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What is the value of s?

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by rajataga » Sat Jan 03, 2009 1:08 am
IMO B

s = 1

after every 90degrees the co-ordinates will get inverted with the sign assigned according to the quadrant in which the point lies.

For eg.,

suppose Point P was at (-2,0), then a point on the same circle 90 degrees away from that point would have been either (0,2) or (0,-2), depending on the quadrant.

Here, Point Q lies in the first quadrant, hence both 's' and 't' will be positive, and the co-ordinates of Point P will be inverted.

Hence, Q = (1,root3)
and s = 1
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by ocean » Sat Jan 03, 2009 2:51 am
s=1 is correct.

Thank you for the explanation!
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by umaa » Sat Jan 31, 2009 2:37 pm
But when you measure it, you'll get sqrroot 3 for x, right? P and Q are in the same distance. Y is same. I don't understand the logic. Can you please explain me?
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by aroon7 » Sat Jan 31, 2009 5:29 pm
umaa wrote:But when you measure it, you'll get sqrroot 3 for x, right? P and Q are in the same distance. Y is same. I don't understand the logic. Can you please explain me?
A really tricky one...
pl refer diagram in attachment

we have,
angle AOB = 90
OD = sqrt(3)
AD = 1
so, triangle OAD is right triangle with sides 1, sqrt(3), 2. this is a 30-60-90 triangle
we can see that OD > AD
so angle DAO should be 60 and angle AOD should be 30

now,
angle DOA + angle AOB + angle BOC = 180
30 + 90 + angle BOC =180
angle BOC = 60

trianlge OBC is another 30-60-90
since angle BOC is larger (=60) than angle OBC (=30),
BC = sqrt(3) and
OC = 1

hence point B is (1, sqrt(3))

HTH
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circle1.jpg
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by umaa » Sat Jan 31, 2009 6:08 pm
aroon7, thanks a lot. I didn't notice these 30-60-90 angles.
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by abcdefg » Mon Jul 20, 2009 1:07 pm
hmmm the above solution makes perfect sense. Thanks. However please take a look at the picture I've uploaded.

Shouldnt the 2 x coordinates add up to the 2*(square root 2)?
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by ogbeni » Fri Aug 21, 2009 10:04 am
Wow this question is such a trick question and just demonstrates that you cannot neglect detail on this exam. My instinct was to assume that the angles on both right triangles were similar and got it wrong. Thanks for the detailed explanation!
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by LalaB » Tue Dec 13, 2011 9:28 am
since PO=OB=2, then OQ=2*sqroot2

the distance between two points P and Q is -

sqroot((s+sqroot3)^2+(t-1)^2)= 2*sqroot2
sqroot3*S=t

S^2+t^2=OQ^2
3*S^2+S^2=4
S=1
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by Anurag@Gurome » Tue Dec 13, 2011 6:41 pm
ocean wrote:Hi,
can somebody explain the attached question to me?
Thx and regards
ocean

Point P and Q lies on the same circle with center at (0, 0).
Thus, (s² + t²) = (-√3)² + 1² = 3 + 1 = 4

Again line segments OP and OQ are perpendicular.
Thus (slope of OP)*(slope of OQ) = -1

Slope of OP = 1/(-√3) = -(1/√3)
=> Slope of OQ = (t - 0)/(s - 0) = t/s = (-1)/(-1/√3) = √3
=> t = √3s

Thus, (s² + (√3s)²) = 4
=> (s² + 3s²) = 4
=> s² = 1
=> s = ±1

As point Q lies in the first quadrant s = 1.

The correct answer is B.
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