## Task 6 - 240736

A quadrilateral $A B C D$ has the following properties:

(1) $A B \| D C$ and $A D \nVdash B C$,

(2) $A D=B C=3 \cdot D C=a$, where $a$ is a given length,

(3) $\angle B A D=60^{\circ}$.

Determine the perimeter $u$ of this quadrilateral in terms of $a$!
Due to (1) and (2), \(ABCD\) is an isosceles trapezoid. Therefore, from (3) it follows that

(4) \(\angle CBA = 60^\circ\).

The line parallel to \(BC\) through \(D\) intersects \(AB\) at \(E\). Since alternate interior angles formed by intersecting parallel lines are equal, it follows from this and (4) that

(5) \(\angle DEA = 60^\circ\).

From this and (3) (as well as the angle sum theorem), it follows that triangle \(AED\) is equilateral. Therefore, due to (2),

(6) \(AE = AD = a\).

Since \(BC \parallel ED\) and due to (1) also \(EB \parallel DC\), \(EBCD\) is a parallelogram. From this and (2) it follows that

(7) \(EB = DC = \frac{a}{3}\).

For the perimeter \(u\) of quadrilateral \(ABCD\), it follows that

\[
u = AE + EB + BC + CD + DA = a + \frac{a}{3} + a + \frac{a}{3} + a = \frac{11}{3} a
\]

Solutions from the third round of 1984 taken from [5]

### 4.27 25th Olympiad 1985

### 4.27.1 First Round 1985, Class 7