Consider a triangle with sides $a, a+1, a+2$; the smallest angle is half of the largest angle. Calculate the sides and angles.
According to the Law of Sines:

$$
a: a+2=\sin \alpha: \sin 2 \alpha
$$

or

$$
a: a+2=1: 2 \cos \alpha
$$

from which

$$
\cos \alpha=\frac{a+2}{2 a}
$$

Based on Carnot's Theorem:

$$
a^{2}=(a+1)^{2}+(a+2)^{2}-2(a+1)(a+2) \cos \alpha
$$

from which

$$
\cos \alpha=\frac{a^{2}+6 a+5}{2(a+1)(a+2)}=\frac{a+5}{2(a+2)}
$$

From (3) and (5) we get:

$$
\frac{a+2}{2 a}=\frac{a+5}{2(a+2)}
$$

from which \(a=4\).

Thus, the sides of the triangle are: 4, 5, 6. We calculate one of the angles of the triangle using (3); substituting the value of \(a\):

$$
\cos \alpha=\frac{3}{4}
$$

from which \(\alpha=41^{\circ} 24^{\prime} 35^{\prime \prime}\), \(2 \alpha=82^{\circ} 49^{\prime} 10^{\prime \prime}\); finally, the third angle: \(55^{\circ} 46^{\prime} 15^{\prime \prime}\).

(Geist Emil, 8th grade student, Györ.)

The problem was also solved by: Berger Hugó, Feuer Mór, Freund Antal, Friedmann Bernát, Goldstein Zsigmond, Goldziher Károly, Grünhut Béla, Hofbauer Ervin, Kántor Nándor, Klein Mór, Riesz Frigyes, Schneider Béla, Szabó István.