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Tough Coordinate Geometry problem from the gmac test

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Tough Coordinate Geometry problem from the gmac test [#permalink]

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New post 10 Jul 2010, 19:17
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A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

36% (06:06) correct 64% (01:37) wrong based on 41 sessions

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Can anyone help with this?

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-the-figure-above-points-p-and-q-lie-on-the-circle-with-85154.html
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problem.GIF
problem.GIF [ 5.22 KiB | Viewed 3286 times ]

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Re: Tough Coordinate Geometry problem from the gmac test [#permalink]

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New post 12 Jul 2010, 10:29
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Hence we see that the correct answer is 1.

Hope this helps.

PS: In the triangle on the right the angles are wrong. 30 is 60 and 60 is 30. Sorry for that.
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Re: Tough Coordinate Geometry problem from the gmac test [#permalink]

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New post 10 Jul 2010, 21:53
The answer is B, because of the tg 45 = 1. Look, OQ is 45 degrees to Ox.
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Re: Tough Coordinate Geometry problem from the gmac test [#permalink]

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New post 11 Jul 2010, 01:22
Well, this took me way more than 2 minutes, but knowing the answer I could come up with the following method (not sure if its recommended):

PO sqare + OQ square = PQ square

Equation of circle with centre 0,0 is x square + y square = radius square

putting in coordinates of P into the equation we get radius of 2 (thus PO = OQ = 2)

=> 2SQ. + 2SQ. = PQ sq.
=> root 8 = 2 root 2 = PQ = approx 2 X 1.4 = 2.8

Now PQ = distance from x coordinate of P to center + distance of x coordinate of Q to center

=> 2.8 = root 3 (or 1.7) + s
=> 2.8 -1.7 = ~1 = s

Guys/Gals, is there a more elegant way to do this? there must be...
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New post 12 Jul 2010, 08:20
Can someone explain this please!!!
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Re: Tough Coordinate Geometry problem from the gmac test [#permalink]

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New post 12 Jul 2010, 11:02
Great diagram! Thanks and Kudos!
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Re: Tough Coordinate Geometry problem from the gmac test   [#permalink] 12 Jul 2010, 11:02
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