Saturday, August 15, 2009

Fox 127

The ones whose shoulders you rise on, may shape who you are.


  1. For a solution with pictures, see:

    First we give names:
    The big triangle is ABC with the right angle at C. The big square is CEDF, with D on AB, E on BC and F on AC.
    P, Q and R are vertices of the left 4x4-square, with P on AD, Q on DF and R on AF.
    S, T and U are points on the right 3x3-square, with S on BD, T on DE and U on BE.
    SD = x.
    The triangles DST, FRQ, PQD and ETU are similar right-angled triangles:
    - With Pythagoras: TD = sqrt(9+x^2)
    - FQ = 3*4/sqrt(9+x^2)=12/sqrt(9+x^2)
    - QD = 4/x*sqrt(9+x^2)
    - TE = x*3/sqrt(9+x^2)
    DE = DT + TE = sqrt(9+x^2) + 3x/sqrt(9+x^2)
    DF = QD + FQ = 4/x*sqrt(9+x^2) + 12/sqrt(9+x^2) = 4/x * DE
    But CEFD is a square and DE = DF.
    So 4/x = 1 or x = 4.
    This gives: DE = DF = 5 + 12/5 = 37/5.
    The area of the biggest square CEFD = (37/5)^2 = 54,76
    The answer is A.

    1. How you found QD = 4/x*sqrt(9+x^2) ?
      Should be: QD = 4*sqrt(9+x^2)/x... Surprisingly: the answer is the same but you have to solve the 3rd degree equation...

  2. Yes, the answer is A. I solved went about it at a different approach however. First I showed that the triangle above the 16 area square and the triangle above the 9 area square were similar, and the shared the same length on the side that was on the yellow square.

    From there, we have 4sinx = 3cosx. And from that point, it's pretty easy.

  3. I first came up with D as an answer but looking at it at a different angle of some sorts i finally got A as my answer in the end it was totally easy.