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

s=1 is correct.

Thank you for the explanation!

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?

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

aroon7, thanks a lot. I didn't notice these 30-60-90 angles.

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)?

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!

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