Problem Solving

in the figure attached, Pints p and Q lie on the circle with center O. what is the value Of s?

1/2
1
square root of 2
square root of 3
(square root of 2) / 2

Try looking at this as a triangle problem, since we can draw lines from the 2 points down to the x axis.
We have two triangles. First, let's look at the left one. We have two sides 1^2+(-rt3)^2=a^2
4=a^2 a=2. This is also our radius, and it is equal to the side from O to (s,t). However, we can also notice that this is a 30-60-90 triangle, therefore the angle between the radius and the x axis is 30 degrees. We know that all angles on the line sum to 180, we have the 90 between the lines and this 30 degrees, therefore the angle from 0 to point (s,t) is 180-30-90=60 degrees.
Again, we have a golden triangle, where the side in front of the 30 angle equals half of the hypotenuse.
The hypotenuse is the radius, 2, thus, the side equals 1, and since it is on the x axis, it is the value of s. Although not asked, but we can also know that the value of t is rt3.

Hope it helped.

IMO B
two point distance formula will give the radius as 2
distance (0,0) and (-rut3,1) can be calculated as
r=sqrt((-rut3-0)^2+(1-0)^2)
=sqrt(3+1)=2

simillarly we can get the eqn for (s,t)as

s^2+t^2=4........eqn 1

since rt angle using pythagoras theorem we can get second eqn as
(s+rut3)^2+(t-1)^2=8......(a^2+b^2=h^2).....eqn2

solving eqn 1 and 2
t=s(rut3)
putting this value in eqn 1

s=1
HTH

thank you guys.

i was doing a Mock test yesterday and this was one of the questions - i could not manage to Solve under 2 min. I am Pretty sure i could have pin this down if i had More time.

Leon do you think you could have solved this 2 min, if You first saw it on the Test?

xcusemeplz2009, i undrestand how u got eqn #1 , but not #2

yes. I actually did get this exact question on some test I did and solved it. The solution is simple, it's only a matter of 3:4:5 triangles.
It may just seem long because of my long explanation.