrt{2}
* [sqrt{3}
* 2/sqrt{3}

shortest and fastest way to solve this ????

mention the time taken also

better way to solve this by using diagram.

mid point of (0,6) and (6,2) is .. (3,4) and slope is -2/3
slope of perpendicular line passing through (3,4 ) 3/2 .=y2-y1/x2-x1
we can find two vertices by decreasing (or increasing) x by 2 and y by 3
other two points are (1,1) and (5,7)

(1,1) is nearest point to (0,0) and distince is sqrt(2).

I took 90 secs .

Suresh ..You got a solution for every problem..
are you a mathematician .??..
man...no kidding..
That is wonderful...I did not think it can be done that way...that is completely new approach for me
I learned something..thanks buddy
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M15 #19 [#permalink]

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05 Oct 2008, 22:05
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?

Answer: Square Root of 2

First, we have to "guess" the coordinates of the vertex of the square closest to the origin. We can do it by making a sketch and keeping in mind that GMAT does not use "difficult" numbers. The coordinates of the vertex is (1, 1) and it is units away from the origin.

Question: Please tell me what I'm "missing" and why the found vertex is 1,1 opposed to 0,2 which would actually make it a rectangle. My "guess" is not working. Please help, thank you.
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Re: Difficult Geometry [#permalink]

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04 Aug 2009, 05:10
yezz wrote:
maliyeci wrote:
Say (a,b) is a vertex of that square that does not makes the end points of the diagonal of that square.

Then, distances to the end points must be exactly 1/srqt2 of the lenght of the diagonal ( diagonal of a square and two sides of the square makes a right triangle)

lenght of diagonal is sqrt52 ( sqrt ((6-0)^2+(6-2)^2) )
so length of the edges of square is sqrt26.

Thus,
lets calculate the distance of (a,b) to end points.

(a-0)^2 + (b-6)^2 = 26
(a-6)^2 + (b-2)^2 = 26

how did you solve these equations??
SOlving these,

(a,b) can be (1,1) or (5,7)

(1,1) has the shortest distance to (0,0) that is sqrt2.

Hi Yezz,
It is a 2 variable equation with 2 equations. So it can be solved easily.
a^2 + b^2 -12b + 36 = 26
a^2 + b^2 -12b-10=0
a^2=12b+10-b^2 (1*)
______________________________________________________

a^2 - 12a +36 + b^2 -4b +4 =26
a^2 - 12a + b^2 -4b +14=0 (putting 1*)
12b+10-b^2-12a+b^2-4b+14=0
8b-12a+24=0
2b-3a+6=0
a=(2b+6)/3 (2*)
____________________________________________
putting 2* in 1*
You can solve.
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Re: Difficult Geometry [#permalink]

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04 Aug 2009, 06:12
maliyeci wrote:
yezz wrote:
maliyeci wrote:
Say (a,b) is a vertex of that square that does not makes the end points of the diagonal of that square.

Then, distances to the end points must be exactly 1/srqt2 of the lenght of the diagonal ( diagonal of a square and two sides of the square makes a right triangle)

lenght of diagonal is sqrt52 ( sqrt ((6-0)^2+(6-2)^2) )
so length of the edges of square is sqrt26.

Thus,
lets calculate the distance of (a,b) to end points.

(a-0)^2 + (b-6)^2 = 26
(a-6)^2 + (b-2)^2 = 26

how did you solve these equations??
SOlving these,

(a,b) can be (1,1) or (5,7)

(1,1) has the shortest distance to (0,0) that is sqrt2.

Hi Yezz,
It is a 2 variable equation with 2 equations. So it can be solved easily.
a^2 + b^2 -12b + 36 = 26
a^2 + b^2 -12b-10=0
a^2=12b+10-b^2 (1*)
______________________________________________________

a^2 - 12a +36 + b^2 -4b +4 =26
a^2 - 12a + b^2 -4b +14=0 (putting 1*)
12b+10-b^2-12a+b^2-4b+14=0
8b-12a+24=0
2b-3a+6=0
a=(2b+6)/3 (2*)
____________________________________________
putting 2* in 1*
You can solve.

Thanks Maliy, i appreciate. it was 3 am here and i tried to solve it for 15 minutes and i couldnt.
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m15 #19 [#permalink]

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06 Aug 2009, 17:14
If 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?

a) 1/sqrt(2)
b) 1
c) sqrt(2)
d) sqrt(3)
e) 2*sqrt(3)

if (0,6) and (6,2) are end points of a diagonal of a square - shouldnt they be two of the vertices of the square too?
also, if it were the vertices, the diagonal length shows that the quadrilateral isnt a square....which is where im most confused.... can anybody help?? thanks in advance.

this is what OA says:

First, we have to "guess" the coordinates of the vertex of the square closest to the origin. We can do it by making a sketch. The coordinates of the vertex are (1,1) and it is units away from the origin.

The correct answer is C.
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Re: Difficult Geometry [#permalink]

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17 Sep 2009, 21:11
[/quote]

Hi Yezz,
It is a 2 variable equation with 2 equations. So it can be solved easily.
a^2 + b^2 -12b + 36 = 26
a^2 + b^2 -12b-10=0
a^2=12b+10-b^2 (1*)
______________________________________________________

a^2 - 12a +36 + b^2 -4b +4 =26
a^2 - 12a + b^2 -4b +14=0 (putting 1*)
12b+10-b^2-12a+b^2-4b+14=0
8b-12a+24=0
2b-3a+6=0
a=(2b+6)/3 (2*)
____________________________________________
putting 2* in 1*
You can solve.[/quote]

Thanks Maliy, i appreciate. it was 3 am here and i tried to solve it for 15 minutes and i couldnt. [/quote]

How do you expect to do this much math in 2 mins. Gotta be a better way
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Re: PS Geometry. [#permalink]

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25 Nov 2009, 04:33
powerka wrote:
I spent an exorbitant amount of time on this problem.

