If we extend the internal angle bisectors of the triangle $ABC$ to the circumference of the circumcircle of the triangle, we obtain the triangle $A^{\prime} B^{\prime} C^{\prime}$; the sides of this triangle intersect the angle bisectors of the original triangle at points $A_{1}, B_{1}$, and $C_{1}$; the angle bisectors of the triangle $A_{1} B_{1} C_{1}$ intersect the circumference of the circumcircle of this triangle at points $A^{\prime \prime}, B^{\prime \prime}, C^{\prime \prime}$; the sides of the triangle $A^{\prime \prime}, B^{\prime \prime}, C^{\prime \prime}$ intersect the angle bisectors of the triangle $A_{1} B_{1} C_{1}$ at points $A_{2} B_{2} C_{2}$, and so on. Continuing this process to infinity, calculate the sum of the perimeters and areas of the triangles

$$
A B C, \quad A_{1} B_{1} C_{1}, \quad A_{2} B_{2} C_{2}, \ldots
$$

if the perimeter of the triangle $ABC$ is $K$ and its area is $T$. Furthermore, show that

$$
T: T^{\prime}=8 \sin \frac{\alpha}{2} \sin \frac{\beta}{2} \sin \frac{\gamma}{2}
$$

and

$$
r: r^{\prime}=a b c\left(a^{\prime}+b^{\prime}+c^{\prime}\right): a^{\prime} b^{\prime} c^{\prime}(a+b+c),
$$

where $T^{\prime}$ is the area of the triangle $A^{\prime} B^{\prime} C^{\prime}$, and $r$ and $r^{\prime}$ are the radii of the inscribed circles of the respective triangles.
$\left(1^{\circ}\right) \mathrm{Az} A B C$ triangle's internal angle bisectors intersect at point $P$.

![](https://cdn.mathpix.com/cropped/2024_05_02_389b60bea1b5421f8c8bg-1.jpg?height=401&width=398&top_left_y=220&top_left_x=847)

The diagram shows that

$$
\begin{gathered}
A^{\prime} C^{\prime} C \varangle+C C^{\prime} B^{\prime} \varangle+B B^{\prime} C^{\prime} \varangle=A^{\prime} A C \varangle+C B B^{\prime} \varangle+B C C^{\prime} \varangle= \\
=\frac{\alpha}{2}+\frac{\beta}{2}+\frac{\gamma}{2}=90^{\circ}
\end{gathered}
$$

and thus

$$
C^{\prime} B_{1} B^{\prime} \varangle=90^{\circ} .
$$

Furthermore, since

$$
B C^{\prime} A^{\prime} \varangle=B A A^{\prime} \varangle=\frac{\alpha}{2}=A^{\prime} A C \varangle=A^{\prime} C^{\prime} C \varangle,
$$

it follows that

$$
B B_{1}=P B_{1},
$$

or

$$
P B=2 P B_{1}
$$

Similarly,

$$
P A=2 P A_{1} \quad \text { and } \quad P C=2 P C_{1} \text {. }
$$

Since the points $A_{1}, B_{1}, C_{1}$ divide the distances $A P, B P, C P$ in the same ratio, it follows that $A B: B C: C A=A_{1} B_{1}: B_{1} C_{1}: C_{1} A_{1} \mathrm{~s}$ and thus $\triangle A B C \sim \triangle A_{1} B_{1} C_{1}$. Therefore,

$$
K=A B+A C+B C=2 A_{1} B_{1}+2 A_{1} C_{1}+2 B_{1} C_{1}=2 K_{1}
$$

from which

$$
K_{1}=\frac{1}{2} K
$$

Similarly,

$$
K_{2}=\frac{1}{2} K_{1}, \quad K_{3}=\frac{1}{2} K_{2}, \ldots
$$

and thus the sum of the perimeters of the triangles $A B C, A_{1} B_{1} C_{1}, A_{2} B_{2} C_{2}, \ldots$ is:

$$
S_{1}=\frac{K}{1-\frac{1}{2}}=2 K
$$

Furthermore,

$$
T: T_{1}=\overline{A B}^{2}:{\overline{A_{1} B_{1}}}^{2}=4{\overline{A_{1} B_{1}}}^{2}:{\overline{A_{1} B_{1}}}^{2}=4: 1
$$

therefore

$$
T_{1}=\frac{1}{4} T
$$

Similarly,

$$
T_{2}=\frac{1}{4} T_{1}, \quad T_{3}=\frac{1}{4} T_{2}, \ldots
$$

and thus the sum of the areas of the mentioned triangles is:

$$
S_{2}=\frac{T}{1-\frac{1}{4}}=\frac{4}{3} T
$$

$\left(2^{\circ}\right)$ Based on problem 275 (K.M.L.IV.100.1.)

$$
T=2 R^{2} \sin \alpha \sin \beta \sin \gamma
$$

therefore

$$
\begin{gathered}
T^{\prime}=2 R^{2} \sin \frac{1}{2}(\beta+\gamma) \sin \frac{1}{2}(\alpha+\gamma) \sin \frac{1}{2}(\alpha+\beta)= \\
=2 R^{2} \cos \frac{\alpha}{2} \cos \frac{\beta}{2} \cos \frac{\gamma}{2}
\end{gathered}
$$

and thus

$$
\begin{gathered}
\frac{T}{T^{\prime}}=\frac{\sin \alpha \sin \beta \sin \gamma}{\cos \frac{\alpha}{2} \cos \frac{\beta}{2} \cos \frac{\gamma}{2}}= \\
=\frac{2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \cdot 2 \sin \frac{\beta}{2} \cos \frac{\beta}{2} \cdot 2 \sin \frac{\gamma}{2} \cos \frac{\gamma}{2}}{\cos \frac{\alpha}{2} \cos \frac{\beta}{2} \cos \frac{\gamma}{2}} \\
=8 \sin \frac{\alpha}{2} \sin \frac{\beta}{2} \sin \frac{\gamma}{2}
\end{gathered}
$$

$\left(3^{\circ}\right)$

$$
\begin{gathered}
r: r^{\prime}=\frac{T}{s}: \frac{T^{\prime}}{s^{\prime}}=\frac{\frac{a b c}{4 R}}{a+b+c}: \frac{\frac{a^{\prime} b^{\prime} c^{\prime}}{4 R}}{a^{\prime}+b^{\prime}+c^{\prime}}= \\
=a b c\left(a^{\prime}+b^{\prime}+c^{\prime}\right): a^{\prime} b^{\prime} c^{\prime}(a+b+c)
\end{gathered}
$$

(Jenő Filkorn, Nyitra.)

The problem was also solved by: Bertrám J., Czank K., Freibauer E., Krausz B., Krisztián Gy., Lukhaub Gy., Lupsa Gy., Messik G., Messik V., Oltay K., Sasvári G., Szabó J.