Monday, April 11, 2011

Fox 338

This is a little related to the recent ones, but much simpler. Here we'd appreciate a pure geometric solution if one found.


9 comments:

  1. Maybe I don't understand. But if tou take a 12x12 square and divide it in a 9x12 rectangle and three 3x4 small rectangles, then the square is divided into 4 similar rectangles.
    Generally, if L is the square side and x,y are the small rectangles width and height, then y=L/3 and x=L/4

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  2. you are right Cesar. what a shame! what was I thinking? let's correct this one way or another.
    -8foxes

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  3. additional information (a<b) is added. Is this enough to make the claim hold?
    Sorry for the earlier error.

    These are what make us human, right? Come on we are no machines, but flesh and blood.

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  4. You mean:
    if L is the square side then:
    a=L/3

    if rectangles must be similars:
    a/b=(L-b)/L.

    These leads to:
    3*b²-3*L*b+L^2=0

    whose discriminant is:
    9*L²-12*L^2=-3*L²<0

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  5. http://imageshack.us/photo/my-images/834/fox338.png/

    jyu

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  6. correction: ACB, not ACD

    jyu

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  7. Hello Yu,
    How can you say that "C is on the outside of the semicircle"? I just couldn't see that.
    Thanks.

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  8. C is 2 units below the centre of the semicircle whilst the radius of the semicircle is less than 2 units. The semicircle does not touch the base of the square and C is a point on the base.

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