Sophie Diagrams

**gunsandroses234** · May 8th 2010, 09:47 PM

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.

**Soroban** · May 9th 2010, 05:31 AM

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

**Swlabr** · May 10th 2010, 03:11 AM

Originally Posted by Soroban

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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