In a right-angled triangle, the two legs are $a, b$; the perpendicular projections of these on the hypotenuse $c$ are $p$ and $q$. Construct the right-angled triangle if $a$ and $q$ are given.
I. Solution: According to a known theorem,

$$
a^{2}=c \cdot(c-q)
$$

from which

$$
c^{2}-q c-a^{2}=0
$$

Thus,

$$
c=\frac{q}{2}+\sqrt{\left(\frac{q}{2}\right)^{2}+a^{2}}
$$

From the perspective of solving our problem, the square root only makes sense with a positive sign (otherwise $c<0$ would be the case).

According to (2), the hypotenuse of the triangle we are looking for is obtained by adding the segment $\frac{q}{2}$ to the hypotenuse of the right triangle constructed with $a$ and $\frac{q}{2}$ as legs (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_7e95e14aa4b9415becd0g-1.jpg?height=360&width=548&top_left_y=850&top_left_x=775)

Figure 1

Our problem always has one solution.

Ligeti Zsolt (Budapest, XIV., Chemical Industry Secondary School II. class)

II. Solution: The construction can be carried out using the tangent and secant from a point outside a circle. Indeed, the relationship (1) used in the first solution can also be written as

$$
a^{2}=p(p+q)
$$

Accordingly, $p$ can be constructed as follows: draw a circle with radius $\frac{q}{2}$ centered at $O$ at the endpoint $Q$ of the segment $P Q=a$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_7e95e14aa4b9415becd0g-1.jpg?height=488&width=631&top_left_y=1773&top_left_x=736)

The ray $P O$ intersects the circle at points $U$ and $V$ respectively.

$$
P U=p
$$

and

$$
P V=p+q=c
$$

Feledy Mária (Esztergom, Dobó Katalin Secondary School I. class)