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# In the figure above, points P

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VP
Joined: 18 May 2008
Posts: 1176
In the figure above, points P  [#permalink]

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13 Nov 2008, 03:35
Help

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Manager
Joined: 08 Aug 2008
Posts: 226

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13 Nov 2008, 08:01
good one really...
at 1st sight, s looks as $$sqrt3$$

now, distance between P and Q=2$$sqrt2$$

now (s+$$sqrt3$$)^2 + (t-1)^2 =8
to solve this equation i need $$sqrt3$$ value for T, hence t=$$sqrt3$$

Hence s=1.
(if u look at the equation closely, it'll be clear....)
Manager
Joined: 23 Jul 2008
Posts: 180

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13 Nov 2008, 12:06
1
prasun84 wrote:
good one really...
at 1st sight, s looks as $$sqrt3$$

now, distance between P and Q=2$$sqrt2$$

now (s+$$sqrt3$$)^2 + (t-1)^2 =8
to solve this equation i need $$sqrt3$$ value for T, hence t=$$sqrt3$$

Hence s=1.
(if u look at the equation closely, it'll be clear....)

it s an interesting method we will have to rely on hit and trial to get the answer
My approach

Starting from the -ve X axix the order of angles formed will be 30,60,30,60
hence applying tan60 in the triangle created in the first quadrant will have sides hypotenuse= 2 base=1 (which is s) perpendicular =root 3(which is t)

hence s is 1
Director
Joined: 14 Aug 2007
Posts: 692

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13 Nov 2008, 20:00
radius = 2 (cant be -ve)

the equation of line passing through point P is,

y = -1/sqrt(3)x [ slope = (1-0)/(-sqrt(3) - 0) and y intercept of the line is 0)

The line passing through point Q is perpendicular to the above line
hence its equation is
y = sqrt(3)x
i.r t = sqrt(3)*s

Now OQ being radius can be expressed as
OQ^2 = s^2 + t^2
2^2 = s^2 + (sqrt(3)*s)^2
4 = S^2 * 4
S^2 = 1

S= 1 (being in Ist quadrant)
VP
Joined: 18 May 2008
Posts: 1176

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13 Nov 2008, 21:42
Thats a very intelligent ans. but shldnt slope of OP be -1/sqrt(3)? (slope=difference in y/difference in x)
alpha_plus_gamma wrote:
radius = 2 (cant be -ve)

the equation of line passing through point P is,

y = -sqrt(3)x [ slope = (-sqrt(3) - 0) / (1-0) and y intercept of the line is 0)
The line passing through point Q is perpendicular to the above line
hence its equation is
y = sqrt(3)x
i.r t = sqrt(3)*s

Now OQ being radius can be expressed as
OQ^2 = s^2 + t^2
2^2 = s^2 + (sqrt(3)*s)^2
4 = S^2 * 4
S^2 = 1

S= 1 (being in Ist quadrant)
VP
Joined: 18 May 2008
Posts: 1176

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13 Nov 2008, 21:48
Wow ! wht a shortcut. Applying angle formula never occured to me.
+1 from me
hibloom wrote:
prasun84 wrote:
good one really...
at 1st sight, s looks as $$sqrt3$$

now, distance between P and Q=2$$sqrt2$$

now (s+$$sqrt3$$)^2 + (t-1)^2 =8
to solve this equation i need $$sqrt3$$ value for T, hence t=$$sqrt3$$

Hence s=1.
(if u look at the equation closely, it'll be clear....)

it s an interesting method we will have to rely on hit and trial to get the answer
My approach

Starting from the -ve X axix the order of angles formed will be 30,60,30,60
hence applying tan60 in the triangle created in the first quadrant will have sides hypotenuse= 2 base=1 (which is s) perpendicular =root 3(which is t)

hence s is 1
Director
Joined: 14 Aug 2007
Posts: 692

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13 Nov 2008, 21:55
ritula wrote:
Thats a very intelligent ans. but shldnt slope of OP be -1/sqrt(3)? (slope=difference in y/difference in x)

Yes you are correct. I have edited my post. thanks
Manager
Joined: 21 Oct 2008
Posts: 131
Schools: Rady School of Management at UC San Diego GO TRITONS

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14 Nov 2008, 12:28
There's a much faster way to do this problem. To find the coordinates of q, all you have to do is switch the x- and y- values of the coordinates of P, then multiply the new y-value (t) by -1.

To understand why, imagine the full circle. Draw any two perpendicular lines through the center. They will intersect the circle at (a, b), (b, -a), (-a, -b), and (-b, a).
Director
Joined: 27 Jun 2008
Posts: 506
WE 1: Investment Banking - 6yrs

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14 Nov 2008, 12:45

http://gmatclub.com/forum/7-t71146

--== Message from GMAT Club Team ==--

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Re: GMATPREP- semicircle &nbs [#permalink] 14 Nov 2008, 12:45
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# In the figure above, points P

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