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In the figure, point P and Q lie on the circle with center

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In the figure, point P and Q lie on the circle with center [#permalink] New post 03 Sep 2007, 08:07
In the figure, point P and Q lie on the circle with center O. What is the value of S?

A) 1/2
B)1
C) sqrt 2
D) sqrt 3
e) sqrt2 / 2

If you draw a line perpendicular from Point P and Q to x-axis, you can see that the two triangle are congruent. My answer is D. The OA is B. Why is that?

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 [#permalink] New post 03 Sep 2007, 08:17
Note: GMAT diagrams are not drawn to scale. You cannot assume you have congruent triangles.

Recognize that this is a 45-45-90 triangle with sides of length 2 and hypotenuse 2sqrt(2).

So QO = s^2 + t^2 = 4

PQ = sqrt(s+sqrt3)^2 + (t-1)^2 = 2sqrt2
2s(sqrt3) = 2t
s(sqrt3) = t

So QO = s^2 + 3s^2 = 4
s^2 = 1
s = +/- 1. (-1 is invalid as Q lies on the positive x plane).

So s = 1
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 [#permalink] New post 03 Sep 2007, 08:35
How did you figure out that

Recognize that this is a 45-45-90 triangle with sides of length 2 and hypotenuse 2sqrt(2).

So QO = s^2 + t^2 = 4
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 [#permalink] New post 03 Sep 2007, 16:37
thanks ywilfred!!

i see what you mean, i believe both triangles are congruent triangles, with 30, 60 and 90. but the triangle from the left has its 30C angle attached to the centre O and the triangle on the left has its 60C angle attached to centre O.
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 [#permalink] New post 03 Sep 2007, 20:43
hey, plz answer beatgmat's Q.
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 [#permalink] New post 03 Sep 2007, 20:51
From the left we have a triangle: 1-sqrt3-2, so radius = 2.
Triangle in the middle is a right triangle with the legs = radius = 2
From the right we have a triangle 1-sqrt3-2, but the leg sqrt3 is situating opposite 60 degree angle and is a vertical line. So S coordinate of this point = 1

Last edited by Whatever on 03 Sep 2007, 20:57, edited 4 times in total.
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 [#permalink] New post 03 Sep 2007, 20:52
somethingbetter wrote:
hey, plz answer beatgmat's Q.


PO=OQ=radius

A triangle that has two sides equal has two adjacent angles equal, since PO=OQ; angle QPR=anglePQO.

The angle between PO and QO is 90, leaving 90 {180 -90} degrees for the other two EQUAL angles.

Thus 90/2 is 45 degrees each angle

45:45:90

sum of angles of any tiangle is 180.
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 [#permalink] New post 04 Sep 2007, 16:48
IrinaOK wrote:
somethingbetter wrote:
hey, plz answer beatgmat's Q.


PO=OQ=radius

A triangle that has two sides equal has two adjacent angles equal, since PO=OQ; angle QPR=anglePQO.

The angle between PO and QO is 90, leaving 90 {180 -90} degrees for the other two EQUAL angles.

Thus 90/2 is 45 degrees each angle

45:45:90

sum of angles of any tiangle is 180.


Don't be tricked by the figure:

"Thus 90/2 is 45 degrees each angle" this is incorrect.

It takes be awhile to see this, if you re-draw the graph yourself and have the up-side-down triangle slightly tited to your left.
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 [#permalink] New post 04 Sep 2007, 16:53
mbunny wrote:
IrinaOK wrote:
somethingbetter wrote:
hey, plz answer beatgmat's Q.


PO=OQ=radius

A triangle that has two sides equal has two adjacent angles equal, since PO=OQ; angle QPR=anglePQO.

The angle between PO and QO is 90, leaving 90 {180 -90} degrees for the other two EQUAL angles.

Thus 90/2 is 45 degrees each angle

45:45:90

sum of angles of any tiangle is 180.


Don't be tricked by the figure:

"Thus 90/2 is 45 degrees each angle" this is incorrect.

It takes be awhile to see this, if you re-draw the graph yourself and have the up-side-down triangle slightly tited to your left.


Yeah, of course,

The the upper part of coordinate system is devided into 30,90 and 60 degrees. and the triagles are not symmetric.
I was saying that the triangle that PQO {that is the one we get when we connect P and Q} is 90:45:45
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 [#permalink] New post 04 Sep 2007, 17:06
IrinaOK wrote:
mbunny wrote:
IrinaOK wrote:
somethingbetter wrote:
hey, plz answer beatgmat's Q.


PO=OQ=radius

A triangle that has two sides equal has two adjacent angles equal, since PO=OQ; angle QPR=anglePQO.

The angle between PO and QO is 90, leaving 90 {180 -90} degrees for the other two EQUAL angles.

Thus 90/2 is 45 degrees each angle

45:45:90



sum of angles of any tiangle is 180.


Don't be tricked by the figure:

"Thus 90/2 is 45 degrees each angle" this is incorrect.

It takes be awhile to see this, if you re-draw the graph yourself and have the up-side-down triangle slightly tited to your left.


Yeah, of course,

The the upper part of coordinate system is devided into 30,90 and 60 degrees. and the triagles are not symmetric.
I was saying that the triangle that PQO {that is the one we get when we connect P and Q} is 90:45:45


I agree that that triangle is a 45:45:90.
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 [#permalink] New post 04 Sep 2007, 19:27
beatgmat wrote:
How did you figure out that

Recognize that this is a 45-45-90 triangle with sides of length 2 and hypotenuse 2sqrt(2).

So QO = s^2 + t^2 = 4


Here you go...

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 [#permalink] New post 18 Oct 2007, 19:20
Ok, I got to the point where now I can see there are two 45 45 90 triangles, how did you get to the 45 45 length of root (3) and 1?

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  [#permalink] 18 Oct 2007, 19:20
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In the figure, point P and Q lie on the circle with center

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