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

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Re: Plane Geometry, Semicircle from GMATPrep [#permalink]

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30 Sep 2010, 23:53
DenisSh wrote:

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

This is how I solved it.. This will take less than a minute...

Drop perpendiculars from P and Q. Mark the points as X and Y.

Now, PXO and QYO are right angled triangles.

We know, PO =1 and XO = sqrt(3). So PO = radius = 2.

XOP + POQ + QOY =180
so, QOY = 30,
This gives, t= sqrt (3) and s = 1.
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05 Oct 2010, 12:58
OP and OR are prependicular to each other. Hence slope of one must be negative inverse of the other.
Slope of OP = -1/sqrt{3}
Slope of OQ = s/r and it should be equal to sqrt{3}.
Hence s = r * sqrt{3} -----------------(1)

Since OP and OQ represent radius of the circle, their lengths must be equal.
Length of OP = sqrt{(sqrt{3})^2 + 1^2}
Length of OQ = sqrt{r^2 + s^2}

OP = OQ, Also substitue the value of s obtained in (1).
Upon solving, we'll get two values of r +1 and -1. Now since r lies in quad I r has to be +ve. Hence +1.

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Re: Plane Geometry, Semicircle from GMATPrep [#permalink]

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04 Jan 2011, 11:14
That's exactly what I needed. I fully understand the problem now. Thanks Bunuel!

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Re: Plane Geometry, Semicircle from GMATPrep [#permalink]

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04 Jan 2011, 18:37
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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/coordinate-plane-90772.html#p807400

P 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 co-ordinates will get interchanged (both x and y co-ordinates will be positive in the first quadrant). Hence s, the x co-ordinate of Q will be 1 and y co-ordinate of Q will be $$\sqrt{3}$$.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17843 [3], given: 235 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7745 Kudos [?]: 17843 [8], given: 235 Location: Pune, India Re: Plane Geometry, Semicircle from GMATPrep [#permalink] ### Show Tags 05 Jan 2011, 11:46 8 This post received KUDOS Expert's post 2 This post was BOOKMARKED 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 13482 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 co-ordinates 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? _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Plane Geometry, Semicircle from GMATPrep [#permalink]

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13 Feb 2011, 09:40
jullysabat wrote:
BalakumaranP wrote:
DenisSh wrote:

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

This is how I solved it.. This will take less than a minute...

Drop perpendiculars from P and Q. Mark the points as X and Y.

Now, PXO and QYO are right angled triangles.

We know, PO =1 and XO = sqrt(3). So PO = radius = 2.

XOP + POQ + QOY =180
so, QOY = 30,
This gives, t= sqrt (3) and s = 1.

Yeah this is the quickest way I think...
But I think the underlined portion is stated wrongly....

The method with the angles is the quickest one. However, I solved it using the slopes. PO and QO have negative reciprocal slopes. The slope of PO=$$\frac{-1}{\sqrt{3}}$$
The slope QO is the native reciprocal to PO:$$\sqrt{3}$$
This means that we have to solve the system:
$$\frac{t}{s}=\sqrt{3}$$
$$r^2+t^2=4$$

Solve for s and we have $$s=1$$

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Re: GMAT prep question - Geometry [#permalink]

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15 Apr 2011, 03:55
The equation of the line PO is y = -1/sqrt(3) x

The slope of the perpendicular line QO is sqrt(3). Hence the equation of line QO => y = sqrt(3) x

One way is finding the intersection of the circle and the line QO. We know that radius = QO = 2

The equation of the circle is x^2 + y^2 = 2^2 ----(1)
y = sqrt(3) x ----(2)

Solving 1 and 2 we get x = 1, -1 and y = sqrt(3) and -sqrt(3)

Hence x = 1 since we are looking in the first quadrant. Answer B

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Re: Plane Geometry, Semicircle from GMATPrep [#permalink]

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15 Apr 2011, 05:27
t/s * -1/root(3) = -1

t/s = root(3)

