In a certain right-angled triangle, the sum of the hypotenuse and one of the legs is $c+b=14 \text{~cm}, \operatorname{tg} \frac{\alpha}{2}=\frac{1}{2}$, where $\alpha$ is the angle opposite the leg $a$. Calculate the lengths of the sides of the triangle without using trigonometric tables.
I. solution: Using the known formula

$$
\operatorname{tg} \frac{\alpha}{2}=\sqrt{\frac{1-\cos \alpha}{1+\cos \alpha}}
$$

we have

$$
\operatorname{tg}^{2} \frac{\alpha}{2}=\frac{1}{4}=\frac{1-\cos \alpha}{1+\cos \alpha}=\frac{1-\frac{b}{c}}{1+\frac{b}{c}}=\frac{c-b}{c+b}=\frac{c-b}{14}
$$

from which

$$
c-b=\frac{14}{4}=\frac{7}{2}
$$

According to the problem,

$$
c+b=14
$$

From (1) and (2),

$$
c=\frac{35}{4}=8.75 \mathrm{~cm} \quad \text { and } \quad b=\frac{21}{4}=5.25 \mathrm{~cm}
$$

and thus

$$
\begin{aligned}
& a=\sqrt{\frac{35^{2}-21^{2}}{16}}=\sqrt{\frac{56 \cdot 14}{16}}=\sqrt{\frac{16 \cdot 49}{16}}=7 \mathrm{~cm} . \\
& \text { Piroska Gerencsér (Bp., VIII., Zrínyi Ilona lg. IV. o. t.) }
\end{aligned}
$$

II. solution: We can also use the formula

$$
\operatorname{tg} \alpha=\frac{2 \operatorname{tg} \frac{\alpha}{2}}{1-\operatorname{tg}^{2} \frac{\alpha}{2}}
$$

$$
\operatorname{tg} \alpha=\frac{a}{b}=\frac{2 \cdot \frac{1}{2}}{1-\frac{1}{4}}=\frac{4}{3}
$$

Thus, the sides of the triangle are of the form:

$$
a=4 \lambda, \quad b=3 \lambda, \quad \text { and } \quad c=\sqrt{a^{2}+b^{2}}=\sqrt{16 \lambda^{2}+9 \lambda^{2}}=5 \lambda .
$$

According to the problem,

$$
b+c=8 \lambda=14
$$

from which

$$
\lambda=\frac{7}{4}
$$

István Kovács (Bp., VIII., Piarista g. IV. o. t.)

III. solution: We can also reach the goal without using trigonometric formulas. The angle bisector of $\alpha$ divides the side $a$ into segments $x$ and $y$. (See the figure.)

![](https://cdn.mathpix.com/cropped/2024_05_02_0b4814175b0b7619390cg-1.jpg?height=358&width=594&top_left_y=1943&top_left_x=752)

According to a known theorem,

$$
\frac{x}{y}=\frac{b}{c}
$$

Since $\operatorname{tg} \frac{\alpha}{2}=\frac{1}{2}=\frac{x}{b}$, we have $x=\frac{b}{2}$, and thus from (1),

$$
y=\frac{c}{2}
$$

Therefore,

$$
a=x+y=\frac{b+c}{2}=\frac{14}{2}=7
$$

By the Pythagorean theorem,

$$
7^{2}+b^{2}=(14-b)^{2}=196-28 b+b^{2}
$$

from which

$$
28 b=147, \quad \text { and thus } \quad b=\frac{147}{28}=\frac{21}{4} \text { etc. }
$$

Bálint Benkő (Sárospatak, Rákóczi g. III. o. t.)

IV. solution: We can even do without the angle bisector theorem. Rotate the hypotenuse $AB=c$ around $A$ into the extension of $AC=b$: $AB'=c$. (See the figure.) Then clearly

$$
B'C=b+c=14
$$

and

$$
\angle BB'C=\frac{\alpha}{2}
$$

So from the right triangle $BB'C$,

$$
a=(b+c) \operatorname{tg} \frac{\alpha}{2}=14 \cdot \frac{1}{2}=7
$$

etc., as in the III. solution.

Tamás Rázga (Bp. II., Rákóczi g. III. o. t.)