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In the figure above, points P and Q lie on the circle with
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12 Sep 2005, 09:02
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In the figure above, points P and Q lie on the circle with center O. What is the value of s? A. \(\frac{1}{2}\) B. \(1\) C. \(\sqrt{2}\) D. \(\sqrt{3}\) E. \(\frac{\sqrt{2}}{2}\) Attachment:
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Re: In the figure above, points P and Q lie on the circle with
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27 Apr 2010, 16:02
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).
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Re: In the figure above, points P and Q lie on the circle with
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06 Sep 2009, 13:14
There's a simple reason and misconception as to why people find this question confusing: > the 90 degree angle between P and Q seems like it is cut perfectly right down the centre >> to make two 45 degree angles. > that's why your brain instinctively thinks the answer is r = sq(3), since the distance seems the same from the left and to the right of the origin! However, that's the trick to this question and that's why it is CRUCIAL that you train yourself to be highly critical and weary of accepting the simple answer! Upon closer inspection of the question: P(sq(3),1) Which means the triangle beneath point P has: height = 1 base = sq(3) meaning >> radius = 2 (306090 triangle) This follows the ideal case of the 306090 triangle, which means that opposite to the height of 1, there is an angle of 30 degrees from the xaxis to point P. This means that from the yaxis to point P, there is an angle of 60 degrees (to make a 90 degrees total from xaxis to yaxis). Therefore on the right side, since the figure shows there is a 90 degree angle between point P and Q, 60 degrees have been taken up from the left of the yaxis, which leaves 30 degrees on the right of the yaxis. Furthermore, this leaves 60 degrees from Point Q to the positive xaxis. Therefore, making the exact same 306090 triangle once again, since the radius is the same (the two points lie on the semicircle). The 60 degree angle is now where the 30 degree angle was on the other triangle, which makes the height now sq(3) and base now 1 >> therefore, base r = 1! **Have a look at the attachment and then it should make perfect sense!
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gmatprepPSgeoQ.jpg [ 20.74 KiB  Viewed 165390 times ]




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Re: In the figure above, points P and Q lie on the circle with
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17 Sep 2005, 22:41
there is no need for much calculation,
simply switch their X and Y
P and Q are always perpendicular and same distance from O
OP is perpendicular to OQ means you rotate OQ 90 degrees, you will
get OP, in terms of X,Y value, that is just to switch them and flip the sign
to have (Y,X)



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Re: In the figure above, points P and Q lie on the circle with
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18 Sep 2005, 12:20
Not sure if this is simpler or even different approach from the others(I maybe just formulizing qpoo's approach!)u.....
Lets say slope of OP =m1 and OQ =m2
Since OP perpendicular to OQ => m1m2=1
Since m1= 1/sqrt(3)
=>m2=sqrt3
=>s=1,t=sqrt(3)



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Re: In the figure above, points P and Q lie on the circle with
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27 Apr 2008, 22:49
jbpayne wrote: Thats what I thought, but the answer is B, 1. Anyone able to figure this one out? There are more than one way to solve this. I like this way The line PO has slope =1/√3 The line QO has slope = t/s PO and QO is perpendicular so [1/√3]*t/s = 1 ==> t/s =√3 or √3/1 so s = 1
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Re: In the figure above, points P and Q lie on the circle with
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27 Apr 2008, 23:08
bsd_lover wrote: Could you please elaborate the bolded part ? sondenso wrote: There are more than one way to solve this. I like this way
The line PO has slope =1/√3 The line QO has slope = t/s PO and QO is perpendicular so [1/√3]*t/s = 1 ==> t/s =√3 or √3/1 so s = 1
I know I write like this [the colored] will raise the disscussion. In general slope of PO = a/b slope of QO = t/s PO and QO is perpendicular so [a/b]*[t/s]=1, so t/s = b/a In this case, b = √3, a = 1 , so t/s = [√3]/[1] = √3/1, so s =1
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Re: In the figure above, points P and Q lie on the circle with
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29 Apr 2008, 09:36
sondenso wrote: bsd_lover wrote: Could you please elaborate the bolded part ? sondenso wrote: There are more than one way to solve this. I like this way
The line PO has slope =1/√3 The line QO has slope = t/s PO and QO is perpendicular so [1/√3]*t/s = 1 ==> t/s =√3 or √3/1 so s = 1
I know I write like this [the colored] will raise the disscussion. In general slope of PO = a/b slope of QO = t/s PO and QO is perpendicular so [a/b]*[t/s]=1, so t/s = b/a In this case, b = √3, a = 1 , so t/s = [√3]/[1] = √3/1, so s =1 From the picture t^2 +s^2 = 4(radius = 2) By substituting t =√3s we get s=1



