Geometry GMAT Prep question

Question

gmat367.jpg

maliyeci · Answer

length of OP = sqrt{sqrt(3)^2+1^2}=2 
This means, length of OQ=radius of circle= 2 
i.e.  s^2+t^2=4 
on the other hand
when you draw a line from p to q you get a triangle OPQ. This is a right triangle. length of PQ (hypotenus) =  2sqrt2 
that is equal to  sqrt{(s+sqrt3)^2+(t-1)^2} 
i.e.
 (s+sqrt3)^2+(t-1)^2=8 
 s^2+3+2.s.sqrt3 + t^2-2t+1=8 
 s^2+2.s.sqrt3 + t^2-2t=4 
since  s^4+t^2=4 
 2.s.sqrt3-2t=0 
 2.s.sqrt3=2t 
 t=2.s.sqrt3/2 
 t=s.sqrt3 
putting it in first equation
 s^2+(s.sqrt3)^2=4 
 4s^2=4 
 s=1

defoue · Answer

Any onther way to solve it?

maliyeci · Answer

Exactly, yes! 
As you can see in the figure below.
The angles noted as x are equal. As you notice from the aux. lines, there are two similar triangles in the figure. 
Soooo
 s/t  is equal to  1/sqrt3 . As hypotenuses of those two triangles are same in length. s becomes 1 and t becomes  sqrt3

LenaA · Answer

and another approach:

observe that lines OP and QO are perpendicular => the product of theri slopes=-1
Hence,  \frac{1}{ \sqrt{3}} 	imes \frac{t}{s}=-1 
and from the above  s=1  and  t=\sqrt{3}

gmate2010 · Answer

another way;

using pythagorus theorem, we can find the radius of the circle, r =  \sqrt{( 3})^2 + 1   = 2

Hence, equation of circle becomes:  x^2 + y^2 = 4

any points on the circle can be taken as(2cosA,2sinA) Let (s,t) = (2cosA,2sinA)

equation of OQ can be written as:-  y = \sqrt{3}*x

r.,s lies on OQ as well..

=> 2SinA = \sqrt{3}* 2CosA => A = 60 degrees

=> r = 1

Geometry GMAT Prep question

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