# Sophie Diagrams

• May 8th 2010, 09:47 PM
gunsandroses234
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.
• May 9th 2010, 05:31 AM
Soroban
Hello, gunsandroses234!

Quote:

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

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

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 $\displaystyle DXY\!:\;\;XD = 20,\;XY = 15 \quad\Rightarrow\quad DY = 25$

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

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

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

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

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

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

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

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

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

• May 10th 2010, 03:11 AM
Swlabr
Quote:

Originally Posted by Soroban
Hello, gunsandroses234!

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

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

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

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

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

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

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

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

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

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

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

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

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.