Monday, March 12, 2012

Fox 349

This is just one step forward to the general case. Should not be very hard unless the claim is wrong :)
Yes, a geometric solution will be appreciated...

5 comments:

  1. hint: two additional straight lines on #348 solution picture help.

    bleaug

    ReplyDelete
  2. See my full worked out solution:
    http://home.kpn.nl/henkreuling/solutions8foxes/349_8foxes_solution.pdf

    ReplyDelete
  3. similar to henkie's solution:
    http://bleaug.free.fr/8foxes/8foxes349.png

    by construction:
    a(ADE)=a(BCE)=a(CFE)=a(DGE)

    let O = midpoint of DC. Symmetry centered in O sends DGH in CG'H'. Hence a(DGE)=a(DHE)+a(DGH)=a(DHE)+a(CG'H')

    Therefore:
    a(CDE)=a(CFE)+a(DHE)+a(CG'H')+a(FHH'G')=2*a(ADE)+a(FHH'G') > 2*a(ADE) > 1.96*a(ADE). Claim is wrong.

    bleaug

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  4. http://imageshack.us/photo/my-images/850/fox349.jpg/
    jyu

    ReplyDelete
  5. Good solutions...
    We have another one with totally analytic expressions. Some day, we may compile all of these into a single post. May be later...
    -8foxes

    ReplyDelete