Let $A_{1}, B_{1}$, and $C_{1}$ be the points on the sides of triangle $A B C$ such that

$$
\frac{A C_{1}}{C_{1} B}=\frac{B A_{1}}{A_{1} C}=\frac{C B_{1}}{B_{1} A}=\lambda
$$

$A_{2}, B_{2}$, and $C_{2}$ are the points on the sides of triangle $A_{1} B_{1} C_{1}$ such that

$$
\frac{A_{1} C_{2}}{C_{2} B_{1}}=\frac{B_{1} A_{2}}{A_{2} C_{1}}=\frac{C_{1} B_{2}}{B_{2} A_{1}}=\lambda
$$

For what value of $\lambda$ will the area of triangle $A_{2} B_{2} C_{2}$ be half the area of triangle $A B C$?
If the areas of triangles $ABC$, $A_1B_1C_1$, and $A_2B_2C_2$ are $T$, $T_1$, and $T_2$, respectively, then clearly $T : T_1 = T_1 : T_2$, since the triangle $A_1B_1C_1$ is obtained from the triangle $ABC$ using the same procedure as the triangle $A_2B_2C_2$ is obtained from the triangle $A_1B_1C_1$. First, let's determine the ratio $T : T_1$.

![](https://cdn.mathpix.com/cropped/2024_05_02_3027a988f235d6e31900g-1.jpg?height=412&width=550&top_left_y=276&top_left_x=777)

First, we calculate the area $T_A$ of triangle $AC_1B_1$. Since $\frac{C_1B}{AC_1} = \frac{1}{\lambda}$, we have $\frac{AB}{AC_1} = \frac{AC_1 + C_1B}{AC_1} = 1 + \frac{1}{\lambda} = \frac{\lambda + 1}{\lambda}$. Similarly, we get $\frac{CA}{CB_1} = \frac{\lambda + 1}{\lambda}$, or $\frac{CA}{B_1A} = \frac{1}{\frac{B_1A}{CA}} = \frac{1}{\frac{CA - CB_1}{CA}} = \frac{1}{1 - \frac{\lambda}{\lambda + 1}} = \lambda + 1$. According to the theorem of parallel chords, this ratio is equal to the ratio of the heights corresponding to sides $AB$ and $AC_1$ in triangles $ABC$ and $AC_1B_1$. Therefore,

\[
\frac{T}{T_A} = \frac{AB}{AC_1} \cdot \frac{CA}{B_1A} = \frac{\lambda + 1}{\lambda} \cdot (\lambda + 1)
\]

so $T_A = T \cdot \frac{\lambda}{(\lambda + 1)^2}$.

By similar calculations, we find that the areas $T_B$ and $T_C$ of triangles $BA_1C_1$ and $CB_1A_1$ are also $T \cdot \frac{\lambda}{(\lambda + 1)^2}$. Therefore, $T_1$ depends only on $T$ and $\lambda$:

\[
T_1 = T - (T_A + T_B + T_C) = T \cdot \left(1 - \frac{3\lambda}{(\lambda + 1)^2}\right)
\]

For the second inscribed triangle,

\[
T_2 = T_1 \left(1 - \frac{3\lambda}{(\lambda + 1)^2}\right) = T \cdot \left(1 - \frac{3\lambda}{(\lambda + 1)^2}\right)^2
\]

and if $T_2 = \frac{1}{2} T$, then

\[
\frac{1}{2} = \left(1 - \frac{3\lambda}{(\lambda + 1)^2}\right)^2
\]

or

\[
\frac{\sqrt{2}}{2} = \pm \left(1 - \frac{3\lambda}{(\lambda + 1)^2}\right)
\]

![](https://cdn.mathpix.com/cropped/2024_05_02_3027a988f235d6e31900g-1.jpg?height=707&width=588&top_left_y=1999&top_left_x=755)

If the sign on the right side is positive, then after rearranging, we get the equation

\[
(\sqrt{2} - 1) \lambda^2 - (2 + \sqrt{2}) \lambda + \sqrt{2} - 1 = 0
\]

or

\[
\lambda^2 - (3\sqrt{2} + 4) \lambda + 1 = 0
\]

The solutions to this equation are

\[
\begin{aligned}
& \lambda_1 = \frac{1}{2}(3\sqrt{2} + 4 + \sqrt{30 + 24\sqrt{2}}) \approx 8.119 \\
& \lambda_2 = \frac{1}{2}(3\sqrt{2} + 4 - \sqrt{30 + 24\sqrt{2}}) \approx 0.123
\end{aligned}
\]

It can be easily verified that for these values of $\lambda$, $T_2 = \frac{1}{2} T$ holds.

If, however, the sign on the right side of equation (2) is negative, then after rearranging, we get the equation $\lambda^2 + (3\sqrt{2} - 4) \lambda + 1 = 0$, which has no real solutions.

Thus, the conditions of the problem are satisfied by $\lambda_1$ and $\lambda_2 = \frac{1}{\lambda_1}$.

Remarks. 1. It was evident from the start that if $\lambda$ is a solution to the problem, then $\frac{1}{\lambda}$ is also a solution, since it does not matter which side the longer segment falls on when placing the division points.

2. Using the known area formula $T = \frac{1}{2} ab \sin \gamma$, equation (1) can also be obtained with simpler calculations.