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

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In the figure above, points P and Q lie on the circle with [#permalink] New post 18 Nov 2012, 07:19
you can break the triangle into 2 individual triangles, find the height and base for the one the left i.e. h=1 and b=-root(3)

therefore b^2+h^2=r^2---->1^2+root(-3)^2=r^2=4

r=2

Use this now to find the Height on the other side, the BASE will remain the same as root(-3)
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Re: Coordinate System and Geometry [#permalink] New post 19 Nov 2012, 22:30
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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.
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Re: Coordinate System and Geometry [#permalink] New post 19 Nov 2012, 22:47
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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.
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solution.png [ 13.5 KiB | Viewed 3933 times ]


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Re: Coordinate System and Geometry [#permalink] New post 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. :-D
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Re: Coordinate System and Geometry [#permalink] New post 19 Nov 2012, 23:55
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agarboy wrote:
Can someone explain how to do this problem.. I have no clue how to even start on this.

In the attached picture, from question OA = \(\sqrt{3}\)
PA =1
Thus radius, OP = 2

Thus OPA is a 30, 60, 90 triangle. where angle POA is 30.
Also angle POQ is 90, given in question.

therefore, angle QOB = 60
And hence, angle OQB =30

trianlge QOB is also 30, 60, 90 triangle. in which we know OQ =OP =2 (radius)
Therefore BQ = \(\sqrt{3}\)
and OB = 1

We need value of r which is BQ = \(\sqrt{3}\)

Ans D it is!
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Re: Coordinate System and Geometry [#permalink] New post 19 Nov 2012, 23:59
Expert's post
Vips0000 wrote:
agarboy wrote:
Can someone explain how to do this problem.. I have no clue how to even start on this.

In the attached picture, from question OA = \(\sqrt{3}\)
PA =1
Thus OA = 2


Thus OPA is a 30, 60, 90 triangle. where angle POA is 30.
Also angle POQ is 90, given in question.

therefore, angle QOB = 60
And hence, angle OQB =30

trianlge QOB is also 30, 60, 90 triangle. in which we know OQ =OP =2
Therefore BQ = \(\sqrt{3}\)
and OB = 1

We need value of r which is BQ = \(\sqrt{3}\)

Can you please explain the red part. It should OP i sppose.
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Re: gmatprep geo [#permalink] New post 19 Oct 2013, 08:01
chetanojha wrote:
Check another approach for this problem.


Can we assume that coordinates of the center O are (0,0) here, can we always assume so?
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Re: gmatprep geo [#permalink] New post 20 Oct 2013, 19:35
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obs23 wrote:
chetanojha wrote:
Check another approach for this problem.


Can we assume that coordinates of the center O are (0,0) here, can we always assume so?


From the figure it is clear that O lies at (0, 0). O is the point where x axis and y axis intersect in the figure.
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Re: Plane Geometry, Semicircle from GMATPrep [#permalink] New post 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..
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Re: Plane Geometry, Semicircle from GMATPrep [#permalink] New post 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 \((-1, \sqrt{3})\).
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Re: In the figure above, points P and Q lie on the circle with [#permalink] New post 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.

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

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