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Math Help - Framed!

  1. #1
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    Jun 2006
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    NSW Australia
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    Framed!

    Kevin had five rectangular photographs. The ten side lengths of the photographs were 1 cm, 2 cm, 3 cm, …, 10 cm, (with each occurring once).

    Kevin assembled the five photographs with no gaps or overlaps into a square frame of side length 11 cm. Three of the photographs had size 1 cm x 6 cm, 2 cm x 10 cm and 3 cm x 9cm.
    a. Find the side lengths of the other two photographs and show how the photographs can be put in the frame.

    b. Katie also had five rectangular photographs with side lengths 1 cm, 2 cm, 3 cm, … , 10 cm (with each occurring once ). She assembled hers into a rectangular frame of size 15 cm x 10 cm.

    Find the side lengths of each photograph and show how the photographs can be put in the frame.

    Kevin and Katie noticed that, whenever they put the photographs into a rectangular frame, they were always arranged with one photograph not touching any side of the frame.

    A frame of this kind is shown.

    The letters a, b, c, d, f, g, h, I, j and k represent the lengths of the photographs in centimetres and these are the numbers 1, 2, 3, …., 10 in some order.

    c. Show that one of j and k is even and the other odd.
    d. Show that at least one of the lengths or the width of the frame must be odd.
    Attached Thumbnails Attached Thumbnails Framed!-framed.gif  
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  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
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    Hello, lalji!

    Here is the first part . . .

    Kevin had five rectangular photographs. The ten side lengths of the photographs were:
    1 cm, 2 cm, 3 cm, …, 10 cm, (with each occurring once).

    Kevin assembled the five photographs with no gaps or overlaps into a square frame of side length 11 cm.
    Three of the photographs had size 1 cm x 6 cm, 2 cm x 10 cm and 3 cm x 9cm.
    a) Find the side lengths of the other two photographs
    and show how the photographs can be put in the frame.
    The square frame has an area of: 11^2 = 121 cm².

    The three given photos have a total area of: (1\times6) + (2\times10) + (3\times9) \:=\:53 cm².

    This leaves: 121 - 53 \:=\:68 cm² to be covered by the other two photos.

    The two photos have dimensions \{4,5,7,8\}

    (a) The only combination which totals 68 cm² is: . (4\times7) + (5\times8)

    Code:
                          9                       2
          * - - - - - - - - - - - - - - - - - * - - - *
          |                                   |       |
          |   .   .   .   .   .   .   .   .   |   .   |
        3 |                                   |       |
          |   .   .   .   .   .   .   .   .   |   .   |
          |                                   |       |
          * - - - - - - - - - * - - - - - - - *   .   |
          |                   |               |       |
          |   .   .   .   .   |   .   .   .   |   .   |
          |                   |               |       |
          |   .   .   .   .   |   .   .   .   |   .   | 10
          |                   |               |       |
          |   .   .   .   .   |   .   .   .   |   .   |
          |                   | 7             |       |
        8 |   .   .   .   .   |   .   .   .   |   .   |
          |                   |               |       |
          |   .   .   .   .   |   .   .   .   |   .   |
          |                   |               |       |
          |   .   .   .   .   |   .   .   .   |   .   |
          |                   |       4       |       |
          |   .   .   .   .   * - - - - - - - * - - - *
          |                   |                       | 1
          * - - - - - - - - - * - - - - - - - - - - - *
                    5                     6
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  3. #3
    Member
    Joined
    Nov 2005
    From
    Wethersfield, CT
    Posts
    92
    Hi:

    I haven't much time at present, but a couple of things stand out which may be helpful.

    1.First, note that the perimeter of shown 15X10 rectangle is (1/2)sum of
    perimeters of four peripheral boxes, or, sum of half perimeters for peripheral boxes

    2. Perimeter of 15X10 is 50 cm.

    3. Since dimensions of all pics are taken on {1,2,3,...,10} without replacement, the sum: l1+w1+l2+w2+...+w5 = (1+2+3+...+10)/2 = 10(11)/2 = 55. Thus, j+k = 55 - outside perimeter of 2(15+10)= 5. Among other things, this settles your one odd, one even quandry regarding j and k.

    4. Think system of equation perhaps.

    5. Think hypotenuse via Pyth. Thm.

    6. 5sqrt(13) = diagonal length of rectangular frame.

    7. sum areas, etc.

    sorry no more time. moving furnature today -- not nearly as much fun as math...or a heart attack for that matter

    Good Luck.

    Rich B.
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  4. #4
    Newbie
    Joined
    Jun 2006
    From
    NSW Australia
    Posts
    12

    Smile Thanks for the reply

    Thanks Soroban and Rich.

    Can any one please help me out with c and d.

    Thanks
    Lalji
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