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GMAT Prep: Geometry

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Manager
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GMAT Prep: Geometry [#permalink]

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23 Jul 2008, 18:16
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Question in attachment.
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GMAT Prep- Geometry_part2.doc [59.5 KiB]

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Re: GMAT Prep: Geometry [#permalink]

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24 Jul 2008, 02:48
OP = 2
because of circle radius, OP = OQ

therefore s^2+t^2 = 4 --- 1

now using Pyth Thm

OP^2+OQ^2 = PQ^2

s^2+t^2+4=(s+sq.root(3))^2 + (t-1)^2

solving this we get t^2 = 3s^2

hence using (1) we get s= 1

what is OA?
thanks

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Re: GMAT Prep: Geometry [#permalink]

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24 Jul 2008, 04:17
I also got 1 as the result.

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Re: GMAT Prep: Geometry [#permalink]

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24 Jul 2008, 07:01
OA is 1. Thanks for the solution!

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Re: GMAT Prep: Geometry [#permalink]

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24 Jul 2008, 07:20
And what is the explanation for OA?

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Re: GMAT Prep: Geometry [#permalink]

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24 Jul 2008, 07:37
Draw a line from P to Q

Let the intercept of line segment PQ and y axis be A

PO^2 = 1^2 + (-3^0.5)^2
PO = 2
so, QO = 2

consider triangle POA:
COS angle POA = AO/PO = 1/2
so, angle POA = 60

So angle QOA = 30
s/QO = SIN angle QOA = SIN 30 = 1/2
As QO = 2
s = 1

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Re: GMAT Prep: Geometry [#permalink]

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26 Jul 2008, 03:05
Not to be a spoiler, but I think the use of COS, SIN creates more complications especially for those of us that has forgotten these rules.

I think this question really comes down to knowledge and usage of the Pyth Thm. Once you can find out OP, OA (created by test taker), PQ (PA, AQ) and OQ. Then we are quickly are able to find the coordinates of s and t. And therefore establish that s is 1.

Correct me if I am wrong.

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Re: GMAT Prep: Geometry [#permalink]

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08 Aug 2008, 10:45
Is the angle between PO and OQ 90 degrees? If so, can't we just do the following:

PO = sqrt(sqrt(3)^2+1^2) = 2 = QO since both PO and OQ represent the radii of the circle
since PO = QO and the angle between them is 90 degrees, then the other 2 angles in the POQ triangle are 45 degrees and thus PQ = 2*sqrt(2)

since we know that PQ comprises of the x-portion of PO and the x-portion of OQ and we know the x-portion of PO = sqrt(3) than the x-portion of OQ is 2*sqrt(2)-sqrt(3) which is approximately 1.09 (not exactly 1, but close).

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Re: GMAT Prep: Geometry [#permalink]

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08 Aug 2008, 11:12
balboa wrote:
Question in attachment.

Here is my approach.

OP ^2= (-sqrt(3))^2 + 1^2
-- OP =2 and OQ=2

SLOPE OF OP = Y2-Y1/X2-X1 = (1-0)/(-SQRT(3) -1) = -1/SQRT(3)
SLOPE OF OQ = SQRT(3)
becaue OP and OQ are perpendicular.

t-0/s-0 = sqrt(3) ---> t=s *sqrt(3)
s*s + t*t = 2*2
s*s (3+1)=4 --> s=1
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Re: GMAT Prep: Geometry [#permalink]

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08 Aug 2008, 11:21
from the figure we get radius=2
also s^2+t^2=4 ------->I
(s+3^1/2)^2 +(t-1)^2=8 => s*(3 ^ 1/2)=t => s=1 or s=-1 =>
s=1 is the correct solution 1t coordinate

hence IMO B
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Re: GMAT Prep: Geometry [#permalink]

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09 Aug 2008, 06:24
balboa wrote:
Question in attachment.

Just my 2 cents without using pen and paper.

Right angle triagle two sides -sqrt(3), 1 So other side OP =2
OQ=OP =2
OP and OQ are perpendicular lines
point (Q-(x,y)) should be inthe ratio 1:sqrt(3)
obvisously other side is 2
clearly x=1 y=sqrt(3)
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Re: GMAT Prep: Geometry   [#permalink] 09 Aug 2008, 06:24
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GMAT Prep: Geometry

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