Thursday, August 20, 2009

Fox 130

http://www.8foxes.com/

Not terribly hard...

2 comments:

  1. It's not a short proof... I will not give every detail, so some details you must fill in yourself.
    Or see for a full proof:
    http://home.kpn.nl/henkreuling/solutions8foxes/130%20-%208foxes%20-%20solution.pdf

    It's not hard to see that every time you cut off a square from a triangle similar to triangle ABC then the remaining triangle on the right side is similar with triangle ABC with factor 1/3; and the remaining triangle on the left side is similar with triangle ABC with factor 2/3;
    The first square has sides 2/3, and area 4/9.
    On the right side, because of the factor 1/3, the sides of the squares give a geometric sequence with side xn = 2/3*(1/3)^n n=0,1,2,...
    The area give a geometric sequence (xn)^2 = 4/9*(1/9)^n.
    The total area is sum(n from 0 to eternity) 4/9*(1/9)^n = 4/9*(1/(1-1/9))=4/9*9/8 = 1/2.
    Now the left side: the sides of the squares give a geometric sequence with side yn=2/3*(2/3)^n n=0,1,2,...
    The area give a geometric sequence (yn)^2 = 4/9*(4/9)^n, with sum 4/9*(1/(1-4/9))=4/9*9/5 = 4/5.
    But the first square with side 2/3 (and area 4/9) is counted on both sides.
    The total area of all squares is also: 1/2 + 4/5 - 4/9 = 77/90.
    Because the area of triangle ABC = 1, is the ratio of the area of ABC tot the sum of all squares equal to 90/77.
    The correct answer is E.
    (I'm Dutch, so sorry for my poor English...)

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