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Math Help - Sophie Diagrams

  1. #1
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    Sophie Diagrams

    The diagram attached consists of a rectangle ABCD and a triangle DXY so that X and Y are points on the line segments AB and BC respectively and angle DXY = 90 degrees. If all the line segments in the diagram have integer lengths, then we call it a Sophie Diagram.


    a) A particular sophie diagram has AX = 12, XY = 15 and XD = 20. Find the length and width of rectangle ABCD.

    b) Another sophie diagram has DX = 429 and DY = 845. Find the length and width of rectangle ABCD.
    Attached Thumbnails Attached Thumbnails Sophie Diagrams-005.jpg  
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  2. #2
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    Hello, gunsandroses234!

    The diagram attached consists of a rectangle ABCD and a triangle DXY,
    so that X and Y are on AB and BC respectively, and \angle DXY = 90^o.

    a) A particular Sophie diagram has: . AX = 12,\;\;XY = 15,\;\;XD = 20.
    Find the length and width of rectangle ABCD.


    Code:
             12   X       a
        A o - - - o - - - - - - - o B
          |      *90 *    15     |
          |     *         *       | b
          |    * 20           *   |
       16 |   *                   o Y
          |  *              *     |
          | *         * 25        | 16-b
          |*    *                 |
        D o - - - - - - - - - - - o C
                     12+a

    In right triangle DXY\!:\;\;XD = 20,\;XY = 15 \quad\Rightarrow\quad DY = 25

    In right triangle XAD\!:\;\;AX = 12,\;XD = 20\quad\Rightarrow\quad AD = 16

    Let: . XB = a ,\; BY = b \quad\Rightarrow\quad YC \:=\:16-b


    In right triangle XBY\!:\;\;a^2+b^2 \:=\:15^2 \quad\Rightarrow\quad a^2+b^2\:=\:225 .[1]


    In right triangle YCD\!:\;\;(12+a)^2 + (16-b)^2 \:=\:25^2

    . . \text{which simplifies to: }\;24a - 32b + \underbrace{a^2+b^2}_{\text{This is 225}} \:=\:225

    We have: . 24a-32b \:=\:0 \quad\Rightarrow\quad b \:=\: \frac{4}{3}a

    Substitute into [1]: . a^2 + \left(\frac{4}{3}a\right)^2 \:=\:225 \quad\Rightarrow\quad \frac{25}{9}a^2 \:=\:225

    . . . . . . . . . . . . . . a^2 \:=\:81 \quad\Rightarrow\quad a \:=\:9


    Therefore: . \begin{Bmatrix}\text{Length} &=& 21 \\ \text{Width} &=& 16 \end{Bmatrix}

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  3. #3
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, gunsandroses234!


    In right triangle DXY\!:\;\;XD = 20,\;XY = 15 \quad\Rightarrow\quad DY = 25

    In right triangle XAD\!:\;\;AX = 12,\;XD = 20\quad\Rightarrow\quad AD = 16

    Let: . XB = a ,\; BY = b \quad\Rightarrow\quad YC \:=\:16-b


    In right triangle XBY\!:\;\;a^2+b^2 \:=\:15^2 \quad\Rightarrow\quad a^2+b^2\:=\:225 .[1]


    In right triangle YCD\!:\;\;(12+a)^2 + (16-b)^2 \:=\:25^2

    . . \text{which simplifies to: }\;24a - 32b + \underbrace{a^2+b^2}_{\text{This is 225}} \:=\:225

    We have: . 24a-32b \:=\:0 \quad\Rightarrow\quad b \:=\: \frac{4}{3}a

    Substitute into [1]: . a^2 + \left(\frac{4}{3}a\right)^2 \:=\:225 \quad\Rightarrow\quad \frac{25}{9}a^2 \:=\:225

    . . . . . . . . . . . . . . a^2 \:=\:81 \quad\Rightarrow\quad a \:=\:9


    Therefore: . \begin{Bmatrix}\text{Length} &=& 21 \\ \text{Width} &=& 16 \end{Bmatrix}

    I may be wrong, but I don't think that is correct.

    Pythagoras gives you that AD = \sqrt{20^2-12^2}=16.

    We now want to find AB. However, there is only one pythagorean triple with hypotenuse 15, (9, 12, 15). We therefore need to find out if BY is 9 or 12. This is equivalent to showing whether AB=CD=12+9=21 or 12+12=24

    To do this, note that DY = 25 by Pythagoras (the triple (15, 20, 25) ) and note that there are only two triples with hypotenuse 25,

    (7, 24, 25) and (15, 20, 25).

    Clearly, as 21 is not in either of these triples AB=CD=24, and we are done.

    AB=24, AD=16.
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