Ken's POTW


A Square and a Straightedge
  1. Divide a square into 5 regions of equal area using only a straightedge.

    To build up to the above problem, here are some [slightly] shorter ones that can be done first (again, using only a straightedge.) You may or may not need to do each of these in the process of solving the above problem.

  2. Bisect a side of the square.
  3. Trisect a side of the square.
  4. Divide the square into four smaller squares.
  5. Divide a side of the square into 8 equal segments.

    Source: 1. rec.puzzles, 2-5. Original. Used as Ken's Puzzle of the Day 4/13/94.


    Solutions were received from many sources. Denis Borris pointed out that this puzzle is in the rec.puzzles archive under square.five problem. That solution is similar to the one I have:
      1. Label square ABCD clockwise.
      2. Extend AD and BC to lines ADE and BCF.
      3. Draw AC and BD and label center O.
      4. Choose any point G on CD. Reflect G to H on AD: Draw AG, intersect BD at J, and extend CJ to H.
      5. Reflect G to I on BC.
      6. Choose another point K on CD. Reflect K to L on AD and M on BC.
      7. Extend HG to N on BF. Extend IG to P on AE.
      8. Extend LK to Q on BF. Extend MK to R on AE.
      9. Draw NP and QR, intersecting at X.
    1. OX bisects CD at S.
    2. AX and BX trisect CD.
    3. Reflecting S to the other sides creates the four squares.
    4. Drawing diagonal of the smaller squares leads to dividing one side into eighths.
    Draw a trapezoid with one base 5/8 of the square side and the other base one side of the square, find the intersection T of the trapezoid's "diagonals", and reflect the five 1/8 sections through T to 1/5 sections on the opposite side. Reflect the 1/5 sections through O and draw five rectangles (solving part 1).
    Thane Larson provided this different construction. (I transcribe his drawing here):
    1. Draw AC and BD to find center of the square O.
    2. Extend AC beyond C and pick two points (E,F) on AC outside the square.
    3. Draw BF and DE to cross at G. Draw BE and DF to cross at H.
    4. GH is parallel to BD (similar triangles). Extend GH to cross (extended) BC at I and AD at J.
    5. BIJD is a parallelogram. BJ and ID cross at center K.
    6. OK bisects CD.
    Thane and several others then disected the square into five equal areas in this way. Bisect all four sides, and connect the midpoints to non-adjacent corners, clockwise. This results in a central square and four right triangles, all with the same area.
    Other solutions were also received from Nick Baxter and Phillipe Fondanaiche.
    Denis Borris also provided this slightly humorous solution:
    Sticking to the wording of the problem (what is not explicitly
    prohibited is allowed: per James F Fixx in his book "Solve It"),
    my square is a one piece square.
    
    Using it to trace squares of same size, I patiently make a 9 by 9 grid
    on a large sheet of paper (at least 81 times the size of my square!).
    
    I then use the middle square to be subdivided in 5 equal areas.
    
    This is easily done by joining a corner of my square to the top
    corner of the squares in first and last columns of the grid; 
    let me illustrate using part of the 9 by 9 grid, my square being ABCD:
    L-----------------U
    | | | | | | | | | |
    M-----------------T
    | | | | | | | | | |
    N-----------------S
    | | | | | | | | | |
    O-------D-C-------R
    | | | | | | | | | |
    P-------A-B-------Q
    Join AR AS AT AU, then BO BN BM BL;
    this will "mark" CB and DA in beautiful "fifths".
    So the job is completed.
    
    Why do I use a 9 by 9 grid?
    Well, the "joining" can be similarly continued,
    marking DC and AB in similar "fifths", hence
    resulting in my square being subdivided in
    twenty-five (25) equal areas!
    

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