# Sophie Diagrams

#### gunsandroses234

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.

#### Attachments

• 23.3 KB Views: 145

#### Soroban

MHF Hall of Honor
Hello, gunsandroses234!

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

#### Swlabr

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.