9. In triangle $ABC$, the bisectors $AL, BM$, and $CN$ are drawn, and $\angle ANM = \angle ALC$. Find the radius of the circumcircle of triangle $LMN$, two sides of which are equal to 3 and 4.
Answer: 2

Solution.

The following statement is true:

Let $A L, B M$ and $C N$ be the bisectors of the internal angles of a triangle. If $\angle A N M = \angle A L C$, then $\angle B C A = 120^{\circ}$.

Proof. Draw $M D \| A L$, then $\angle C M D = \frac{\alpha}{2}$. The angle $\angle A L C$ is an external angle of triangle $A L B$, so $\angle A N M = \angle A L C = \frac{\alpha}{2} + \beta$. The angle $\angle B M C$ is an internal angle of triangle $B M C$, so $\angle B M C = \pi - \frac{\beta}{2} - \gamma = \alpha + \frac{\beta}{2}$. Therefore, $\angle B M D = \alpha + \frac{\beta}{2} - \frac{\alpha}{2} = \frac{\alpha + \beta}{2}$. The angle $\angle N M C$ is an external angle of triangle $A N M$, so $\angle N M C = \frac{3 \alpha}{2} + \beta$. Therefore, $\angle B M N = \frac{3 \alpha}{2} + \beta - \frac{\alpha + \beta}{2} - \frac{\alpha}{2} = \frac{\alpha + \beta}{2}$. This means triangles $B M N$ and $B M D$ are congruent, so $N M = M D$ and $\angle M D N = \frac{\pi - (\alpha + \beta)}{2} = \frac{\gamma}{2}$. Therefore, points $N, M, C$ and $D$ lie on the same circle, and since $N M$ and $N D$ are chords subtending equal arcs, $N M = N D$, i.e., triangle $N M D$ is equilateral, so $\frac{\gamma}{2} = \frac{\pi}{3}$. Therefore, $\angle B C A = 120^{\circ}$ (see Fig. 4).

![](https://cdn.mathpix.com/cropped/2024_05_06_024d7be6fbdc002fb8c6g-6.jpg?height=429&width=663&top_left_y=962&top_left_x=228)

Fig. 4

![](https://cdn.mathpix.com/cropped/2024_05_06_024d7be6fbdc002fb8c6g-6.jpg?height=560&width=654&top_left_y=959&top_left_x=924)

Remark. The converse statement is also true: if $\angle B C A = 120^{\circ}$ and $A L, B M$ and $C N$ are the bisectors of the internal angles of the triangle, then $\angle A N M = \angle A L C$.

The line $C B$ is the bisector of angle $N C F$ ( $\Rightarrow$ point $L$ is equidistant from lines $C N$ and $C F$ ), and the line $A L$ is the bisector of angle $B A F$ ( $\Rightarrow$ point $L$ is equidistant from lines $A N = N B$ and $A F = C F$). Therefore, point $L$ is equidistant from lines $N C$ and $N B$, i.e., line $N L$ is the bisector of angle $B N C$. Similarly, the line $C A$ is the bisector of angle $E C N$ ( $\Rightarrow$ point $M$ is equidistant from lines $C N$ and $C E$ ), and the line $B M$ is the bisector of angle $A B C$ ( $\Rightarrow$ point $M$ is equidistant from lines $B A = N A$ and $B E = C E$). Therefore, point $M$ is equidistant from lines $N A$ and $N C$, i.e., line $N M$ is the bisector of angle $A N C$. Therefore, $\angle M N L = 90^{\circ}$ (see Fig. 5).

Let's estimate the values that the tangents of the acute angles of $\triangle L M N$ can take.

From $\triangle A N C$ we get that $\angle A N C = 120^{\circ} - \alpha \Rightarrow \angle M N C = 60^{\circ} - \frac{\alpha}{2} \Rightarrow \angle N M C = 60^{\circ} + \frac{\alpha}{2}$.

From $\triangle B N C$ we get that $\angle B N C = 120^{\circ} - \beta \Rightarrow \angle L N C = 60^{\circ} - \frac{\beta}{2} \Rightarrow \angle N L C = 60^{\circ} + \frac{\beta}{2}$.

By the Law of Sines for $\triangle N M C$ we get: $\frac{N C}{\sin \angle N M C} = \frac{N M}{\sin \angle N C M} \Rightarrow N C = \frac{2 N M \sin \left(60^{\circ} + \frac{\alpha}{2}\right)}{\sqrt{3}}$.

By the Law of Sines for $\triangle N L C$ we get: $\frac{N C}{\sin \angle N L C} = \frac{N L}{\sin \angle N C L} \Rightarrow N C = \frac{2 N L \sin \left(60^{\circ} + \frac{\beta}{2}\right)}{\sqrt{3}}$.

Considering the equality $\frac{\alpha}{2} + \frac{\beta}{2} = 30^{\circ}$ from the last two relations, we get that

$$
\operatorname{tg} \angle N L M = \frac{N M}{N L} = \frac{\sin \left(90^{\circ} - \frac{\alpha}{2}\right)}{\sin \left(60^{\circ} + \frac{\alpha}{2}\right)}
$$

Since $\frac{\alpha}{2} \in (0^{\circ}, 30^{\circ})$ and the function $y = \sin t$ is increasing in the first quadrant, then $\operatorname{tg} \angle N L M \in \left(\frac{\sqrt{3}}{2} ; \frac{2}{\sqrt{3}}\right)$. If 3 and 4 are the lengths of the legs of $\triangle N L M$, then either $\operatorname{tg} \angle N L M = \frac{3}{4}$ or $\operatorname{tg} \angle N L M = \frac{4}{3}$. But since $\frac{3}{4} < \frac{\sqrt{3}}{2} = \sqrt{\frac{3}{4}}$, this case is impossible. Therefore, only the case where $M L = 4$ (hypotenuse) and 3 is the length of one of the legs is possible. The length of the other leg by the Pythagorean theorem is $\sqrt{7}$. Since the inequalities $\frac{\sqrt{3}}{2} < \frac{\sqrt{7}}{3} < \frac{2}{\sqrt{3}} \Leftrightarrow \frac{3}{4} < \frac{7}{9} < \frac{4}{3}$ hold, this case is realized. Since in a right triangle the radius of the circumscribed circle is half the hypotenuse, then

$$
R = \frac{M L}{2} = 2
$$