Saturday, February 20, 2010

Fox 250
Bleaug submitted this one, saying that he got inspired by Yu's elegant solution to Fox 246. More is on the way...

1 comment:

  1. Say base of trapezoid B is x.
    Say trapezoid D is a multiple of trapezoid B by factor k.
    Then base trapezoid D = k*x and A(D) = k^2*A(B).
    The base of trapezoid C is then x + kx = x*(k+1).
    So trapezoid C is a multiple of B by factor k+1 and A(C)=(k+1)^2*A(B).
    The shaded area = A(C)-A(B)-A(D) = ((k+1)^2-1-k^2)*A(B) = 2k*A(B).
    Area E is the result of Area(B) first horizontally stretched by factor k and then vertically multiplied by factor 2, so A(E) = A(B)*k*2 = 2k*A(B).
    (Sorry for my poor English... I'm Dutch...)