Wednesday, November 11, 2009

Fox 184


  1. Let the centre of the circle be O, its radius = r and the point vertically below O on the circle P, the bottom left square vertex, Q and the concerned point of tangency, R. Now OQ = V2*r and PQ = r/V2 ; hence tan(QOP) = (r/V2)/r =1/V2 or angle QOP = 35.2644 and therefore angle ROP = 2*35.2644 = 70.5288 deg. Now angle PQR = 180-70.5288 = 109.4712 deg. since OPQR is concyclic. Finally, Φ = 109.4712 - 90 = 19.4712 deg and sin(Φ)= 0.3333.. or = 1/3

  2. Good work again. But you should be able to compute sin(Φ) without computing the angles.
    Thanks Ajit.

  3. Let ABCD the square with A the vertex of angle phi. Let T the tangency point on side AD.
    Let r=1 the radius of circle with center O.
    Then AT=x=1/sqrt(2) because parallel to side AB passing through the point T, also pass through point O, and the parallel to side AD passing through O meet on side AB at point P, and by construction OD=AT=x=DB and the triangle DBO is rectangle with x^2+x^2=1.
    Now we draw segment AO. The angle TAO = alpha, and sin(alpha) = 1/sqrt(1+x^2)= sqrt(2/3).
    We've theta = pi/2-alpha. Then phi+theta=alpha.
    cos(theta)=sin(alpha), and finally:
    sin(phi+theta)=sin(phi)cos(theta)+cos(phi)sin(theta)=sqrt(2/3) -> sqrt(2)*sin(phi)+cos(phi)=sqrt(2) -> 3sin^2(phi)-4sin(phi)+1=0 -> sin(phi)=1/3 or sin(phi)=1, but phi=90 is the tangent segment AD.

    Note: sqrt = square root

    Answer B.

  4. Erratum

    Where say OD=AT=x=DB, must say: