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PS # half circle

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PS # half circle [#permalink] New post 02 Nov 2005, 13:26
00:00
A
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D
E

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(N/A)

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0% (00:00) correct 0% (00:00) wrong based on 1 sessions
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please explain...
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 [#permalink] New post 02 Nov 2005, 13:39
it looks like it is sqrt(3) as the angle POQ is 90 degrees and therefore it is a reflection.

another way to think about it
what is the result of rotating a radius 90 degrees along the circumference.
in that case what is the effect in the cosine?

make sense?
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 [#permalink] New post 02 Nov 2005, 13:49
can you repost the question..not sure what we are asking for s, t?
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 [#permalink] New post 02 Nov 2005, 13:56
I get 2sqrt2 - sqrt3... which is neither of the choices...

We know that the radii must be the same for both. We have two triangles...
one with legs of length 1 and sqrt3.... a similar triangle, thus the 3rd length, or the radius, must be 2.
The second triangle with the 90┬░ angle is then a 45-45-90. So the line connecting points P and Q should have a length of sqrt2.

Deduct the length of the x coordinate from the first point, gives you the x coordinate of the second.

:EDIT:
just realized, that this only works when the line PQ is parallel to the x axis... which must not be the case.
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 [#permalink] New post 02 Nov 2005, 17:29
Wow, this is a great one. Don't let the diagram fool you here.

First, as was previously mentioned you can see that the radius is 2.

Now calculate the distance between the two points by drawing a line between them. Obviously the distance between them is equal to the hypotenuse of the triangle with the two radii as its legs.

So distance between them = sqrt(2^2+2^2) = sqrt(8)

Now the distance formula between to points says that:
DF = (s+sqrt(3))^2+(t-1)^2 had better equal sqrt(8)^2 = 8 as well.

We need one more equation that relates s and t and we are ready to go.
That equation of course is that s^2+t^2 = 2^2 (eqn of the circle)

Now solve for t^2 = 4 - s^2 and t = sqrt(4-s^2)

Now go back to the distance formula above (DF), multipy it out and then plug in the expressions above for t^2 and t. Now solve for s and you will see that s = 1 and t = sqrt(3).

So the drawing is playing a little trick as the y axis isn't really bisecting the 90 deg angle. As a quick check if it were the other way around and s = sqrt(3) and y = 1, then the distance between the points would simply be sqrt(3) + sqrt(3) = 2sqrt(3) which does not equal sqrt(8) (or 2sqrt(2)).
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 [#permalink] New post 02 Nov 2005, 21:23
first of all, we should find the function of the two lines OP and OQ
from the coordinates provided we can easily find:
OP: y=-1/sqrt3 *x
OQ: y= t/s*x
Becoz OP and OQ are perpendicular --> t/s* ( -1/sqrt3)= -1
---> t/s= sqrt3--> t= s*sqrt3-->t^2= 3*s^2
We have:
s^2+t^2= (-sqrt3)^2+1^2= radius^2 = 4
--> s^2+3*s^2= 4 ---> s=- or + 1
becoz Q belongs to the II portion of the coordination --> s=1

Last edited by laxieqv on 02 Nov 2005, 21:30, edited 1 time in total.
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 [#permalink] New post 02 Nov 2005, 21:30
S = sqrt3 by "eyeballing" and under 10 seconds 8-)

OA?
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 [#permalink] New post 02 Nov 2005, 21:34
first of all, we should find the function of the two lines OP and OQ
from the coordinates provided we can easily find:
OP: y=-1/sqrt3 *x
OQ: y= t/s*x
Becoz OP and OQ are perpendicular --> t/s* ( -1/sqrt3)= -1
---> t/s= sqrt3--> t= s*sqrt3-->t^2= 3*s^2
We have:
s^2+t^2= (-sqrt3)^2+1^2= radius^2 = 4
--> s^2+3*s^2= 4 ---> s=- or + 1
becoz Q belongs to the II portion of the coordination --> s=1
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 [#permalink] New post 03 Nov 2005, 11:29
good job guys... OA is 1.

how long did it take you guys to do it... for me it took over 4 minutes... is it okay?
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  [#permalink] 03 Nov 2005, 11:29
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