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Coordinate Geometry/Triangles Problem from GMAT Prep

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Coordinate Geometry/Triangles Problem from GMAT Prep

by tonebeeze » Sun Jan 02, 2011 6:23 pm
I got this problem wrong on my practice CAT. Can someone please walk me through this problem. Thanks!


Image

In the figure above, points P and Q lie on the circle with center O. What is the value of S?

a. 1/2
b. 1
c. square root of 2
d. square root of 3
e. (square root of 2) / 2
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by anshumishra » Sun Jan 02, 2011 6:38 pm
p(-sqrt3, 1), O(0,0), Q(s,t)
slope of OP, m1 = -1/sqrt3
slope of OQ, m2 = t/s

Since OP and OQ make 90 degree.
m1*m2 = -1 => -1/sqrt3 * t/s = -1
=> t/s = sqrt3 => t=s*sqrt3

OP = OQ
s^2+t^2 = 3+1 = 4
=> 4s^2 = 4
=> s = 1


Hence b
Last edited by anshumishra on Sun Jan 02, 2011 7:13 pm, edited 1 time in total.
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by Night reader » Sun Jan 02, 2011 6:51 pm
tonebeeze wrote:I got this problem wrong on my practice CAT. Can someone please walk me through this problem. Thanks!


Image

In the figure above, points P and Q lie on the circle with center O. What is the value of S?

a. 1/2
b. 1
c. square root of 2
d. square root of 3
e. (square root of 2) / 2
you need to keep in mind that center O of the circle is at origin (0,0) then s is x-coordinate of P with the opposite sign OR sqrt(3)

answer D.
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by tonebeeze » Sun Jan 02, 2011 7:09 pm
The official answer is (b) 1. Please explain.
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by anshumishra » Sun Jan 02, 2011 7:10 pm
Night reader wrote:
tonebeeze wrote:I got this problem wrong on my practice CAT. Can someone please walk me through this problem. Thanks!


Image

In the figure above, points P and Q lie on the circle with center O. What is the value of S?

a. 1/2
b. 1
c. square root of 2
d. square root of 3
e. (square root of 2) / 2
you need to keep in mind that center O of the circle is at origin (0,0) then s is x-coordinate of P with the opposite sign OR sqrt(3)

answer D.
Night reader,
What if P was not (-sqrt 3, -1) ? In fact select some point further towards left for P. Do you still think the x co-ordinate of P and Q will still be the same ? NO, right ?
It was luckily a symmetrical condition.
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Anshu

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by tonebeeze » Sun Jan 02, 2011 7:10 pm
Night reader wrote:
tonebeeze wrote:I got this problem wrong on my practice CAT. Can someone please walk me through this problem. Thanks!


Image

In the figure above, points P and Q lie on the circle with center O. What is the value of S?

a. 1/2
b. 1
c. square root of 2
d. square root of 3
e. (square root of 2) / 2
you need to keep in mind that center O of the circle is at origin (0,0) then s is x-coordinate of P with the opposite sign OR sqrt(3)

answer D.
Your logic was the same as my logic.
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by anshumishra » Sun Jan 02, 2011 7:14 pm
tonebeeze wrote:The official answer is (b) 1. Please explain.
Hi,
Edited my solution. Please check it now. I had made a calculation mistake where instead of t/s = sqrt3, i had written 1/sqrt3.
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by anshumishra » Sun Jan 02, 2011 7:30 pm
tonebeeze wrote:I got this problem wrong on my practice CAT. Can someone please walk me through this problem. Thanks!


Image

In the figure above, points P and Q lie on the circle with center O. What is the value of S?

a. 1/2
b. 1
c. square root of 2
d. square root of 3
e. (square root of 2) / 2
Alternate Solution :

The angle which OP makes with -ve side of x-axis = 30
Angle POQ = 90
So, angle which QO makes with +ve x-axis = 60 => t/s = sqrt 3
Now since OQ = 2, So t = sqrt3, and s=1.
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Anshu

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by Night reader » Sun Jan 02, 2011 7:39 pm
ok, i first went on with min-max for s and t, then switched to hypotenuse PQ sqrt(8) where x-coordinate would be not more than sqrt(2)...

now I see the obvious solution in applying trigonometric relationship for 1-sqrt(3)-2 on the left, around point P => angle PO-x-abscess is 30` and angle QO-x-abscess must be 60` (180- POQ 90`), so the angle POQ is skewed to the left => from sin of 60` on the right side around Q we may determine radius=2 (hypotenuse) and the other two sides x-coordinate=1 and y-coordinate=sqrt(3).

hope this satisfies the possible complexities of such kind of problems on GMAT...
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