Thursday, December 10, 2009

Fox 206

We have some work on this, but not the final solution.
Please post your result here if you can find one.


  1. I have a proof but.... forget it:)

  2. Don't be shy share with us. This is really hard!

  3. sqrt(15)/4, of course i am not sure:)

  4. You can ask the biggest area too

  5. sqrt(15)/4 looks reasonable but, I actually don't have the answer. Max area will be a different fox, may be later...

  6. I am sure that the answer is 0.949489745, i will post the solution later if i find a shorter one, now i have too sleep:)

  7. it is interesting that max area-min area=3

  8. Newzad's two last posts are right!

    Let A,B,C the vertixes on inner, middle and outer circles. Put the circles centered on origen O and place A at coordinates [1,0]. Let b=<XOB and c=<XOC. Of course, c depends on b, because of the rectangle angle, so the triangle depends only on angle b.

    We have the A, B, C coordinates:
    A[1,0]; B[2cos(b),2sin(b)];C[3cos(c),3sin(c)]

    and must satisfy:
    vector BA orthogonal vector BC and their scalar product BA.BC is 0. This lead to equation:

    u1 = 2cos(c-b)-cos(c)=(4-2cos(b))/3

    The area of ABC is half the modulus of vectorial product BA x BC. From here we obtain a second equation:

    u2 = 2sin(c-b)-sin(c)=(2Area-2sin(b))/3

    Let's calculate u1^2 + u2^2. It is 5-4cos(b)!!. We are lucky!!!. All c's are gone away!!

    Now whe have:

    (2Area-2sin(b))^2/9+(4-2cos(b))^2/9 = 5-cos(b)

    and, from here, you can obtain Area as function of b:


    Now that my luck is over, I used Maple to find these values (sorry!):
    minimal = 0.9494897434
    maximal = 3.949489742

    César Lozada - Venezuela

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