Tried to tackled it via algebra, and got stuck.

This is what I did:
. midpoint= (3,4)
. slope= -2/3
. then perpendicular slope= 2/3
. plugging in (3,4) into y=3/2x+k got k=-1/2 -> y=3/2x-1/2 .....(eq1)
. distance between points (diagonal)= sqrt(6^2+4^2)= sqrt(52)
. half the distance= sqrt(13)
. then distance between midpoint and vertex is given by: sqrt[(x-3)^2+(y-4)^2]=sqrt(13)->(x-3)^2+(y-4)^2=13 ....(eq2)
. plugging equation1 into equation2 I tried to solve to get vertex (1,1) as said by other posters
. but did not succeed. Reached x^2-3x+5=0, which has no real roots.

Good news is that googling around I found a fantastic shortcut:

From (3,4), on a perpendicular line (with slope 3/2), to find a point the same distance from (3,4) as (0,6) is, we decrease x by 2 and decrease y by 3, thus getting vertex (1,1), or we increase x by 2 and increase y by 3, thus getting vertex (5,7).

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

u mean perpendicular slope= 3/2 right?
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Re: Coordinate plane [#permalink]

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02 Mar 2010, 02:33
Thanks for the detailed explanation.
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Re: coordinate [#permalink]

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03 Aug 2010, 05:31
Hi,

The way I see this is that the vertexes are (0,6), (6,6), (0,2) and (6,2).
The closest from (0,0) is (0,2) and hence the distance is sqrt (0^2+2^2) =2

What am I missing here?

regards,
Jack
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Re: coordinate [#permalink]

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03 Aug 2010, 14:21
jakolik wrote:
Hi,

The way I see this is that the vertexes are (0,6), (6,6), (0,2) and (6,2).
The closest from (0,0) is (0,2) and hence the distance is sqrt (0^2+2^2) =2

What am I missing here?

That is not a square - it's a rectangle. Its length is 6 and its width is 4. In this question, the square is slightly rotated; it does not have sides parallel to the x and y axes.
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Re: Coordinate plane [#permalink]

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04 Aug 2010, 11:44
Hey Bunuel,

I understand how the square is placed and I have attached the same for everyone's benefit.

However I want your help for the below things:
(I have not been able to paste the formulae that you have typed)

(1) Length of the diagonal formula. Should we use that formula every time we want to derive the length of diagonal.

(2) The formula that you used for the "Distance between the unknown vertices to the midpoint is half the diagonal" in the very end.
Attachments

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Re: Coordinate plane [#permalink]

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04 Aug 2010, 11:52
seekmba wrote:
Hey Bunuel,

I understand how the square is placed and I have attached the same for everyone's benefit.

However I want your help for the below things:
(I have not been able to paste the formulae that you have typed)

(1) Length of the diagonal formula. Should we use that formula every time we want to derive the length of diagonal.

(2) The formula that you used for the "Distance between the unknown vertices to the midpoint is half the diagonal" in the very end.

Distance between point $$A(x_1,y_1)$$ and $$B(x_2,y_2)$$ can be found by the formula $$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$.

So the length of the diagonal would be the distance between two points B(0,6) and D(6,2): $$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(6-0)^2+(2-6)^2}=\sqrt{52}$$.

For more on this topic check Coordinate Geometry chapter of Math Book (link in my signature).

Hope it helps.
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Re: Coordinate plane [#permalink]

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05 Aug 2010, 02:18
OMG....completely missed all thse informations
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Re: Coordinate plane [#permalink]

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11 Sep 2010, 13:01
Square: ABCD,
Midpoint: M

A(0;6)
C(6;2)

Midpoint is between A and C: M(3;4)

To get the closest vertex:
1. move the square and make the origin the midpoint. M'(0;0), so C'(3;-2)
2. rotate C' clockwise 90 degrees around M' to get B (or D), so C''(-2;-3) = B(-2;-3)
3. move back the square to the original position, so B'(1;1)
4. calculate distance between B' and Origin: sqrt(2)*1
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Re: Coordinate plane [#permalink]

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11 Sep 2010, 22:38
nonameee wrote:
Is it a real GMAT question? I can't even write the solution under 2 minutes (not to mention to solve it under 2 minutes).

Yes it doesn't seem to be realistic GMAT question as it requires lengthy calculations and even backsolving can not help much to get the right answer quickly. But good for practice though.
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Re: Coordinate plane [#permalink]

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19 Sep 2010, 08:01
I found the following way, it seems faster but not perfect I guess.
1) Denote the coordinates of the point we are looking for as (m;n)
2) Build 2 equations based on the sides of the square: x^2=(6-n)^2+m^2 and x^2=(6-m)^2+(2-n)^2
3) Solving these equations we can get 3m-1=2n
4) We can assume that according to the draft m<6 and n<2
5) Thus, a possible solution is n=1, m=1
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Re: Coordinate plane [#permalink]

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21 Sep 2010, 17:37
thanks for the great explanation Bunuel.
Re: Coordinate plane   [#permalink] 21 Sep 2010, 17:37

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m15q19: (6, 2) and (0, 6) are the endpoints of the diagonal

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