\section*{Task 1 - 191021}

A rectangular picture, whose side lengths are in the ratio \(2: 3\), is to receive a frame of uniform width, whose area is as large as that of the picture.

Determine all values of the ratio of the lengths of the outer edges of the frame that satisfy this requirement!
}

Let \(2a\) be the length of the shorter side of the picture, then \(3a\) is the length of its longer side. Let \(d\) be the width of the frame, then \(2a + 2d\) and \(3a + 2d\) are the lengths of the outer edges of the frame. The requirement stated in the problem is satisfied if and only if the area enclosed by these edges is twice the area of the picture, i.e., if and only if

\[
(2a + 2d)(3a + 2d) = 2 \cdot 2a \cdot 3a
\]

This is equivalent to

\[
\begin{aligned}
6a^2 + 4ad + 6ad + 4d^2 & = 12a^2 \\
4d^2 + 10ad - 6a^2 & = 0 \\
d^2 + \frac{5}{2}ad - \frac{3}{2}a^2 & = 0
\end{aligned}
\]

This quadratic equation has exactly the solutions

\[
d_{1,2} = -\frac{5}{4}a \pm \sqrt{\frac{25}{16}a + \frac{24}{16}a^2}
\]

of which exactly \(d = \frac{a}{2}\) is non-negative. Therefore, the requirement is satisfied if and only if the outer edges have the lengths \(2a + 2\frac{a}{2} = 3a\) and \(3a + 2\frac{a}{2} = 4a\).

If this is the case, they have the length ratio 3:4, and conversely:

If the outer edges have the length ratio \((2a + 2d):(3a + 2d) = 3:4\), then \(8a + 8d = 9a + 6d\), so \(2d = a\) and thus \(d = \frac{a}{2}\). Therefore, exactly the length ratio 3:4 satisfies the conditions of the problem.