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Re: Plane Geometry, Semicircle from GMATPrep [#permalink]
10 Dec 2013, 05:34

Bunuel wrote:

zz0vlb wrote:

I need someone like Bunuel or walker to look at my statement and tell if this is true.

Since OP and OQ are perpendicular to each other and since OP=OQ=2(radius) , the coordinates of Q are same as P but in REVERSE order, keep the sign +/- based on Quandrant it lies. in this case Q(1,\sqrt{3)}.

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

Yes you are right.

Point Q in I quadrant (1, \sqrt{3}); Point P in II quadrant (-\sqrt{3}, 1); Point T in III quadrant (-1, -\sqrt{3}) (OT perpendicular to OP); Point R in IV quadrant (\sqrt{3},-1) (OR perpendicular to OQ).

Is it the case of reflection in Y axis ? I doubt that but still want to clarify..

Re: Plane Geometry, Semicircle from GMATPrep [#permalink]
10 Dec 2013, 05:40

Expert's post

ygdrasil24 wrote:

Bunuel wrote:

zz0vlb wrote:

I need someone like Bunuel or walker to look at my statement and tell if this is true.

Since OP and OQ are perpendicular to each other and since OP=OQ=2(radius) , the coordinates of Q are same as P but in REVERSE order, keep the sign +/- based on Quandrant it lies. in this case Q(1,\sqrt{3)}.

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

Yes you are right.

Point Q in I quadrant (1, \sqrt{3}); Point P in II quadrant (-\sqrt{3}, 1); Point T in III quadrant (-1, -\sqrt{3}) (OT perpendicular to OP); Point R in IV quadrant (\sqrt{3},-1) (OR perpendicular to OQ).

Is it the case of reflection in Y axis ? I doubt that but still want to clarify..

No. (-\sqrt{3}, 1) is not a reflection of (1, \sqrt{3}) around the Y-axis. The reflection of (1, \sqrt{3}) around the Y-axis is (-1, \sqrt{3}).
_________________

Re: In the figure above, points P and Q lie on the circle with [#permalink]
11 Dec 2013, 11:51

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

This problem took be a bit to conceptualize but I got it as soon as I realized that the graph I was looking at wasn't drawn completely accurately.

If you are to accurately graph the point -√3,1 on a piece of graph paper, you see that the angle it forms with regards to the origin (and the x plane) is less than 45 degrees. Because of this, and the fact that both points lie on the circle, and that the two radii form a 90 degree angle, Q must be higher on the circle than P. Upon looking at this graph, I thought that the x and y coordinates of Q would be the same (except positive) which is, I am sure, the intent of the test makers. The answer is simply the positive reciprocal of point P.

B. 1

gmatclubot

Re: In the figure above, points P and Q lie on the circle with
[#permalink]
11 Dec 2013, 11:51