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

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Re: In the figure above, points P and Q lie on the circle with [#permalink]  31 Mar 2012, 23:58
since these two lines are perpendicular, the result of multiplication of their slopes is (-1)

((1-0)/(-sqrt3-0))*((t-0)/(s-0))=-1

t/s=sqrt3/1

s=1
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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.
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In the figure above, points P and Q lie on the circle with [#permalink]  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]  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.
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Re: Coordinate System and Geometry [#permalink]  19 Nov 2012, 22:47
Expert's post
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.
Attachments

solution.png [ 13.5 KiB | Viewed 3985 times ]

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Re: Coordinate System and Geometry [#permalink]  19 Nov 2012, 23:51
Marcab wrote:
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.
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Re: Coordinate System and Geometry [#permalink]  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]  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: 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..
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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 $$(-1, \sqrt{3})$$.
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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.

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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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Re: In the figure above, points P and Q lie on the circle with [#permalink]  24 Jul 2015, 05:28
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).

Hi Bunuel

Cant we do this by applying 45-45-90 getting distance between O and Q as 1/sqroot 3? As OP=OQ (radii)? but with this I get t/s=1/sqroot3. Pls advise
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Re: In the figure above, points P and Q lie on the circle with [#permalink]  24 Jul 2015, 06:10
sinhap07 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).

Hi Bunuel

Cant we do this by applying 45-45-90 getting distance between O and Q as 1/sqroot 3? As OP=OQ (radii)? but with this I get t/s=1/sqroot3. Pls advise

Let me try to answer this.

$$\angle{POA} = 30$$ by following steps:

In triangle POA, OA = $$\sqrt{3}$$, AP = 1, thus this triangle is a 30-60-90 (as the sides are in the ratio $$1:\sqrt{3}:2$$) triangle (either remember this or use trigonometry to figure it out!).

Now, per your question, triangle OPQ is isosceles with 45-45-90 triangle and sides in the ratio $$1:1:\sqrt{2}$$

Thus, PQ should be = $$\sqrt{2}$$ and not 1/sqroot 3
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GPrep.jpg [ 6.59 KiB | Viewed 23 times ]

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Re: In the figure above, points P and Q lie on the circle with   [#permalink] 24 Jul 2015, 06:10

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