Semi circle prob

[JA]

Could someone please explain why B is the correct answer and not D?

Thanks!!!

[TakingGMAT]

The cordinates of P are (- root 3, 1). Drop a perpendicular to X axis from P say it touches at point x. The traingle formed will be 30-60-90 triangle whose sides are in ratio 1:root 3: 2. So angle POX (o being the origin) = 30 degreesSame thing applies to traingle formed by point Q. here angle QOX will be 60 degrees and OQX will be 30. hence s=1

[JA]

It seems so easy now. Thanks for your help!!!

[Guest]

I solved it in a different way.

O is the center, so point OP and OQ are radii of the circle.

OP = 2 ( distance formula, coordinates of point P is provided ) = sqrt((0+sqrt(3)) ^ 2 + ( 0 -1)^2 ) = 2
OQ = 2 (radius- thus equal distance)

Since it is a 90 degrees angle: PQ = 2*sqrt(2) (Phyth. theorem)

Thus, we know s^2 + t^2 = 4 ( Distance formula and we know the radius) - Eq 1

Finding the distance between points P and Q
PQ = (s+sqrt(3))^2 + (t -1)^2 = 4*2 (Distance formula)
s^2 + 3 + 2sqrt(3)s + t^2 + 1 - 2t = 8
s^2 + t^2 + 4 + 2sqrt(3)s -2t = 8
Substituting s^2 + t^2 = 4 from Eq 1
4+4+ 2sqrt(3)s-2t = 8
2sqrt(3)s = 2t
sqrt(3)s = t
s = t/sqrt(3)

Then since we want the distance between P,Q to be 2sqrt(2), i substituted s = sqrt(3), t = 3 -- > which doesn't give you the solution, ie the length of the hypothenus. using s =1, t=sqrt(3), length of QP = 2sqrt(2) which is consistent with what we previously found..

[RonPurewal]

a ridiculously long and detailed thread on this problem can be found here.