\section*{Problem 3B - 111043B}

Dirk explains to Jürgen the usefulness of differential calculus using the solution to the following problem: Let \(A B C D E\) be a planar convex pentagon such that \(A, B, C, E\) are the vertices of a rectangle and \(C, D, E\) are the vertices of an equilateral triangle. The area of the pentagon \(A B C D E\) is now prescribed to be a certain value \(\mathrm{F}\).

Determine whether among all such pentagons, there exists one with the smallest perimeter \(u\)! If this is the case, calculate for all such pentagons with the smallest perimeter the value \(a: b\), where \(A B=a\) and \(B C=b\).

The next day, Jürgen tells Dirk that he has found a solution to this problem without using differential calculus.

Provide a solution that Jürgen might have found.
}

It is \(F=a b+\frac{\sqrt{3}}{4} a^{2}\) and \(u=3 a+2 b\), so \(b=\frac{u-3 a}{2}\) and thus

\[
F=a \cdot \frac{u}{2}-\frac{3}{2} a^{2}+\frac{\sqrt{3}}{4} a^{2} \quad \text { or } \quad u=\frac{1}{a} \cdot\left(2 F+3 a^{2}-\frac{\sqrt{3}}{2} a^{2}\right)=\frac{2 F}{a}+\left(3-\frac{\sqrt{3}}{2}\right) a
\]

For positive real numbers \(x\) and \(y\), the inequality between the arithmetic and geometric means holds: \(\frac{x+y}{2} \geq \sqrt{x y}\) or, after multiplication by 2: \(x+y \geq 2 \sqrt{x y}\), with equality if and only if \(x=y\).

If we set \(x:=\frac{2 F}{a}>0\) and \(y:=\left(3-\frac{\sqrt{3}}{2}\right) a>0\) in this inequality, we obtain

\[
u=x+y \geq 2 \cdot \sqrt{\frac{2 F}{a} \cdot\left(3-\frac{\sqrt{3}}{2}\right) a}=2 \cdot \sqrt{(6-\sqrt{3}) F}
\]

Thus, \(u\) attains its minimum value exactly when

\[
\frac{2 F}{a}=\left(3-\frac{\sqrt{3}}{2}\right) a
\]

or \(4 F=(6-\sqrt{3}) a^{2}\), so \(a=2 \sqrt{\frac{F}{6-\sqrt{3}}}\) and \(\frac{\sqrt{3}}{4} a^{2}=\frac{\sqrt{3}}{6-\sqrt{3}} \cdot F\), which gives us

\[
a b=F-\frac{\sqrt{3}}{4} a^{2}=F \cdot\left(1-\frac{\sqrt{3}}{6-\sqrt{3}}\right)=F \cdot \frac{6-2 \sqrt{3}}{6-\sqrt{3}}
\]

and finally

\[
a: b=\frac{a}{b}=\frac{a^{2}}{a b}=\frac{4 \frac{F}{6-\sqrt{3}}}{F \cdot \frac{6-2 \sqrt{3}}{6-\sqrt{3}}}=\frac{4}{6-2 \sqrt{3}}=\frac{2}{3-\sqrt{3}}=\frac{2(3+\sqrt{3})}{9-3}=\frac{3+\sqrt{3}}{3}=1+\frac{\sqrt{3}}{3}
\]