$A B$ draw a right-angled triangle $A B C$ over the leg $A B$ such that the sum of the hypotenuse $B C$ and the leg $C A$ is equal to twice the length of $A B$.

 

Let's denote the length of $A B$ as $x$. Therefore, we have:

\[ BC + CA = 2x \]

Given that $A B C$ is a right-angled triangle with the right angle at $A$, we can use the Pythagorean theorem:

\[ BC^2 = AB^2 + CA^2 \]

Substituting $AB = x$ and $CA = y$, we get:

\[ BC^2 = x^2 + y^2 \]

We also know:

\[ BC + y = 2x \]

Let's solve these equations step by step. First, express $BC$ in terms of $x$ and $y$:

\[ BC = 2x - y \]

Substitute this into the Pythagorean theorem:

\[ (2x - y)^2 = x^2 + y^2 \]

Expanding and simplifying:

\[ 4x^2 - 4xy + y^2 = x^2 + y^2 \]

Subtract $y^2$ from both sides:

\[ 4x^2 - 4xy = x^2 \]

Subtract $x^2$ from both sides:

\[ 3x^2 = 4xy \]

Divide both sides by $x$ (assuming $x \neq 0$):

\[ 3x = 4y \]

Solving for $y$:

\[ y = \frac{3x}{4} \]

Now substitute $y = \frac{3x}{4}$ back into the equation for $BC$:

\[ BC = 2x - \frac{3x}{4} = \frac{8x}{4} - \frac{3x}{4} = \frac{5x}{4} \]

So, the lengths of the sides of the right-angled triangle $A B C$ are:

\[ AB = x \]
\[ CA = \frac{3x}{4} \]
\[ BC = \frac{5x}{4} \]

These satisfy the condition that the sum of the hypotenuse and one leg is twice the length of the other leg.
Consider the problem as solved and extend $AC$ to $D$ such that $CD = CB$. Then clearly $AD = 2AB$ and $\angle CDB = \angle CBD$. Therefore, on one leg of a right angle, we measure $AB$, and on the other leg, we measure twice this length, thereby obtaining the triangle $ADB$. The perpendicular erected at the midpoint $E$ of $DB$ intersects $AD$ at $C$, because $CBD$ is an isosceles triangle.

(András Földiák, Budapest.)

The problem can also be solved this way: Let $AB = a$, then $CB + AC = 2a$ and $\overline{CB}^2 = \overline{AC}^2 + \overline{AB}^2$; solving these equations, we get

$$
AC = \frac{3}{4}a, \quad CB = \frac{5}{4}a
$$

Therefore, on one leg of a right angle, we measure the given $a$, and on the other leg, we measure $\frac{3}{4}a$, and connect the obtained points.

(Bertalan Ragány, Eger.)

Number of solutions: 16.