3. To construct a right triangle where one leg is twice as large as the other and the sum of the legs and the height dropped to the hypotenuse is known (i.e., $a+b+h_{c}$ is known).
Solution. Let $ABC$ be a right triangle such that $\overline{BC}=2 \overline{AC}$ and let $h$ be the altitude dropped to the hypotenuse. Then $\overline{AB}^{2}=5 \overline{AC}^{2}$, so for the area $P$ we get

![](https://cdn.mathpix.com/cropped/2024_06_05_a9dc3c06af03b935fe82g-5.jpg?height=269&width=658&top_left_y=1899&top_left_x=839)

$$
P=\overline{AC}^{2} \text { and } \quad P=\frac{\sqrt{5} \cdot \overline{AC} \cdot h}{2} .
$$

From this we get

$$
h=\frac{\overline{AC}}{\sqrt{5}} \text { and } \quad \frac{\overline{AC}+\overline{BC}+h}{\overline{AC}}=3+\frac{2}{\sqrt{5}} .
$$

From the discussion, the following construction arises.

We construct a right triangle $A_{1} B_{1} C$ such that $\overline{B_{1} C}=2 \overline{A_{1} C}$ (see diagram). On the ray, we mark points $D$ and $D_{1}$ such that

$$
\overline{CD_{1}}=\overline{A_{1} C}+\overline{B_{1} C}+h \text { and } \overline{CD}=a+b+h
$$

(the given sum). From the ratios

$$
\overline{CD_{1}}: \overline{CA_{1}}=3+\frac{2}{\sqrt{5}}=\overline{CD}: \overline{CA} \quad \text { and } \quad \overline{BC}=2 \overline{AC}
$$

we find the points $A$ and $B$, thus constructing the triangle.

Note that the problem has a unique solution.