3 + 1 = t^2 + s^2

=> 4 = 4s^2

=> s = 1 (as s is +ve)

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Re: Plane Geometry, Semicircle from GMATPrep [#permalink]

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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: Plane Geometry, Semicircle from GMATPrep [#permalink]

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19 Apr 2011, 06:04
I solved it with triangles and not lines.
If we can c that the left triangle have two sides (squrt 3 and 1)
we can immediately know that the hypotenuse = 2 (thats the radii as well)
so we also know very easy that the angels are 30,60,90
and if we know the radii and the angels - we can also know all of that very easy on the right triangle as well - just make sure u put the 30 and 60 deg in the right place.
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Re: Plane Geometry, Semicircle from GMATPrep [#permalink]

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30 Apr 2011, 20:47
for 30 deg triangle sides are in the ratio 3^(1/2) : 1 : 2
for 60 deg triangle sides are in the ratio 1: 3^(1/2) : 2

Hence S = 1.
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Re: ps - value of s [#permalink]

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19 Jul 2011, 22:56
s = 1.
According to the diagram,
we see that OP and OQ are perpendicular.
Let us not assume anything here. The figure might be misleading to some because some may think the points are symmetrical. They are not.

Since, OP is perp. to OQ,
slope of OP * slope of OQ = -1
=> (-1/sqrt3)*(t/s) = -1
=> t/s = sqrt3
=> t = s*sqrt3
Hence, point Q can be written as Q(s,s*sqrt3)

Now, since both points are on the circumference of the circle,
distance to the centre(origin here) will be the same.

Distance from (x,y) to (0,0) is sqrt(x^2 + y^2)
=>dist. from (-sqrt3,1) to (0,0) = sqrt(3+1) = 2 ----(1)
similarly dist. from (s,s*sqrt3) to (0,0) = 2s -----(2)

equating (1) and (2), we get s = 1.
Hope it helps.

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05 Jan 2012, 02:41
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There are some quadrants rules which I kind of forgot something like points on one quadrant are reflection of other. Can somebody explain please
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05 Jan 2012, 03:44
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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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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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05 Jan 2012, 15:39
You can use an alternate approach here.

NOTE: When you have a right angle at origin O and at center of the circle, both intersection edges of the angle (with the circle) will always have the co-ordinates flipped. that is P(x1, y1) = Q(y1, x1), keeping the quadrants into respect.

Imagine a right angle travelling this way inside the circle and how the edges change with the movement.

Therefore, here P(-sqrt(3), 1) for Q will be (1, sqrt(3)). s = 1.
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06 Jan 2012, 11:08
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Is anything wrong in my approach? As first step, I try to find the radius of the circle by finding the distance of OP : Sqrt( 3 + 1) = 2 ; since OP= OQ = 2 ; PQ = Sqrt( 2^2 + 2^2) = 2* Sqrt(2)
our intent is to find "s" - i.e. x co-ordinates of Q -> PQ- (distance of P from O on x axis) -> (2* Sqrt (2)) - Sqrt(3)
Obviously my answer seems wrong ! where am I going wrong ?

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Re: Plane Geometry, Semicircle from GMATPrep [#permalink]

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31 Mar 2012, 22:01
Hi Bunuel,
Request you to provide a detailed reasoning on the below concept mentioned by you.
I am aware of the reflection concepts but it is failing over here. i.e If Point(A,B) is reflected over y axis then the reflected points become (-A,B). I applied this concept and got the answer incorrect. Could you please explain where I am going wrong.
Can't we apply the Reflection concept. If yes, then the angles forming between the lines joining from origin(0,0) to point(A,B) and origin(0,0) to point(-A,B) should be 90.

Thanks
H

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).

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

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01 Apr 2012, 00:58
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]

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29 May 2012, 18:14
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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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Re: In the figure above, points P and Q lie on the circle with   [#permalink] 29 May 2012, 18:14

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