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

  2. 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