In the figure above, points P and Q lie on the circle with : GMAT Problem Solving (PS)
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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]

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12 Sep 2005, 08:02
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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. $$\frac{\sqrt{2}}{2}$$
[Reveal] Spoiler: OA

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17 Sep 2005, 21:41
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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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18 Sep 2005, 11:20
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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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27 Apr 2008, 21:49
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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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27 Apr 2008, 22:08
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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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29 Apr 2008, 08:36
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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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16 Jun 2009, 09:18
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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 $$30\textdegree-60\textdegree-90\textdegree$$triangle with sides $$1-sqrt3-2$$ and angle $$NOP=30\textdegree$$.

Similarly draw another perpendicular from the x-axis to point Q. Lets call the point on the x-axis 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 $$30\textdegree-60\textdegree-90\textdegree$$triangle with sides $$1-sqrt3-2$$ and angle $$OQR=30\textdegree$$

The side opposite to this angle is OR = 1.

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23 Aug 2009, 14:54
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Check another approach for this problem.
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06 Sep 2009, 12:14
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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 (30-60-90 triangle)

This follows the ideal case of the 30-60-90 triangle, which means that opposite to the height of 1, there is an angle of 30 degrees from the x-axis to point P. This means that from the y-axis to point P, there is an angle of 60 degrees (to make a 90 degrees total from x-axis to y-axis). 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 y-axis, which leaves 30 degrees on the right of the y-axis. Furthermore, this leaves 60 degrees from Point Q to the positive x-axis. Therefore, making the exact same 30-60-90 triangle once again, since the radius is the same (the two points lie on the semi-circle). 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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gmatprep-PSgeoQ.jpg [ 20.74 KiB | Viewed 103111 times ]

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

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11 Oct 2009, 11:48
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Since OP and OQ are the radii of the circle, and we know P has co-ordinates (-\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 + (t-1)^2 = PQ^2 = 8. ->(2)

From (1) and (2) you can solve for s.
Hope this helps.
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Re: Plane Geometry, Semicircle from GMATPrep [#permalink]

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13 Oct 2009, 04:29
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, 12:15

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

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09 Nov 2009, 19:25
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Correct me if I'm wrong, (and forgive my english)
the coordinates of a point p(x,y) that lays on a circle can be expressed as p(sen,cos).
we know that sen π/6 = cos π/3,
and that sen π/3 = cos π/6

in this case I just have to invert x with y and y with x -> indeed we have p(y,x)

As we know x and y have to be positive we have to change the sign for x, the new coordinates are p (y,-x).

in this way to solve the problem doesn't take more than 20 secs.
hope I was enough clear.
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13 Nov 2009, 06:32
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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gmat prep line semi circle [#permalink]

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17 Nov 2009, 09:52
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I solved it in the foll manner:

X is point 1 on the y axis

POQ = 90 deg - given
PX = rt 3 - given
XO = 1 - given

Hence PO = 2 (30, 60, 90 triangle)

POQ is 90 deg
90 deg angle rule - a, a, rt2 a
PO = 2
QO = 2
Hence PQ = 2 rt2

XQ = 2 rt2 - rt3
approx - 2.8 - 1.7 = 1

Last edited by study on 02 Dec 2009, 10:43, edited 3 times in total.
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17 Nov 2009, 22:51
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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27 Apr 2010, 14:36
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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: Plane Geometry, Semicircle from GMATPrep [#permalink]

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27 Apr 2010, 15:02
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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: Plane Geometry, Semicircle from GMATPrep [#permalink]

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18 Jun 2010, 03:35
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: Help on a Geometry PS Question [#permalink]

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09 Sep 2010, 15:46
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Attachment:

a.jpg [ 12.91 KiB | Viewed 101963 times ]

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 101901 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 co-ordinate of P = 1
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Last edited by gurpreetsingh on 09 Sep 2010, 16:00, edited 1 time in total.
Re: Help on a Geometry PS Question   [#permalink] 09 Sep 2010, 15:46

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