In a right-angled triangle, the segment of the angle bisector of one of the acute angles that falls within the triangle is $k$ times as large as the segment of the opposite leg that falls on the hypotenuse. Calculate the angles of the triangle.

For which values of $k$ does the problem have a solution?

For which value of $k$ is the triangle isosceles?

Numerical example: $k=7 / 2$.
Solution. I. Let the legs of the right triangle $ABC$ be $CA, CB$ such that for the angle bisector $AD$ from $A$, we have $AD: DB = k$.

![](https://cdn.mathpix.com/cropped/2024_05_02_8662ff9675dfcdfd0a87g-1.jpg?height=393&width=596&top_left_y=243&top_left_x=751)

Denoting the angles in the usual way, from the sine theorem in triangle $ABD$ and the identities for double angles, we get an equation for $\sin \frac{\alpha}{2}$:

$$
\begin{gathered}
k = AD: DB = \sin \beta: \sin \frac{\alpha}{2} = \cos \alpha: \sin \frac{\alpha}{2} = \left(1 - 2 \sin^2 \frac{\alpha}{2}\right): \sin \frac{\alpha}{2} \\
2 \sin^2 \frac{\alpha}{2} + k \sin \frac{\alpha}{2} - 1 = 0
\end{gathered}
$$

and from this, uniquely,

$$
\sin \frac{\alpha}{2} = \frac{1}{4} \left( \sqrt{k^2 + 8} - k \right)
$$

(negative root is immediately discarded).

II. Considering $k$ to be positive, we get a triangle for any value of $k$. Indeed, for $\frac{\alpha}{2} > 0$, it holds that

$$
\sqrt{k^2 + 8} > k \quad \text{and} \quad \sqrt{k^2 + 8} > \sqrt{8}
$$

Thus,

$$
\frac{1}{4} \left( \sqrt{k^2 + 8} - k \right) < \frac{\sqrt{8}}{4} = \frac{1}{\sqrt{2}}
$$

Calculating $\alpha$ uniquely determines $\beta$.

$$
\begin{aligned}
& \text{For } k = \frac{7}{2}, \quad \sin \frac{\alpha}{2} = \frac{1}{4}, \quad \cos \alpha = \frac{7}{8} = 0.875 \\
& \alpha = 28^\circ 57', \quad \beta = 61^\circ 3'
\end{aligned}
$$

III. Triangle $ABC$ is isosceles if $\alpha = 45^\circ$, i.e., according to (1), if

$$
k = \frac{\cos 45^\circ}{\sin 22.5^\circ} = \frac{\sin 45^\circ}{\sin 22.5^\circ} = 2 \cos 22.5^\circ = 2 \sqrt{\frac{1 + \sqrt{2}}{2}} = \sqrt{2 + \sqrt{2}} \approx 1.841
$$

(Triangle $ABD$ is isosceles when $k = 1$, i.e., $\alpha = 60^\circ$.)

László Graffjódi (Miskolc, Földes F. Gymn., III. grade) Péter Körtvélyessy (Szeged, Ságvári E. Pract. Gymn., IV. grade)

Remark. The fact that the positive value of $\sin \frac{\alpha}{2}$ obtained from (2) is always less than $\frac{1}{\sqrt{2}}$ can also be derived as follows: the left side of our equation is negative when $\sin \frac{\alpha}{2} = 0$ and positive when $\sin \frac{\alpha}{2} = \frac{1}{\sqrt{2}}$, so the root lies between these two limits. This relies on the fact that the graph of the left side is a continuous line in the interval $0 \leq \sin \frac{\alpha}{2} \leq \frac{1}{\sqrt{2}}$.