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# GMAT Prep: Geometry

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Manager
Joined: 23 Jun 2008
Posts: 140

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23 Jul 2008, 18:16
Question in attachment.

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GMAT Prep- Geometry_part2.doc [59.5 KiB]

Intern
Joined: 22 Apr 2008
Posts: 35

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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
Senior Manager
Joined: 14 Mar 2007
Posts: 277
Location: Hungary

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24 Jul 2008, 04:17
I also got 1 as the result.
Manager
Joined: 23 Jun 2008
Posts: 140

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24 Jul 2008, 07:01
OA is 1. Thanks for the solution!
SVP
Joined: 17 Jun 2008
Posts: 1502

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24 Jul 2008, 07:20
And what is the explanation for OA?
Senior Manager
Joined: 19 Mar 2008
Posts: 350

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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
Intern
Joined: 19 Feb 2007
Posts: 9

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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.
Intern
Joined: 05 Aug 2008
Posts: 20

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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).
SVP
Joined: 07 Nov 2007
Posts: 1757
Location: New York

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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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VP
Joined: 17 Jun 2008
Posts: 1322

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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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Its Now Or Never

SVP
Joined: 07 Nov 2007
Posts: 1757
Location: New York

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

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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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