Tuesday, May 25, 2010

Fox 290


6 comments:

  1. ACTUALLY IT WAS PRETTY EASY!!!
    PICK AN ARBITARY POINT ON UPPER LINE.NAME IT 'A'. THEN SEGREGATE A 30 DEGREE ANGLE THAT CUTS THE LOWER LINE AT POINT 'B'. LINE BISECTION of 'AB' MEETS THE THIRD LINE WHICH IS THE BETWEEN 2 ELSE ONES. NAME THE LAST POINT 'O'. CONSEQUENTLY IF ONE YOU DRAW A CIRCLE WITH CENTER 'O' AND ALSO LENGTH 'OA' AS RADII; YOU CAN GET YOUR POPOSE, AN EQUILATERAL TRIANGLE.THE ONLY PROBLEM YOU MAYBE STUCK WITH IS HOW TO GET A 30 DEGREE ANGLE? THAT'S ALSO SUCH AN EASE. YOU GOTTA CONSTRUCT AN EQUILATERAL TRIANGLE. NEXT YOU SHOULD BISECT A N ANGLE AS U WISH TO CHOOSE. AFTERWARDS U JUST NEED TO BUILD 2 TRIANGLE CONTAINING THAT 30 ANGLE TOO ON BOTH SIDES, ARBITARY EQUILATERAL TRIANGLE & POINT 'A' WHICH CONCLUDE UPPER LINE AS AN SIDE OF TRIANLE AND SO ON. PROVING THE CONSTRUCTURE ABOVE IIS ON READER. ITS SO STRAIGHT LIGHT.


    GOOD LU(o)CKS, UNLOCK IT PLEASE!

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  2. Another solution:
    Take a perpendicular which crosses top and middle lines in A and B resp. Build equilateral triangle ABC. Orthogonal projection of C on bottom line is P. Take O as middle of AB. Perpendicular to OP in O crosses top and middle lines in Q and R. PQR is equilateral.

    bleaug

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  3. Let's name the paralleles L1, L2, L3 (L2: middle) and A and B the distances among L1,L2 and L2,L3. The side S of the searched triangle is
    S=sqrt(A²+B²+AB)/(sqrt(3)/2)

    If we build a triangle PQR with PQ=A, QR=B and <PQR=120º, then PR=q=sqrt(A²+B²+AB). Build now an equilateral triangle PRZ. The altitudes (heights) of PRZ are
    q/(sqrt(3)/2)=sqrt(A²+B²+AB)/(sqrt(3)/2)=S.

    César Lozada, Venezuela

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  4. Sorry, my mistake in last post.
    1) Draw a perpendicular crossing paralleles in A,B,C
    2) Draw equlateral triangle BCD
    3) Draw triangle ADE, with <ADE=30º and <DAE=90º

    DE is the side of the target triangle.

    César Lozada, Venezuela

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  5. César,
    Your answer is correct. Long ago, I was planning to ask the general case, you've noted. See the last one on page:
    http://www.8foxes.com/Home/brain-teaser

    That may be one of the first foxes ever constructed (That's why it is not numbered). But upon your solution, I will re-design and post it in the following days. Symmetry in the formula deserves it!

    Gracias!

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  6. Another solution. Take a point Pon the lowest line and erect PX at 60 to it meeting the second line at x. draw angle XPY=60 meeting the third line at Y. draw angle XYZ=60 meeting the first line at Z. Then XYZ is the required triangle. proof: YZPX is cyclic as external angle=int opp angle, so YZX=YPX=60
    Murali

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