Wednesday, November 28, 2012

Fox 354


Roland Sampy has submitted this problem.
It looks right, but not confirmed yet.

3 comments:

  1. Two equilateral triangles OPQ and ORS of respective sides a and b.

    Case 1: a=b
    then OP=OQ=OR=OS=a=b=r, i.e. (P,Q,R,S) lie on the same circle, irrespective of triangles' orientation.
    Hence a^2+a.b+b^2=r^2+r.r+r^2=3.r^2

    Case 2: a<>b
    Assume PQ and RS are not parallel. (P,Q,R,S) on the same circle implies that circle's center I is the intersection of PQ and RS bisectors, which is O by construction. Since OP=a<>b=OR we have a contradiction. Therefore PQ//RS.

    PQ and RS share a common bisector as depicted on the following figure:
    http://bleaug.free.fr/8foxes/8foxes354.png

    Since P,J,S belong to circle centered in I we have QPR=QJR=QSR=QIR/2=60°. Hence QR=r.sqrt(3)

    PQ//RS implies QOR=120°, therefore in triangle OQR:
    QR^2=OQ^2+OR^2-2.OQ.OR.cos(QOR)
    which can be rewritten as:
    3.r^2=a^2+b^2+a.b

    bleaug

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  2. to be complete, we should also consider the (sub-)case where PQ and RS lie on the same side w.r.t. O as depicted in:
    http://bleaug.free.fr/8foxes/8foxes354b.png

    If a and b represent positive length values, formula becomes:
    3.r^2=a^2+b^2-a.b

    If we adopt the convention that (a.b) = scalar product (OQ.OS) including sign, the above formula holds in all cases.

    bleaug

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  3. Two equilateral triangles OPQ and ORS of respective sides a and b.
    then arc(PQ) + arc(RS) = 2 . 60 degree =120 degree
    but arc(PR) = arc(QS) .
    Then arc(PR) = arc(QS) = 120 degree.
    and PR = QS = r.sqrt(3) .
    In the triangle OPR angle(POR) = 120 degree.
    Then OP^2 + OR^2 + OP.OR = PR^2
    and a^2 + b^2 + a.b = r.sqrt(3) όπερ έδει δείξαι.
    Philip Pydnaios , from Hellas

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