\section*{Problem 4 - 021044}

A rectangle is to be constructed that has the same area as a given square with side \(a\) and whose perimeter is twice as large as that of the given square.

How many solutions does this problem have?
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There is exactly one rectangle that satisfies these conditions.

Let \(c\) and \(d\) be the sides of the rectangle. Then \(c \cdot d = a^2\) and \(2(c + d) = 2 \cdot 4a\).

From the first equation, we get \(c = \frac{a^2}{d}\). Substituting this into the second equation, we obtain \(2\left(\frac{a^2}{d} + d\right) = 8a\) and thus \(d^2 - 4ad + a^2 = 0\).

The quadratic equation has the two solutions \(d_1 = (2 + \sqrt{3})a\) and \(d_2 = (2 - \sqrt{3})a\). Accordingly, the solutions are a rectangle with side lengths \((2 + \sqrt{3})a\) and \(\frac{a}{2 + \sqrt{3}}\), and a rectangle with side lengths \((2 - \sqrt{3})a\) and \(\frac{a}{2 - \sqrt{3}}\).

Since \(\frac{a}{2 + \sqrt{3}} = a(2 - \sqrt{3})\) and \(\frac{a}{2 - \sqrt{3}} = a(2 + \sqrt{3})\), both rectangles are identical. Therefore, there is exactly one such rectangle up to congruence.