In the figure shown, points P and Q lie on the circle

[Harish Dorai]

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




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

[givemeanid]

OP = OQ
(1-0)^2 + (-sqrt3-0)^2 = (s-0)^2 + (t-0)^2
1+3 = s^2+t^2

s^2+t^2 = 4 ---------(1)


Since the lines are perpendicular to each other, product of slopes = -1
[(1-0)/(-sqrt3-0)]*[(t-0)/(s-0)] = -1
t/s = sqrt3
t = s*sqrt3
From (1), s^2 + 3s^2 = 4
s^2 = 1
Since (s,t) is in first quadrant, s=1

[Harish Dorai]

Good explanation! 1 is the correct answer.

[Guest]

I don't understand this explanation at all. Can someone help out?

[bag]

GMAT Prep says the answer to this one is radical 3. I'm assuming they screwed up on this one and the correct answer is indeed "1". Can anyone help corroborate this? Thx!

[bag]

Ignore the previous post. GMAT Prep says the answer is indeed "1." If someone can punch out an explanation that makes sense to those of us not fluent in PASCAL, it would be most appreciated!

[RonPurewal]

Ignore the previous post. GMAT Prep says the answer is indeed "1." If someone can punch out an explanation that makes sense to those of us not fluent in PASCAL, it would be most appreciated!

there is a very thorough treatment of this problem here.

[mahesh_kumarg]

I am new to the forums. Seems this thread is quite old, so this query might have been further dicussed in some new thread.

I solved this problem using simple trigonometry. We know the radii = 2 = hypotenuse for the triangles. From given inormation we can find that the angle split is 60:30, so the side ratio ( x - cordinates) is sqrt(3):1

Thanks
Mahesh

[Ben Ku]

You can do it that way. Although it might involve trig, you should be familiar enough with 30-60 right triangles to see how this works.

[borhan11]

My first post, attempting to give an explanation for those not fluent in PASCAL ;) I actually didn't understand this fully until I read that PASCAL code above, but then I realized a simple shortcut can be applied.

Here's the shortcut:

Since we have a right triangle we can leverage the relationship between the slop of the perpendicular lines PO and OQ, which is inverse negative.

Slope of PO is -1/Root(3) - (no need to simplify this.)
Thus the slope of OQ becomes Root(3)/1 or just Root(3)

Because we did not simplify PO's slope we know that our derived slope for OQ is exactly in terms of the units of the (x,y) points it belongs to as well (because radius PO = radius OQ)

So based on our new slope we have Root(3) increase per 1 unit of x... which puts Q (s,t) at (1,Root(3)) Thus s=1

[RonPurewal]

here's another thread on this problem, which has about a zillion trillion posts:
gmat-prep-geometry-2-t2493.html

i'm locking this thread; if you have any more questions on this problem (probably unlikely, after you read through all of the million zillion posts in that thread), please post them on that thread.