\section*{Task 1 - 171021}

For four circles \(k_{1}, k_{2}, k_{3}, k_{4}\), it is required that they have the following two properties (1), (2):

(1) The diameter of \(k_{4}\) is \(1 \mathrm{~cm}\) larger than the diameter of \(k_{3}\), whose diameter is \(1 \mathrm{~cm}\) larger than that of \(k_{2}\), and whose diameter is \(1 \mathrm{~cm}\) larger than that of \(k_{1}\).

(2) The area of \(k_{4}\) is as large as the sum of the areas of the other three circles. Investigate for which length of the diameter of \(k_{1}\) these two requirements (1), (2) are satisfied!
}

If (1) and (2) are satisfied for a diameter length \(d \mathrm{~cm}\) of \(k_{1}\), then \(k_{2}, k_{3}\) and \(k_{4}\) have diameters of length \((d+1) \mathrm{~cm}, (d+2) \mathrm{~cm}\) and \((d+3) \mathrm{~cm}\) respectively, and it holds that

\[
\frac{\pi}{4} d^{2}+\frac{\pi}{4}(d+1)^{2}+\frac{\pi}{4}(d+2)^{2}=\frac{\pi}{4}(d+3)^{2}
\]

From this, we get \(d^{2}+d^{2}+2 d+1+d^{2}+4 d+4=d^{2}+6 d+9\), thus \(2 d^{2}=4\) and since \(d>0\), we obtain \(d=\sqrt{2}\).

Therefore, only the length \(\sqrt{2} \mathrm{~cm}\) can satisfy the requirements (1) and (2).

A check of the sum of the areas of the circles confirms the result.