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Re: In the figure above, points P and Q lie on the circle with
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16 Jun 2009, 10:18
The answer is B.
Lets see how: First draw a perpendicular from the xaxis to the point P. Lets call the point on the xaxis N. Now we have a right angle triangle \(\triangle NOP\). We are given the coordinates of P as \((\sqrt{3}, 1)\). i.e. \(ON=\sqrt{3}\) and \(PN=1\). Also we know angle \(PNO=90\textdegree\). If you notice this is a \(30\textdegree60\textdegree90\textdegree\)triangle with sides \(1sqrt32\) and angle \(NOP=30\textdegree\).
Similarly draw another perpendicular from the xaxis to point Q. Lets call the point on the xaxis R. Now we have a right angle triangle \(\triangle QOR\). We know angle \(QRO=90\textdegree\).
Now angle NOP + angle POQ (= 90  given) + angle QOR = 180
\(\Rightarrow 30\textdegree+90\textdegree+angleQOR=180\)
\(\Rightarrow angleQOR=60\)
If you notice \(\triangle QOR\) is also a \(30\textdegree60\textdegree90\textdegree\)triangle with sides \(1sqrt32\) and angle \(OQR=30\textdegree\)
The side opposite to this angle is OR = 1.
Hence answer is B.



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Re: In the figure above, points P and Q lie on the circle with
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23 Aug 2009, 15:54
Check another approach for this problem.
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Re: In the figure above, points P and Q lie on the circle with
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11 Oct 2009, 12:48
Since OP and OQ are the radii of the circle, and we know P has coordinates (\sqrt{3}, 1), from the distance formula we have OP = OQ = 2. Therefore, s^2+t^2=4. > (1) Also from the right triangle, we know the length of PQ=2\sqrt{2}. Again, using the distance formula, (s+\sqrt{3})^2 + (t1)^2 = PQ^2 = 8. >(2)
From (1) and (2) you can solve for s. Hope this helps.



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Re: In the figure above, points P and Q lie on the circle with
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27 Apr 2010, 15:36
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)}.



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Re: In the figure above, points P and Q lie on the circle with
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Updated on: 09 Sep 2010, 17:00
Attachment:
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Line OP => y=mx => 1 = \(sqrt{3} * m\) => slope of OP = \(\frac{1}{sqrt{3}}\) Since OP is perpendicular to OQ => m1*m2 = 1 => slope of OQ = \(sqrt{3}\) equation of line OQ => y = \(sqrt{3}x\) since s and t lie on this line t= \(sqrt{3}s\) also radius of circle = \(sqrt{ 1^2+ 3}\) = 2 = \(sqrt{ s^2+ t^2}\) => \(4 = s^2+ t^2 = s^2 + 3s^2 = = 4s^2\) => s=1  Attachment:
b.jpg [ 13.36 KiB  Viewed 163905 times ]
Using Similar triangles. In triangle APO , angle APO + AOP = 90 1 In triangle BOQ , angle BOQ + BQO= 90 2 Also SINCE angle POQ = 90 as given. angle AOP + angle BOQ = 90 3 Using equations 1 and 3 we get Angle APO = Angle BQO 4 using equation 2 ,3, 4 we get Angle APO = Angle BQO and AOP = OQB Since hypotenuse is common => both the triangle are congruent => base of triangle APO = height of triangle AQO and vise versa. Thus s = y coordinate of P = 1
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Originally posted by gurpreetsingh on 09 Sep 2010, 16:46.
Last edited by gurpreetsingh on 09 Sep 2010, 17:00, edited 1 time in total.



