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# geometry Q

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Senior Manager
Joined: 07 Feb 2008
Posts: 301

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20 May 2009, 06:40
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Please don't forget the explanations. Thanks.

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Manager
Joined: 10 May 2009
Posts: 63

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20 May 2009, 08:38
First find distance between P and O.

Dist OP=2
Therefor PQ=2 *2^1/2

(2 *2^1/2)^2 = (s+3^1/2)^2 + (t-1)^2

Since t=1, Therefore s=1

Hence B.
Manager
Joined: 28 Jan 2004
Posts: 201
Location: India

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22 May 2009, 21:51
Can you pls. tell why t is 1?
Intern
Joined: 19 May 2009
Posts: 5

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22 May 2009, 22:47
t is NOT equal to 1 . Otherwise distance OQ should be (2)^0.5 and that´s out of the question since OP=OQ=2 .

I guess many people incorrectly assume from the drawing that the Y axis bisects the right angle POQ when it is not really the case.

First solution (The long one):

Draw perpendicular segments to the X axis from points P and Q, call the intersection points of these segments with the X axis P´ and Q´ respectively. You will now have two right-angle triangles POP´ and QOQ´

Applying basic trigonometry: Sin(POP´) = |PP´| / |OP| = 1/2 so

angle POP´=arcsin(1/2) = 30° , since angle POP´ + angle QOQ´=90° then angle QOQ´= 60°

s= |OQ| * cos (angle QOQ´) = 2*cos(60°) = 2*0.5 =1

Second Solution (The short one)= Since |OQ|=|OP| and they form a right angle at the origin, points P(x1,y1) and Q(x2,y2) will have the following symmetry(symmetry of sine and cosine values for complementary angles) : y1 = x2=s = 1 (Our answer! ) and |x1|=y2=t

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Re: geometry Q   [#permalink] 22 May 2009, 22:47
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