## Task 6 - 150836

For a quadrilateral $A B C D$, it is required that the sum of the lengths of the two diagonals $A C$ and $B D$ is $11 \mathrm{~cm}$, that the side $A B$ has the length $a=6 \mathrm{~cm}$, and the side $A D$ has the length $d=1 \mathrm{~cm}$.

Determine a length $x$ and a length $y$ such that for the perimeter $u$ of any quadrilateral that meets the given requirements, the inequality $x \leq u \leq y$ holds, where the equality sign holds exactly when the quadrilateral $A B C D$ degenerates into a line segment, i.e., when the points $A, B, C, D$ lie on the same straight line!

Hint: $A B C D$ can also be non-convex. Furthermore, in the degenerate case, points can also coincide.
Let $BC = b, CD = c, AC = e, BD = f$. By the triangle inequality, applied to the triangles $ABD, ABC$, and $ACD$, we have (all length measurements in centimeters)

\[
\begin{gathered}
5 < f < 7 \\
6 - e < b < 6 + e \\
e - 1 < c < e + 1
\end{gathered}
\]

From (1) and $e + f = 11$, it follows that $4 < e < 6$ (4). From (2), (3), and (4), it follows that $5 < b + c < 2e + 7 < 19$. From this and $u = b + c + 7$, it follows that $12 < u < 26$. As the figure shows, these lengths $x = 12$ (cm) and $y = 26$ (cm) actually occur in the degenerate case, and are therefore the desired lengths. Hence, $12 \leq u \leq 26$.
![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0681.jpg?height=92&width=1080&top_left_y=2466&top_left_x=492)

Solutions from the third round of 1975 taken from [5]

### 5.18 16th Olympiad 1976

### 5.18.1 First Round 1976, Class 8