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cattalk
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GMATPrep Semi-circle on coordinates 19 Feb 2006, 00:53
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I feel really stupid now, but I simply don't get the right answer no matter how many times I do it. It doesn't even look difficult......
Thanks for any help!



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GMATT73
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GMATPrep Semi-circle on coordinates 19 Feb 2006, 02:11
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This is a visual problem that can be easily misleading.

Whatever angle in Quadrant 4 (within the right triangle) must equal the same angle in Quadrant 1 (outside the right triangle). Use similiar triangles and the P-theorum to get 1.

(C)
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GMATPrep Semi-circle on coordinates 19 Feb 2006, 02:56
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You can use angles.
The angle formed by x-axis and OP is equal to 30° because we have a right triangle with sides length ratio 1:sqr(3):2.
You can easily deduce that the angle formed by y-axis and OQ is also 30°. So the sides length ratio is the same as above: s=1 and t=sqr(3).
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GMATPrep Semi-circle on coordinates 19 Feb 2006, 12:14
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Thanks, I see....
But can you tell me how am I thinking through this wrong?
I got to PO and QO equals to 2 alright, then I drew a line connecting P and Q. The length of that line is 2*Sqrt(2) since triangle POQ is now an isosceles. Then I take 2*Sqrt(2), subtract Sqrt(3) from it. (Sqrt(3) came from the x coordinate of point P). I get an answer of
2*Sqrt(2) - Sqrt(3)

I know it's not even a choice, but why is this incorrect? Did I wrongly assume something?
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GMATPrep Semi-circle on coordinates 19 Feb 2006, 13:27
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I think that you assumed that PQ is parallel to x-axis.
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GMATPrep Semi-circle on coordinates 19 Feb 2006, 14:00
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Thank you Petefrom, I understand now.

The ironic thing is that even though I assumed it wrong, and realized my answer was not one of the choices, I estimated the value of 2*Sqrt(2) - Sqrt(3), which happens to be very close to 1.

So I got the right answer, except I was doing it all wrong.... :oops:



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