In a right-angled triangle with legs $a, b$ and hypotenuse $c$, the following relationship holds. Construct the triangle if $c$ is given.

$$
\frac{1}{a}+\frac{1}{b}=\frac{3}{c}
$$
I. solution. We will cut out the vertex of the right angle from the Thales circle above the hypotenuse, using a line parallel to the hypotenuse and at a distance of $m_{c}$, the altitude perpendicular to the hypotenuse. Two expressions for twice the area, and based on (1),

$$
m_{c}=\frac{a b}{c}=\frac{a+b}{3}
$$

thus

$$
a+b=3 m_{c}, \quad a b=c m_{c}
$$

And based on the Pythagorean theorem,

$$
a^{2}+b^{2}=(a+b)^{2}-2 a b=9 m_{c}^{2}-2 c m_{c}=c^{2}
$$

whose only positive root is

$$
m_{c}=c \cdot \frac{1+\sqrt{10}}{9}(\approx c \cdot 0.46)
$$

which is smaller than the radius $c / 2$ of the Thales circle, so there is a solution, and due to symmetry, essentially there is only one solution.

A possible construction of $m_{c}$: Construct a right triangle with legs $P R=3 P Q=3 c$ and $R S=c$, extend the hypotenuse beyond $S$ by $S T=c$, and extend $P R$ beyond $R$ by $R U=2 P R$. Finally, intersect $P S$ with the line through $Q$ parallel to $T U$ at $V$, then $P V=m_{c}$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_2ab1806882c48663234dg-1.jpg?height=336&width=1248&top_left_y=1011&top_left_x=428)

Figure 1

Remark. Another possible construction of the segment $m_{c}$: From the endpoint $A$ of the segment $A B=c$ in any direction, measure a segment of 9 units, with endpoints $P_{1}, P_{2}, \ldots, P_{9}$. At $P_{4}$, measure $P_{4} F=1$ perpendicularly, and rotate $F$ around $P_{1}$ onto $P_{1} P_{9}$, to $G$. Then the line through $G$ parallel to $B P_{9}$ cuts off the segment $A H=m_{c}$ from $A B$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_2ab1806882c48663234dg-1.jpg?height=496&width=593&top_left_y=1649&top_left_x=755)

Figure 2

Péter Komjáth (Budapest, Fazekas M. Gyak. Gimn., II. o. t.)

II. solution. We can use the fact that a right triangle can be constructed from the length of the hypotenuse and the length of the angle bisector of the right angle, as the said $f_{c}$ bisector can be easily expressed from the general relationship valid for all triangles:

$$
f_{c}=\frac{2 a b \cos \gamma / 2}{a+b}=\frac{2 c}{3} \cos \frac{\gamma}{2}=\frac{2}{3} \cdot \frac{c}{\sqrt{2}}
$$

which is the $2/3$ part of the leg of an isosceles right triangle with hypotenuse $c$.[^0]

Let the angle bisector of $\angle A C B$ intersect the hypotenuse $A B$ at $D$ and the Thales circle of the hypotenuse at $E$. Then $E$ bisects the arc of $A B$ not containing $C$, $\angle E A D = \angle E C A$, $\triangle E A D \sim \triangle E C A$, and with $x$ denoting $E D$, $x\left(x+f_{c}\right)=E A^{2}=c^{2} / 2$, from which

$$
x=\sqrt{\frac{c^{2}}{2}+\frac{f_{c}^{2}}{4}}-\frac{f_{c}}{2}
$$

Based on this, $D$ can be cut out, and finally $E D$ cuts out $C$ from the circle. (Figure 3, the mirror image of $A E$ in $B E^{\prime}$ is divided into thirds at $F$, then $x=E F-F B$.)

![](https://cdn.mathpix.com/cropped/2024_05_02_2ab1806882c48663234dg-2.jpg?height=632&width=599&top_left_y=500&top_left_x=728)

Figure 3

Sándor Szamosújvári (Debrecen, KLTE Gyak. Gimn., III. o. t.)


[^0]:    ${ }^{1}$ See, for example, Problem 1123, K. M. L. 24 (1962) p. 201; any angle can be placed at the right angle.