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Re: In the figure above, points P and Q lie on the circle with
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04 Jan 2011, 18:37
tonebreeze: As requested, I am giving you my take on this problem (though I guess you have already got some alternative solutions that worked for you) I would suggest you first go through the explanation at the link given below for another question. If you understand that, solving this question is just a matter of seconds by looking at the diagram without using any formulas. http://gmatclub.com/forum/coordinateplane90772.html#p807400P is (\(\sqrt{3}\), 1). O is the center of the circle at (0,0). Also OP = OQ. When OP is turned 90 degrees to give OQ, the length of the x and y coordinates will get interchanged (both x and y coordinates will be positive in the first quadrant). Hence s, the x coordinate of Q will be 1 and y coordinate of Q will be \(\sqrt{3}\).
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Re: In the figure above, points P and Q lie on the circle with
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05 Jan 2011, 11:46
The method I suggested above is very intuitive and useful. I have been thinking of how to present it so that it doesn't look ominous. Let me try to present it in another way: Imagine a clock face. Think of the minute hand on 10. The length of the minute hand is 2 cm. Its distance from the vertical and horizontal axis is shown in the diagram below. Let's say the minute hand moved to 1. Can you say something about the length of the dotted black and blue lines? Attachment:
Ques1.jpg [ 7.87 KiB  Viewed 32512 times ]
Isn't it apparent that when the minute hand moved by 90 degrees, the green line became the black line and the red line became the blue line. So can I say that the black line \(= \sqrt{3}\) cm and blue line = 1 cm So if I imagine xy axis lying at the center of the clock, isn't it exactly like your question? The x and y coordinates just got inter changed. Of course, according to the quadrants, the signs of x and y coordinates will vary. If you did get it, tell me what are the x and y coordinates when the minute hand goes to 4 and when it goes to 7?
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Re: In the figure above, points P and Q lie on the circle with
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17 Apr 2011, 17:38
ravsg wrote: had it been a DS question, was it okay to assume O as origin ?? It is mentioned in the question that it is a circle with center O. If it is obviously the center, it will be given. If there are doubts and it is not mentioned, you cannot assume it. You cannot assume that the figure will be to scale.
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Re: In the figure above, points P and Q lie on the circle with
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05 Jan 2012, 03:44
Note that this is not a reflection problem. The angle POQ is given as = 90, but angle POY is not the same as angle QOY. To solve, set the product of the slopes of OP and OQ as 1 => 1/sqrt(3) * t/s =1 => t = s*sqrt(3) Also OQ=OP => s^2 + t^2 = 4 => s^2 + 3*s^2 = 4 => s^2 = 1 => s = 1 Option (B)
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Re: In the figure above, points P and Q lie on the circle with
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05 Jan 2012, 05:47
This is not a question on reflection. This is a question on finding the coordinates of a given point within some stated conditions. It would have been a reflection question if angle POY was the same as angle QOY. It is not, so this is not a reflection question. Solve it according to the rules explained in my last post. I took the product of the slopes as 1 because if two lines are perpendicular, the product of their slopes must be 1. OP and OQ are clearly perpendicular as given in the figure, so their slopes must have a product of 1.
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Re: In the figure above, points P and Q lie on the circle with
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19 Oct 2013, 09:01
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: In the figure above, points P and Q lie on the circle with
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20 Oct 2013, 20:35
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: In the figure above, points P and Q lie on the circle with &nbs
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