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Re: Plane Geometry, Semicircle from GMATPrep [#permalink]
13 Oct 2009, 04:29

DenisSh wrote:

Sorry, didn't get that part:

badgerboy wrote:

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

Re: coordinate geometry [#permalink]
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

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

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.

Re: Plane Geometry, Semicircle from GMATPrep [#permalink]
13 Feb 2011, 08:40

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: \(\frac{t}{s}=\sqrt{3}\) \(r^2+t^2=4\)

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. _________________

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. _________________

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.

Re: Coordinate Geometry and radius - Gmatprep [#permalink]
05 Jan 2012, 04:47

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. _________________

Re: Coordinate Geometry and radius - Gmatprep [#permalink]
05 Jan 2012, 14:39

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. _________________

I am the master of my fate. I am the captain of my soul. Please consider giving +1 Kudos if deserved!

DS - If negative answer only, still sufficient. No need to find exact solution. PS - Always look at the answers first CR - Read the question stem first, hunt for conclusion SC - Meaning first, Grammar second RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min

Re: Coordinate Geometry and radius - Gmatprep [#permalink]
06 Jan 2012, 10:08

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 ?

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.

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

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...