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Re: coordinate geometry [#permalink]
 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]
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: Plane Geometry, Semicircle from GMATPrep [#permalink]
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]
30 Sep 2010, 23:53
DenisSh wrote: Please help me to solve this problem.
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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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]
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]
13 Feb 2011, 09:40
jullysabat wrote: BalakumaranP wrote: DenisSh wrote: Please help me to solve this problem.
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=4Solve for s and we have s=1
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Re: GMAT prep question - Geometry [#permalink]
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]
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) Answer - B
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Re: Plane Geometry, Semicircle from GMATPrep [#permalink]
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]
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]
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]
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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Re: Plane Geometry, Semicircle from GMATPrep [#permalink]
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. Please explain. 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]
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]
29 May 2012, 18:14
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.
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In the figure above, points P and Q lie on the circle with [#permalink]
18 Nov 2012, 08:19
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)
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Coordinate System and Geometry [#permalink]
19 Nov 2012, 22:47
Can someone explain how to do this problem.. I have no clue how to even start on this.
Attachments
Geometry.PNG [ 19.18 KiB | Viewed 612 times ]
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Re: Coordinate System and Geometry [#permalink]
19 Nov 2012, 23:30
are you sure about \sqrt{3} as the answer. As per my calculations i got \sqrt{2} as the answer. If I am correct, then I will surely post my answer.
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Re: Coordinate System and Geometry [#permalink]
19 Nov 2012, 23:47
Step 1: Measure the radius. 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.
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solution.png [ 13.5 KiB | Viewed 594 times ]
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Re: Coordinate System and Geometry
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19 Nov 2012, 23:47
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