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But this doesnt seem to be a square... It could have been one if one of the endpoints were (6,0) instead of (6,2)
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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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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]
 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]
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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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]
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]
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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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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I cannot understand why do we have to do such complex calculations. If the vertices of the diagonal are 0,6 and 6,2 then the closest vertex will be (0,2). So the shortest distance from 0,0 should be 2.
Or may be I am missing something !
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Ans is C, \sqrt{2}for line 1 cross these two points: y=(-2/3)x+6 the middle point of line 1 is (3,4), and half of the diagonal is \sqrt{13}therefore line 2 is: y=(3/2)x-(1/2) suppose the vertex of line 2 is (x,y) then 13=(3-x)^2+(4-y)^2x=1, 5 so the closest vertex to (0,0) is (1,1) and the distance is \sqrt{2}
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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).
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Re: Difficult Geometry [#permalink]
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: Difficult Geometry [#permalink]
18 Sep 2009, 07:13
Some questions can be solved over 2 minutes. There are 75 minutes and 37 questions. Some can be done under 1 minute even under 30 seconds. Today I scored 51q and although 2 problems got my 6-7 minutes I finished exam in 64 minutes  That will give help you I think. So the easier under 1 that will give you a chance to solve hards in 5-6 minutes.
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Why to do these calculations and why we can not assume end points of diagonals as the vertices of square? Please explain flyingbunny wrote: Ans is C, \sqrt{2}
for line 1 cross these two points:
y=(-2/3)x+6
the middle point of line 1 is (3,4), and half of the diagonal is \sqrt{13}
therefore line 2 is:
y=(3/2)x-(1/2)
suppose the vertex of line 2 is (x,y)
then
13=(3-x)^2+(4-y)^2
x=1, 5
so the closest vertex to (0,0) is (1,1) and the distance is \sqrt{2}
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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]
02 Mar 2010, 02:33
Thanks for the detailed explanation.
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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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Exactly, tht's what i did. But, the OE is way diffn. Just waiting for someone to come up with that explanation
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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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