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Re: In the figure above, points P and Q lie on the circle with [#permalink]
29 May 2012, 17:14

Beautiful solution by Karishma! Turns out a strong imagination is all it takes to solve this problem. Turn the radius joined by origin and (- root3,1) 90 degrees and you will notice the values will be swapped, with their signs dependent on the quadrant into which the shift takes place. If still not clear, you can think of any point on the x axis, say (5,0). what happens if you turn it 90 degrees upwards? It ends up on the Y-axis at (0,5). Once you see it, you can't unsee it. I am floored! Thank you Karishma! Following you from here on.

Re: Coordinate System and Geometry [#permalink]
19 Nov 2012, 22:30

Expert's post

are you sure about \(\sqrt{3}\) as the answer. As per my calculations i got \(\sqrt{2}\) as the answer. If I am correct, then I will surely post my answer. _________________

Re: Coordinate System and Geometry [#permalink]
19 Nov 2012, 22:47

Expert's post

Step 1: Measure the radius. That will come out to be 2. Step 2: Observe the diagram. Since both OP and OQ are the radii, their measure will be 2 only. Step 3: Draw a line parallel to x axis, from P to Q. Step 4: Angle OQP and angle OPQ= 45 In the diagram attached, the angles marked X are 45 and 45 degrees. Step 5: Draw a perpendicular from Q to x axis. The point at which it meets the X axis, mark it as A. Step 5: The triangle OPA is a 45:45: 90 degree tiangle. This triangle has the property that the hypotenuese:side 1:side 2 will be in the ration \(\sqrt{2}\):1:1. Step 6: Since side 1(OA) and side 2(AQ) are equal, therefore let them be x. These will be equal to \(2/\sqrt{2}\). Step 7: The length of side 1 will come out to be \(2/\sqrt{2}\) or simply \(\sqrt{2}\).

Please let me know what my mistake, if my solution is wrong.

Re: Coordinate System and Geometry [#permalink]
19 Nov 2012, 23:51

Marcab wrote:

Step 1: Measure the radius. That will come out to be 2. Step 2: Observe the diagram. Since both OP and OQ are the radii, their measure will be 2 only. Step 3: Draw a line parallel to x axis, from P to Q. Step 4: Angle OQP and angle OPQ= 45 In the diagram attached, the angles marked X are 45 and 45 degrees. Step 5: Draw a perpendicular from Q to x axis. The point at which it meets the X axis, mark it as A. Step 5: The triangle OPA is a 45:45: 90 degree tiangle. This triangle has the property that the hypotenuese:side 1:side 2 will be in the ration \(\sqrt{2}\):1:1. Step 6: Since side 1(OA) and side 2(AQ) are equal, therefore let them be x. These will be equal to \(2/\sqrt{2}\). Step 7: The length of side 1 will come out to be \(2/\sqrt{2}\) or simply \(\sqrt{2}\).

Please let me know what my mistake, if my solution is wrong.

your solution is wrong in Step 3. PQ is parallel to x axis is an assumption in the solution and has not been stated in question. Figures are not drawn to scale. _________________

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.

Re: In the figure above, points P and Q lie on the circle with [#permalink]
12 Dec 2014, 22:19

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: Coordinate Geometry and radius - Gmatprep [#permalink]
06 May 2015, 00:51

gmat is basic

there are two triangle which are the same. from this we can see the value of s.

this is most simple way

if you like my explanation, pls, give e a kudos _________________

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Re: Coordinate Geometry and radius - Gmatprep
[#permalink]
06 May 2015, 00:51

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