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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]

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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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20 Oct 2013, 20:35
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Expert's post
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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17366 [1], given: 232 Intern Joined: 06 May 2016 Posts: 18 Kudos [?]: 12 [1], given: 5 WE: Education (Education) Re: In the figure above, points P and Q lie on the circle with [#permalink] ### Show Tags 16 May 2016, 13:37 1 This post received KUDOS Hello all, I just got this question in my GMATprep practice test, and came looking here for an explanation afterwards. I understand the math in the comments above, and why the answer is what it is. What concerns me, is that the image is not drawn to scale. It did not say the image was not drawn to scale, and this is an official question, so why could I not assume the image was drawn to scale? (When one has time to look at the question properly it is obvious that it is not drawn to scale - but I have up to now just been assuming that geometry questions are drawn to scale unless clearly indicated that they are not.) Kudos [?]: 12 [1], given: 5 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7676 Kudos [?]: 17366 [1], given: 232 Location: Pune, India Re: In the figure above, points P and Q lie on the circle with [#permalink] ### Show Tags 17 May 2016, 01:39 1 This post received KUDOS Expert's post etienneg wrote: Hello all, I just got this question in my GMATprep practice test, and came looking here for an explanation afterwards. I understand the math in the comments above, and why the answer is what it is. What concerns me, is that the image is not drawn to scale. It did not say the image was not drawn to scale, and this is an official question, so why could I not assume the image was drawn to scale? (When one has time to look at the question properly it is obvious that it is not drawn to scale - but I have up to now just been assuming that geometry questions are drawn to scale unless clearly indicated that they are not.) You cannot solve the questions by assuming that the diagram is drawn to scale. Two angles which look equal may not actually be equal. You can just assume the basics - straight lines that look straight are straight; points are in the order in which they are drawn. But don't take calls based on relative length of lines, relative measure of angles according to the diagram drawn etc. _________________ 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 Oct 2009, 05:29
DenisSh wrote:
Sorry, didn't get that part:
Again, using the distance formula, (s+\sqrt{3})^2 + (t-1)^2 = PQ^2 = 8. ->(2)

The method used to find the length of OP or OQ from origin i.e. the distance formula has been used here as well to find the length of PQ with P as (-[square_root]3,1) and Q as (s,t) which gives us
{s - (-[square_root]3)}^2 + {t-1}^2 = PQ^2 = 8 ->equation 2

from Right angle Triangle POQ we had got the length of PQ^2 as 8

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

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13 Oct 2009, 13:15

S^2 + T^2 = 4 --> equation 1

(t-1)^2 + (s-(-sqrt(3))^2 = 2^2 + 2^2 = 8 --> equation 2

solving for s gives 1
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13 Nov 2009, 07:32
IndianGuardian wrote:
How would you solve it by the distance formula kp1811?

from the figure we can see that OP=OQ = radius of the semicircle. Also Angle POQ = 90degree

thus PQ^2 = OP^2 + OQ^2-------eqn1

using distance formula (O is origin)
OP = sqrt [(1-0)^2 + (sqrt -3 -0)^2] = sqrt [4] = 2 = OQ

also OQ = sqrt [ (s-0)^2 + (t-0)^2] = sqrt[s^2 + t^2]
OQ^2 = s^2+t^2 =4 ----eqn2

now PQ = 2sqrt 2 when we put values of OP and OQ in eqn 1

using distance formula length of PQ = sqrt [ (s+ sqrt3)^2 + (t-1)^2]
PQ^2 = (s+sqrt3)^2 + (t-1)^2 = 8--------eqn3

soving eqn 2 and 3 we get sqrt 3 s = t.Equating this in eqn 2 we get s=1

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17 Nov 2009, 23:51
Because the explanation by IndianGuardian seems incorrect:

Lets see how:
First draw a perpendicular from the x-axis to the point P. Lets call the point on the x-axis N. Now we have a right angle triangle \triangle NOP. We are given the co-ordinates 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\textdegree-60\textdegree-90\textdegreetriangle with sides 1-sqrt3-2 and angle NOP=30\textdegree.

The coordinates of point P is (-\sqrt{3}, 1) ie ON = 1 and PN = root3 - not the other way round as explained in his posting quoted in blue above.

I get PO = QO = 2 (RADIUS)
POQ is a rt angled Triangle.
Hence, isn't PN = 2 sqrt2 ?
So wouldn't s = 2 sqrt2 - sqrt3

Can someone explain how to solve this. Thanks.

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

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18 Jun 2010, 04:35
I understood bunuel's solution ok but i have a problem with the other one.

I got every step of maths but how did you get (t-1) too. However How did you decide that "t" is bigger than 1. Because it can be less then one then it should became (1-t). Can anyone explain it plz just that part??

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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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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, 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: Coordinate Geometry and radius - Gmatprep   [#permalink] 05 Jan 2012, 05:47

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