## Wednesday, November 4, 2009

### Fox 174

1. Let the circle with radius R have its centre at O and let the right upper vertex of the rectangle be R from where we draw a tangent RS to the circle. Let the lower right vertex of the rectangle be P. Now OS =RP =R. Since areas A and B are equal, we've Pi*R^2/4 -(Commonn Area) =R*OP -(Common Area). In other words, OP=Pi*R/4 ad RS^2=(R-R(1-Pi^2/16)^(1/2))*(R+R(1-Pi^2/16)^(1/2))=(Pi*R/4)^2 or RS =Pi*R/4. If OS and RP intersect at T then triangles OPT and TRS are congruent and angle SRP = ang. POT. Now let S be (a,b). We've R as (Pi*R/4,R) Since S is a point on the circle we can say: a^2 + b^2 = R^2
((pR/4)-a)^2+(R-b)^2=p^2*R^2/16 and t=b/a where t is the necessary tangent required which turns out to be = 2/Pi - Pi/8
Ajit: ajitathle@gmail.com

2. Looks good to me!

3. Here's another way to do this if you wish to avoid analytical geometry.
Assume the radius of the circle to be Ras b4. By the given condition, π R^2/4 = OP*R or OP=πR/4
But OP = RS Hence RS = πR/4 (Triangle congruence)
Now tan(ORS)=R/(πR/4)= 4/π and tan(PRO)=(πR/4)/R = π/4 Now, tan(SRP) = tan(ORS - PRO)
=[tan(ORS) - tan(PRO)]/(1+ tan(ORS)* tan(PRO))
= [4/π - π/4]/[1+(4/π)*(π/4)] = (4/π - π/4)/2
= 2/π - π/8
Ajit