What are the lengths of the legs of a right triangle if the segments of the angle bisectors of the acute angles are 1 and 2 units long, respectively?
We denote the halves of the acute angles by $\alpha$ and $\beta$. Using the notation of Figure 1, from the sine theorem in the triangle $A A_{1} B$ we have

$$
2 \cdot \cos \beta - \sin \alpha = \frac{\sin \alpha}{\sin 2 \beta}
$$

This is equivalent to

$$
2 \cdot \sin 2 \beta \cdot \cos \beta = \sin \alpha (1 + \sin 2 \beta)
$$

From this, using the relationship $\alpha + \beta = 45^{\circ}$, we get

$$
2 \cdot \sin 2 \beta \cdot \cos \beta = \sin \left(45^{\circ} - \beta\right) (1 + \sin 2 \beta)
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_f1d4cae9cec929010ab9g-1.jpg?height=569&width=900&top_left_y=569&top_left_x=599)

We first try to solve this equation graphically.

After plotting the factors on both the left and right sides, we perform a "graphical multiplication." Since $\beta$ can only vary in the interval $\left(0 ; 45^{\circ}\right)$, it is sufficient to draw the graphs in this interval (Figure 2). The sketch shows a solution near $10^{\circ}$. By substitution, it can be determined that for $\beta_{1} = 11.5^{\circ}$, the value on the left side is smaller, and for $\beta_{2} = 12^{\circ}$, it is larger than the value on the right side. Therefore, $\beta_{3} = \frac{\beta_{1} + \beta_{2}}{2} = 11.75^{\circ}$ is a more accurate approximate value of the root than $\beta_{1}$ and $\beta_{2}$, and as can be determined by substitution, the root lies between $\beta_{1}$ and $\beta_{3}$. Continuing this so-called bisection method, we get the approximate value $\beta_{4} = \frac{\beta_{1} + \beta_{3}}{2} = 11.625^{\circ}$ (2) for the root. After a few steps, using a pocket calculator, we arrive at the fairly accurate approximate value $\beta = 11.52903^{\circ}$, with which the value on the left side is 0.7675235, and the value on the right side is 0.7675226. With this value of $\beta$, $\alpha = 33.47097^{\circ}$, and the two legs are $\cos \alpha = 0.834165$ and $2 \cdot \cos \beta = 1.959647$.

![](https://cdn.mathpix.com/cropped/2024_05_02_f1d4cae9cec929010ab9g-1.jpg?height=544&width=550&top_left_y=1647&top_left_x=777)

Figure 2

Gusztáv Reményi

Remark. Eliminating $\beta$ from equation (1), we arrive at the cubic equation

$$
\operatorname{tg}^{3} \alpha + \operatorname{tg}^{2} \alpha + (\sqrt{2} - 1) \operatorname{tg} \alpha - 1 = 0
$$

The solution method for such equations can be found, for example, in László Fuchs: Introduction to Algebra and Number Theory, Part II, p. 86. Solving this equation using the so-called Cardano formula, the two legs are $b = 0.8341$ and $a = 1.9596$.