Let's construct a rectangle, if we know one of its sides and we know that the other side is the geometric mean between the given side and the perimeter.
I. solution: Let the shorter side of the rectangle be $a$, and the longer side be $b$. According to the problem,

$$
b=\sqrt{a(2 a+2 b)}
$$

which means

$$
b^{2}=2 a^{2}+2 a b, \quad b^{2}-2 a b-2 a^{2}=0
$$

from which (ignoring the negative root)

$$
b=\frac{2 a+\sqrt{4 a^{2}+8 a^{2}}}{2}=\frac{2 a+2 a \sqrt{3}}{2}=a+a \sqrt{3}
$$

The distance $a \sqrt{3}$ can be constructed as the geometric mean between $a$ and $3 a$, or based on $a \sqrt{3}=\sqrt{4 a^{2}-a^{2}}$, as one leg of a right triangle whose other leg is $a$ and hypotenuse is $2 a$.

$$
\text { Lajos Almási (Bp. II., Rákóczi g. II. o. t.) }
$$

II. solution: Suppose we have successfully constructed a right triangle whose altitude to the hypotenuse is $b$, and the projections of the legs on the hypotenuse are $a$ and $2 a+2 b$. (See the 1st figure, which also shows the labeling.)

Then $b$ is the geometric mean between the sides $a$ and $2 a+2 b$.

Raise perpendiculars to the hypotenuse at points $E$ and $F$. The resulting

$$
\triangle B F H \cong \triangle C D A
$$

because both have one side $b$, one right angle, and the acute angles adjacent to the $b$ leg are equal (as angles with perpendicular sides). Thus,

$$
F H=D A=a
$$

and therefore

$$
E G=2 a
$$

since $\triangle B E G \sim \triangle B F H$ and $B E=2 B F$.

![](https://cdn.mathpix.com/cropped/2024_05_02_7ef85ed3e4913f1c556fg-1.jpg?height=274&width=567&top_left_y=1392&top_left_x=497)

1st figure

![](https://cdn.mathpix.com/cropped/2024_05_02_7ef85ed3e4913f1c556fg-1.jpg?height=320&width=380&top_left_y=1363&top_left_x=1206)

2nd figure

After this, the construction can be completed (2nd figure). First, construct the right triangle $\triangle A G E$ (with legs $A E=3 a$ and $E G=2 a$), then the point $C$. $C$ lies on the perpendicular to $A E$ at $D$, and also on the Thales circle drawn outward from the hypotenuse $A G$, since $\angle A C G$ is a right angle.

Gergely Krammer (Bp. II., Rákóczi g. II. o. t.)