Given a triangle with two sides $a=20 \mathrm{~m}, b=17 \mathrm{~m}$ and the height $m_{a}=14 \mathrm{~m}$ corresponding to side $a$. Find the radius of the circumcircle of the triangle.
It is known that the radius of the circumcircle is

$$
R=\frac{a b c}{4 t}=\frac{a b c}{2 a m_{a}}=\frac{b}{2 m_{a}} \cdot c
$$

According to Carnot's theorem,

$$
c=\sqrt{a^{2}+b^{2} \pm 2 a b_{a}}
$$

where $b_{a}$ represents the projection of side $b$ onto side $a$. However,

$$
b_{a}=\sqrt{b^{2}-m_{a}^{2}}
$$

thus

$$
R=\frac{b}{2 m_{a}} \sqrt{a^{2}+b^{2} \pm 2 a \sqrt{b^{2}-m_{a}^{2}}}
$$

Therefore, our problem has two solutions. The positive sign is used if $\gamma$ is an obtuse angle, and the negative sign if $\gamma$ is an acute angle.

In our case: $1^{\circ} . \quad \gamma>90^{\circ}:$

$$
R_{1}=19.9 \mathrm{~m}
$$

$2^{\circ} . \quad \gamma<90^{\circ}:$

$$
R_{2}=10.5 \mathrm{~m}
$$

(Alfréd Haar, Budapest.)

The problem was also solved by: Ádámffy E., Bánó L., Bartók I., Braun J., Dömény E., Dömény I., Eckhart F., Enyedi B., Füstös B., Fekete M., Hirschfeld Gy., Harsányi Z., Heimlich P., Jánosy Gy., Kertész G., Kiss J., Kürti I., Liebner A., Pazsiczky G., Pám M., Pivnyik J., Popoviciu M., Pető L., Rássy P., Riesz K., Rosenberg J., Szombathy J., Söpkéz Gy., Szécsi J., Szúcs A., Schwarz Gy., Schwemmer J., Székely J., Tandlich E.