Saturday, November 28, 2009

Fox 196

The problem has been modified for the answer was not precise.
Let us know if you see a problem.


  1. See Fox 60 for definition of angle phi.

  2. Let’s assume the circle centre to be O and the right top triangle vertex to be A with triangle side=2a. A is (a,V3a/2) & the circle: x^2+y^2= 3a^2/4 while the tangent is:xx1+yy1=3a^2/4 where (x1,y1) is the point of tangency to be found out. We’ve x1^2+y1^2=3a^2/4 and the tangent passes thru. A. So ax1+V3ay1/2 =3a^2/4. This gives us x1=6a/7 and y1=-V3a/14 while the slope of the tangent is 12/V3. Hence tan(Φ)=(12/V3–V3)/(1+(V3*12/V3))=9/(13V3) and sin(Φ)=tan(Φ)/(1+(tan(Φ))^2)^(1/2)~ 0.37115.
    Option A is close but not exact. Is there some error in my calculation?

  3. Ajit, I think you are making a mistake when you are writing the equation of the tangent line:
    ax1+V3ay1/2 =3a^2/4 does not look right.

    It should be:
    ax1 - ay1 / V3 = a^2/2 OR

    -aV3 x1 + ay1 = -V3 a^2 /2 OR

    Let me know if that leads to an naswer.

  4. [IMG][/IMG]

  5. OK. This question needs a correction.
    Ajit you're right. sinΦ = 0.371
    newzad, yours is correct too but, you've calculated tanΦ, instead of sinΦ.

    Let me revise the question and post again.
    This time cosΦ will be asked for it looks better without sqrt roots.

    Sorry about that.

  6. Let the corner angle between the two tangents be θ. It is easy to prove that tan(θ)=4V3 and tan(Φ)=tan(θ-60)=(4V3-V3)/(1+4V3*V3)=3V3/13 whence sin(Φ)=3V3/14=0.3711537, exactly the same answer as b4.