\section*{Problem 3 - 251013}

The cross-section of a channel has the shape of an isosceles trapezoid. Its parallel sides are horizontal, the longer one on top, the shorter one at the bottom (bottom of the ditch). The sloping side walls are inclined at an angle of \(30^{\circ}\) to the vertical.

Due to the required flow rate, the cross-sectional area \(F\) must have a specified value. Additionally, the length \(a\) of the shorter of the parallel sides of the cross-section (bottom of the ditch) is given.

Determine the depth \(t\) of the channel as a function of \(a\) and \(F\)!

This problem was initially solved using trigonometric methods. The next day, Jörg presented a solution without using trigonometry.

Provide such a solution.
}

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1498.jpg?height=294&width=534&top_left_y=1780&top_left_x=767)

The mentioned trapezoid is \(A B C\) with \(A B \| D C\) and \(A B < D C\).

The feet of the perpendiculars from \(A\) and \(B\) to \(D C\) are \(L\) and \(N\), respectively. According to the problem statement,

\[
A D = B C \quad ; \quad \angle D A L = \angle C B N = 30^{\circ} \quad ; \quad A B = L N = a \quad ; \quad A L = B N = t
\]

and the trapezoid \(A B C D\) has the area \(F\).

If the triangle \(B C N\) is moved along \(B A\) such that the image of \(B N\) coincides with \(A L\), then an isosceles triangle is formed, whose angle between the legs is \(60^{\circ}\).

Thus, the base angles are also each \(60^{\circ}\). The considered triangle is therefore equilateral. If we denote its side length by \(x\), then (by the Pythagorean theorem or the formula for the height in an equilateral triangle)

\[
t = \frac{x}{2} \sqrt{3}
\]

From this, it follows that

\[
\begin{gathered}
D L + N C = x = \frac{2 t}{\sqrt{3}} \quad \text { so } \\
D C = D L + L N + N C = a + \frac{2 t}{\sqrt{3}}
\end{gathered}
\]

This results in

\[
\begin{array}{r}
F = \frac{1}{2}(A B + D C) t = \frac{1}{2}\left(2 a + \frac{2 t}{\sqrt{3}}\right) t = a t + \frac{t^{2}}{\sqrt{3}} \\
t^{2} + a t \sqrt{3} - F \sqrt{3} = 0
\end{array}
\]

Of the two solutions to this quadratic equation, the negative one is discarded, and it follows that

\[
t = \frac{a}{2} \sqrt{3} + \sqrt{\frac{3}{4} a^{2} + F \sqrt{3}}
\]