We consider the $\lambda$-ratio inscribed triangle $H^{\prime}=A^{\prime} B^{\prime} C^{\prime}$ of a triangle $H=A B C$ (where $\lambda>0$) to be the triangle whose vertices $A^{\prime}, B^{\prime}, C^{\prime}$ lie on the segments $B C, C A, A B$ respectively, and satisfy the following condition (1). Let the perimeter of the triangle $H_{0}=A_{0} B_{0} C_{0}$ be 1 unit, and let $H_{0}^{\prime}$ be the $\lambda$-ratio inscribed triangle of $H_{0}$. Starting from these, we construct the sequence of triangles $H_{n}, H^{\prime} n$ such that $H_{n+1}$ is the $\frac{1}{\lambda}$-ratio inscribed triangle of $H_{n}^{\prime}$, and $H_{n+1}^{\prime}$ is the $\lambda$-ratio inscribed triangle of $H_{n+1}$ $(n=0,1,2, \ldots)$.

$$
\frac{A C^{\prime}}{C^{\prime} B}=\frac{B A^{\prime}}{A^{\prime} C}=\frac{C B^{\prime}}{B^{\prime} A}=\lambda
$$

a) Calculate the sum of the perimeters of the triangles $H_{n}$ $(n=0,1,2, \ldots)$.

b) Prove that the ratio of the sum of the perimeters of the triangles $H_{n}$ and $H_{n}^{\prime}$ is equal to the ratio of the perimeters of the triangles $H_{0}$ and $H_{0}^{\prime}$.
Let the $1 / \lambda$ inscribed triangles of triangles $H, H^{\prime}$ be denoted by $H^{*}$ and $\bar{H}$ (read: H with a bar over it). We will show that $\bar{H}$ is the $\lambda$ inscribed triangle of $H^{*}$, and can also be obtained from $H$ by a similarity transformation with center $S$ and ratio

$$
q=\frac{\lambda^{2}-\lambda+1}{\lambda^{2}+2 \lambda+1}
$$

where $S$ is the centroid of $H$.

Let the vertices of $H^{*}$ be denoted by $A^{*}$, $B^{*}$, $C^{*}$, and the midpoints of segments $B^{*} C^{\prime}$, $B^{\prime} C^{*}$, $B C$ be denoted by $P$, $Q$, $R$ respectively.

![](https://cdn.mathpix.com/cropped/2024_05_02_8a9c10dd0fa3cf05003fg-1.jpg?height=548&width=599&top_left_y=474&top_left_x=752)

Since

$$
\begin{aligned}
\frac{A C^{\prime}}{A B} & =\frac{A C^{\prime}}{A C^{\prime}+C^{\prime} B}=\frac{\lambda}{\lambda+1}=p \\
\frac{A B^{*}}{A C} & =\frac{A B^{*}}{A B^{*}+B^{*} C}=\frac{\lambda}{\lambda+1}=p
\end{aligned}
$$

the triangle $A B^{*} C^{\prime}$ can be obtained from $H$ by a similarity transformation with center $A$ and ratio $p$, hence $B^{*} C^{\prime} \| B C$, and $A P = p \cdot A R$. Due to the construction of points $B^{*}$ and $C^{*}$, we have $C^{*} B = A C^{\prime}$ and $A B^{*} = B^{\prime} C$, so $A B^{\prime} C^{*}$ can be obtained from $H$ by a similarity transformation with center $A$ and ratio $(1-p)$, hence $B^{\prime} C^{*} \| B C$, and $A Q = (1-p) \cdot A R$. Let $\bar{A}$ be the point on segment $P Q$ such that

$$
\frac{P \bar{A}}{\bar{A} Q}=\frac{p}{1-p}=\lambda
$$

then the similarity transformation with center $\bar{A}$ and ratio $\lambda$ maps $P$ to $Q$, and $B^{*} C^{\prime}$ to $B^{\prime} C^{*}$, so

$$
\frac{B^{\prime} \bar{A}}{\bar{A} C^{\prime}}=\frac{\bar{A} C^{*}}{B^{*} \bar{A}}=\frac{1}{\lambda}
$$

Thus, $\bar{A}$ is a vertex of the $1 / \lambda$ inscribed triangle of $H^{\prime}$, and the $\lambda$ inscribed triangle of $H^{*}$, and lies on the median $A R$ of $H$. From (2),

$$
A \bar{A}=p \cdot A Q+(1-p) \cdot A P=2 p(1-p) \cdot A R=\frac{2 \lambda}{(\lambda+1)^{2}} A R
$$

so $\bar{A}$ can also be obtained from $A$ by a similarity transformation with center $S$ and ratio

$$
q=\frac{\bar{A} S}{A S}=1-\frac{A \bar{A}}{A S}=1-\frac{3 \lambda}{(\lambda+1)^{2}}=\frac{\lambda^{2}-\lambda+1}{\lambda^{2}+2 \lambda+1}
$$

Similar statements can be shown for the other two vertices of $\bar{H}$, thus completing the proof of our previous remark.

According to the statement we have just proven, $H_{n+1}$ can be obtained from $H_{n}$ by a similarity transformation with ratio $q$, so the perimeter of $H_{n+1}$ is $q$ times the perimeter of $H_{n}$. Therefore, the perimeters of the triangles $H_{n}$ form a geometric sequence where the first term is 1 and the ratio of consecutive terms is $q$, with $|q|<1$. It is known that in this (infinite) sequence, the sum of the terms is $\left(1-q^{n}\right) /(1-q)$, so the sum $K$ of the perimeters of the triangles $H_{n}$, as $n$ grows without bound,

$$
K=\frac{1}{1-q}=\frac{(\lambda+1)^{2}}{3 \lambda} \quad \text { length units. }
$$

We have already seen that the $1 / \lambda$ inscribed triangle of a triangle's $\lambda$ inscribed triangle is the same as the $\lambda$ inscribed triangle of the triangle's $1 / \lambda$ inscribed triangle. Therefore, the perimeters of the triangles $H_{n}^{\prime}$ also form a geometric sequence where the ratio of consecutive terms is $q$. The statement in part b) of the problem thus follows from the fact that the sums of infinite geometric series with the same ratio are in the same ratio as the first terms of these series.