7. A rectangle $A B C D$ with a diagonal of length $20 \text{ cm}$ is circumscribed by a circle. The side $\overline{C D}$ of the rectangle $A B C D$ is the base of an isosceles triangle whose third vertex $E$ is on the shorter arc determined by the chord $\overline{C D}$ of the circle circumscribed around the rectangle. What is the length of the side $\overline{A D}$ of the rectangle if the area of the rectangle $A B C D$ is equal to the area of triangle DCE?

SCHOOL/CITY COMPETITION IN MATHEMATICS21st January 2016.8th grade-elementary school

Problems worth 6 points:
First method:

![](https://cdn.mathpix.com/cropped/2024_05_30_494875a1129ddb904645g-18.jpg?height=597&width=620&top_left_y=1609&top_left_x=318)

Let $|AB| = |CD| = a$, $|BC| = |AD| = b$, and let $v_a$ be the length of the height to the base of triangle $DCE$.

The radius of the circle circumscribed around the rectangle is $r = 20 / 2 = 10 \text{ cm}$.

1 POINT

Triangle $DCE$ is isosceles with base $\overline{DC}$, so $|SE| = r = v_a + \frac{b}{2}$.

1 POINT

Thus, $\frac{b}{2} + v_a = 10$ or $v_a = 10 - \frac{b}{2}$.

1 POINT

From the equality of the areas of rectangle $ABCD$ and triangle $DCE$, we get

$ab = \frac{1}{2} a v_a \Rightarrow b = \frac{1}{2} v_a \Rightarrow v_a = 2b$

2 POINTS

The equation $2b = 10 - \frac{b}{2} \cdot 2$ holds.

$4b = 20 - b$

$5b = 20$

$b = 4$

Therefore, $|AD| = 4 \text{ cm}$.

Note: A guessed (measured) solution without the procedure is worth 2 points.