For the points $C_{1}, B_{1}, A_{1}$ on the sides $A B, A C$, and $C B$ of triangle $A B C$ respectively,

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

Let the radii of the incircles of triangles $A C_{1} B_{1}$, $B C_{1} A_{1}$, $C A_{1} B_{1}$, $A_{1} B_{1} C_{1}$, and $A B C$ be $r_{1}, r_{2}, r_{3}, r_{4}$, and $r$ respectively. For what values of $\lambda$ does the equality

$$
\frac{1}{r_{1}}+\frac{1}{r_{2}}+\frac{1}{r_{3}}=\frac{1}{r_{4}}+\frac{4}{r}
$$

hold?
Let the areas and perimeters of triangles $A C_{1} B_{1}$, $B C_{1} A_{1}$, $C A_{1} B_{1}$, $A_{1} B_{1} C_{1}$, and $A B C$ be denoted by $T_{1}, K_{1}$; $T_{2}, K_{2}$; $T_{3}, K_{3}$; $T_{4}, K_{4}$, and $T, K$ respectively. We know that twice the area of a triangle equals the product of its perimeter and the radius of its inscribed circle. The obvious equality for the perimeters of the five triangles, $\frac{K_{1}+K_{2}+K_{3}}{2}=\frac{K_{4}+K}{2}$, can thus be written as

$$
\frac{T_{1}}{r_{1}}+\frac{T_{2}}{r_{2}}+\frac{T_{3}}{r_{3}}=\frac{T_{4}}{r_{4}}+\frac{T}{r}
$$

Let $\alpha = \angle BAC$. Then $2T = AB \cdot AC \cdot \sin \alpha$ and $2T_{1} = AB_{1} \cdot AC_{1} \cdot \sin \alpha$. Since $\frac{AC_{1}}{C_{1}B} = \lambda$ and $AC_{1} + C_{1}B = AB$, it follows that $AC_{1} = \frac{\lambda}{1+\lambda} \cdot AB$; from $\frac{CB_{1}}{B_{1}A} = \lambda$ and $CB_{1} + B_{1}A = CA$, we get $AB_{1} = \frac{1}{1+\lambda} \cdot AC$. Therefore,

$$
2T_{1} = AB_{1} \cdot AC_{1} \cdot \sin \alpha = \frac{\lambda}{(1+\lambda)^{2}} \cdot AB \cdot AC \cdot \sin \alpha = \frac{\lambda}{(1+\lambda)^{2}} \cdot 2T
$$

Similarly, it can be shown that $T_{2} = T_{3} = \frac{\lambda}{(1+\lambda)^{2}} \cdot T$. From this, it follows that $T_{4} = T - (T_{1} + T_{2} + T_{3}) = \left(1 - \frac{3\lambda}{(1+\lambda)^{2}}\right) \cdot T$. Substituting these into the (1) equality:

$$
\frac{\lambda \cdot T}{(1+\lambda)^{2}}\left(\frac{1}{r_{1}}+\frac{1}{r_{2}}+\frac{1}{r_{3}}\right) = \left(1 - \frac{3\lambda}{(1+\lambda)^{2}}\right) \frac{T}{r_{4}} + \frac{T}{r}
$$

Rearranging and using the condition $\frac{1}{r_{1}}+\frac{1}{r_{2}}+\frac{1}{r_{3}}=\frac{1}{r_{4}}+\frac{4}{r}$, we get

$$
\frac{\lambda}{(1+\lambda)^{2}}\left(\frac{1}{r_{4}}+\frac{4}{r}\right) = \frac{1-\lambda+\lambda^{2}}{(1+\lambda)^{2}} \cdot \frac{1}{r_{4}} + \frac{1}{r}
$$

Multiplying by $(1+\lambda)^{2}$ and rearranging again:

$$
\begin{aligned}
\lambda \cdot \frac{1}{r_{4}} + 4\lambda \cdot \frac{1}{r} & = (1-\lambda+\lambda^{2}) \cdot \frac{1}{r_{4}} + (1+2\lambda+\lambda^{2}) \cdot \frac{1}{r} \\
0 & = (1-\lambda)^{2} \cdot \left(\frac{1}{r_{4}} + \frac{1}{r}\right)
\end{aligned}
$$

This can only be true if $\lambda = 1$.

In the case $\lambda = 1$, the points $A_{1}$, $B_{1}$, and $C_{1}$ are the midpoints of the sides of the triangle, so each of the four smaller triangles is similar to the triangle $ABC$, with a similarity ratio of $1:2$. Therefore, in this case, $r_{1} = r_{2} = r_{3} = r_{4} = \frac{1}{2} r$, and indeed the relationship $\frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}} = \frac{1}{r_{4}} + \frac{4}{r}$ holds.

![](https://cdn.mathpix.com/cropped/2024_05_02_012389fba330272f8d91g-1.jpg?height=420&width=545&top_left_y=1926&top_left_x=801)