In the figure above, points P and Q lie on the circle with c

[victorgsiu]

GMATPrep CD

Picture: Point P is (-Sqrt(3), 1) and Point Q is (s,t). Point P is in quandrant II and Point Q is in quadrant I. Both P and Q lie on the semi circle. One line connects OP and another line connects OQ. Angle POQ is 90 degrees.

In the figure above, points P and Q lie on the circle with center O. What is the value of s?

A) 1/2
B) 1
C) Sqrt (2)
D) Sqrt (3)
E) Sqrt (2)/2

OA: B

My approach:
Make triangle OXP. OX = -Sqrt (3) long and XP = 1 high. Therefore, by pythagorean theorem, PO = 2.

If PO=2, then OQ=2 (both are radii).

Semi-circle = 180 degrees
POQ = 90 degress
Remaining = 90 degrees.

How do you split the remaining? You can eye-ball and see that the split is about half and half, but gmac pictures are deceiving and the assumption is dangerous.

--
Edit: Figured it out.

Answer:
You know that OXP is a 30-60-90 triangle. Hint: 1, Sqrt(3), 2.

The semi-circle as stated above is 180 degrees.
Subtract the 90 degrees from angle POQ.
Subtract the 30 degrees from angel POX.

This leaves you with a 60-30-90 triangle other side, corresponding to Sqrt(3), x, 2x.

Therefore, x MUST equal 1, what what.

[Ben Ku]

As you stated, the key is using the right angle to find out the angle that OQ makes with the axes. Once you have this, it's just using the 30-60 right triangles to determine the coordinates of point Q.

Let me know if you have additional questions on this problem.

[vijaykumar.kondepudi]

Hi,
I am still not clear regarding the solution for this problem.
Let X be the point of intersection of PQ and Y-axis.

Do we assume the PQ is parallel to the X axis?
I think it isn't. Then which is the 30-60-90 triangle we are refering to?

Please explain in detail.

Thanks.

[RonPurewal]

look here:

gmat-prep-geometry-2-t2493.html

[acethegmat]

Ron, I went through the link. Have a question: why wouldn't one just use a more direct method other than struggling with 30-60-90 angles? once you know the radius is 2, the triangle formed by the two radii lines is rt angle; hence the third side would be 2 root2. 2root2 - root3 = 1.1 = ~1

is this not the correct way to solve it?

[RonPurewal]

Ron, I went through the link. Have a question: why wouldn't one just use a more direct method other than struggling with 30-60-90 angles? once you know the radius is 2, the triangle formed by the two radii lines is rt angle; hence the third side would be 2 root2. 2root2 - root3 = 1.1 = ~1

is this not the correct way to solve it?

nope, that doesn't work.

that would work if the line between P and Q were horizontal -- in that case, you would be able to just perform addition or subtraction with x-coordinates, which is what you're trying to do here -- but the line between P and Q is not horizontal.
the diagram is not drawn to scale, as remarked in that other, longer thread.

also, you can actually tell that this method doesn't work, since the problem asks for an EXACT solution. (remember, by default, if the problem doesn't ask for an approximate solution, than the solution is exact.)
since your method doesn't give you exactly 1 as an answer, it must be incorrect.

[RonPurewal]

by the way --

please post all future responses to this problem on the other thread (the one linked above), NOT this thread.

thanks.