Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 06 Dec 2013, 17:12

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:
Intern
Joined: 27 Jul 2010
Posts: 20
Location: Bangalore
Followers: 0

Kudos [?]: 2 [0], given: 0

Re: Plane Geometry, Semicircle from GMATPrep [#permalink]  30 Sep 2010, 22:53
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.
_________________

Nothing is free.. You 've to earn it!!!

Manager
Joined: 15 Apr 2010
Posts: 170
Followers: 4

Kudos [?]: 21 [0], given: 3

Re: gmatprep geo [#permalink]  05 Oct 2010, 11:58
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.
Manager
Status: First-Year MBA Student
Joined: 19 Nov 2009
Posts: 127
GMAT 1: 720 Q49 V39
Followers: 6

Kudos [?]: 37 [0], given: 209

Re: Plane Geometry, Semicircle from GMATPrep [#permalink]  04 Jan 2011, 10:14
That's exactly what I needed. I fully understand the problem now. Thanks Bunuel!
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 3741
Location: Pune, India
Followers: 804

Kudos [?]: 3175 [2] , given: 137

Re: Plane Geometry, Semicircle from GMATPrep [#permalink]  04 Jan 2011, 17:37
2
KUDOS
Expert's post
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}.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 3741 Location: Pune, India Followers: 804 Kudos [?]: 3175 [3] , given: 137 Re: Plane Geometry, Semicircle from GMATPrep [#permalink] 05 Jan 2011, 10:46 3 This post received KUDOS Expert's post 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 6684 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 Save$100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Manager
Joined: 19 Aug 2010
Posts: 78
Followers: 2

Kudos [?]: 8 [0], given: 2

Re: Plane Geometry, Semicircle from GMATPrep [#permalink]  13 Feb 2011, 08:40
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
Director
Status: Matriculating
Affiliations: Chicago Booth Class of 2015
Joined: 03 Feb 2011
Posts: 934
Followers: 10

Kudos [?]: 158 [0], given: 123

Re: GMAT prep question - Geometry [#permalink]  15 Apr 2011, 02:55
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
SVP
Joined: 16 Nov 2010
Posts: 1693
Location: United States (IN)
Concentration: Strategy, Technology
Followers: 26

Kudos [?]: 252 [0], given: 35

Re: Plane Geometry, Semicircle from GMATPrep [#permalink]  15 Apr 2011, 04:27
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)

_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 3741
Location: Pune, India
Followers: 804

Kudos [?]: 3175 [0], given: 137

Re: Plane Geometry, Semicircle from GMATPrep [#permalink]  17 Apr 2011, 16:38
Expert's post
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.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save \$100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Senior Manager
Joined: 08 Nov 2010
Posts: 426
Followers: 6

Kudos [?]: 28 [0], given: 161

Re: Plane Geometry, Semicircle from GMATPrep [#permalink]  19 Apr 2011, 05:04
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.
VP
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1372
Followers: 9

Kudos [?]: 109 [0], given: 10

Re: Plane Geometry, Semicircle from GMATPrep [#permalink]  30 Apr 2011, 19:47
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.
_________________

Visit -- http://www.sustainable-sphere.com/
Promote Green Business,Sustainable Living and Green Earth !!

Manager
Joined: 25 Sep 2010
Posts: 85
Schools: HBS, LBS, Wharton, Kelloggs, Booth
Followers: 0

Kudos [?]: 2 [0], given: 15

Re: ps - value of s [#permalink]  19 Jul 2011, 21:56
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.
Senior Manager
Joined: 07 Sep 2010
Posts: 341
Followers: 2

Kudos [?]: 44 [0], given: 136

Re: Plane Geometry, Semicircle from GMATPrep [#permalink]  31 Mar 2012, 21:01
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).

_________________

+1 Kudos me, Help me unlocking GMAT Club Tests

Senior Manager
Joined: 23 Oct 2010
Posts: 361
Location: Azerbaijan
Followers: 9

Kudos [?]: 95 [0], given: 68

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
_________________

Happy are those who dream dreams and are ready to pay the price to make them come true

Intern
Joined: 07 May 2011
Posts: 42
GMAT 1: Q V
GMAT 2: Q V
Followers: 0

Kudos [?]: 7 [0], given: 11

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.
Intern
Joined: 19 Aug 2012
Posts: 18
Followers: 0

Kudos [?]: 2 [0], given: 10

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)
Verbal Forum Moderator
Status: Preparing for the another shot...!
Joined: 03 Feb 2011
Posts: 1425
Location: India
Concentration: Finance, Marketing
GPA: 3.75
Followers: 98

Kudos [?]: 450 [0], given: 62

Re: Coordinate System and Geometry [#permalink]  19 Nov 2012, 22:30
Expert's post
As per my calculations i got \sqrt{2} as the answer.
If I am correct, then I will surely post my answer.
_________________
Verbal Forum Moderator
Status: Preparing for the another shot...!
Joined: 03 Feb 2011
Posts: 1425
Location: India
Concentration: Finance, Marketing
GPA: 3.75
Followers: 98

Kudos [?]: 450 [0], given: 62

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 1605 times ]

_________________
Director
Status: Done with formalities.. and back..
Joined: 15 Sep 2012
Posts: 626
Location: India
Concentration: Strategy, General Management
Schools: Olin - Wash U - Class of 2015
WE: Information Technology (Computer Software)
Followers: 29

Kudos [?]: 288 [0], given: 23

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.
_________________
Director
Status: Done with formalities.. and back..
Joined: 15 Sep 2012
Posts: 626
Location: India
Concentration: Strategy, General Management
Schools: Olin - Wash U - Class of 2015
WE: Information Technology (Computer Software)
Followers: 29

Kudos [?]: 288 [2] , given: 23

Re: Coordinate System and Geometry [#permalink]  19 Nov 2012, 23:55
2
KUDOS
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 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 (radius)
Therefore BQ = \sqrt{3}
and OB = 1

We need value of r which is BQ = \sqrt{3}

Ans D it is!
Attachments

pqrs.jpg [ 10.93 KiB | Viewed 1159 times ]

_________________
Re: Coordinate System and Geometry   [#permalink] 19 Nov 2012, 23:55
Similar topics Replies Last post
Similar
Topics:
In the attached figure, points P and Q lie on the circle 8 15 Jun 2006, 23:55
In the figure, point P and Q lie on the circle with center 11 03 Sep 2007, 08:07
1 In the figure below, points P and Q lie on the circle with 15 25 Oct 2007, 10:13
In the figure above, points P and Q 1 24 Dec 2008, 15:45
1 In the figure above, points P and Q lie on the circle with 1 29 Jan 2012, 11:41
Display posts from previous: Sort by