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Source: — Problem Solving |
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by live2win » Sat May 22, 2010 5:14 pm
IMO B

LPOQ = 90 => line PO and OQ are perpendicular
If two lines are perpendicular then prodcuct of the slopes is -1
slope of PO = -1 / SQRT3 slope of OQ is SQRT3 hence S = 1 and t = sqrt3
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by Patrick_GMATFix » Sat May 22, 2010 10:27 pm
Nice job live2win.

This one is very tough, but tests whether you realize that:

1) OP = OQ. and
2) If you drew a line straight down from P to the x-axis, you would form a triangle of base 1 and height root(3), so it will be a 30-60-90 triangle.

The angle measures are key because from them you can deduce a lot about s and t. it's tough to explain without drawings but the answer is B.

Have a look at a detailed solution, including a video solution: this is GMATprep question 1082

-Patrick
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by LulaBrazilia » Sun May 23, 2010 8:20 pm
Ii couldn't solve this Q. I got as far as recognizing that the radii must be equal, but I focused on trying to find the length of PQ. The video solution is really helpful. Thanks!
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by albatross86 » Tue May 25, 2010 12:01 am
Slope of line OP = -1/sqrt(3)
So, slope of line OQ = sqrt(3) = t/s

so t = sqrt(3)*s

OP = 2 = OQ

OQ = sqrt(s^2 + t^2)
= sqrt(s^2 + 3*s^2)
= 2s

So 2s = 2

=> s = 1
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by webdk4life » Tue May 25, 2010 8:11 am
You can also use the distance formula to calculate PO. Given that 2 radii are the same and 90 degrees in the middle = 45 45 90 .. - Hypotenuse = PO*SQRT (2). Then you can use S-P = PO*SQRT(2)

and knowing sqrt of 2 ~ 1.4. you can estimate
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