Let $h_1$ be a homothety with center A and ratio -2. Let $h_2$ be a homothety with center B and ratio $\frac{3}{2}$. Let $h_3$ be a homothety with center C and ratio $\frac{2}{3}$. Let $h_4$ be a homothety with center A and ratio $-\frac{1}{2}$. We want to characterize the transformations $f = h_2 \circ h_1$, $g = h_2 \circ h_3$, and $h = h_4 \circ h_1$. ABC is an equilateral triangle and the circle (C) is circumscribed to it. The center of the circle is O, and BC = 4.

1. Problem Description

Let $h_1$ be a homothety with center A and ratio -

2. Let $h_2$ be a homothety with center B and ratio $\frac{3}{2}$.

Let $h_3$ be a homothety with center C and ratio $\frac{2}{3}$.

Let $h_4$ be a homothety with center A and ratio $-\frac{1}{2}$.

We want to characterize the transformations $f = h_2 \circ h_1$, $g = h_2 \circ h_3$, and $h = h_4 \circ h_1$.

ABC is an equilateral triangle and the circle (C) is circumscribed to it. The center of the circle is O, and BC =

4.

2. Solution Steps

Let $f = h_2 \circ h_1$. Then $f$ is a homothety with ratio $k_2k_1 = \frac{3}{2}(-2) = -3$.

The center $\Omega_f$ of $f$ satisfies $\overrightarrow{B\Omega_f} = \frac{k_1}{1-k_2k_1}\overrightarrow{BA} = \frac{-2}{1-(-3)}\overrightarrow{BA} = \frac{-2}{4}\overrightarrow{BA} = -\frac{1}{2}\overrightarrow{BA}$.

Let $g = h_2 \circ h_3$. Then $g$ is a homothety with ratio $k_2k_3 = \frac{3}{2}(\frac{2}{3}) = 1$.

Since the ratio is 1, $g$ is a translation. The translation vector is $\overrightarrow{BG} = k_2\overrightarrow{BC} = \frac{3}{2}\overrightarrow{BC}$.

Let $h = h_4 \circ h_1$. Then $h$ is a homothety with ratio $k_4k_1 = (-\frac{1}{2})(-2) = 1$.

Since the ratio is 1, $h$ is a translation. The translation vector is $\overrightarrow{AH} = k_4\overrightarrow{AA} = 0$. However, we use the formula for the center of the homothety: $\overrightarrow{A\Omega_h} = \frac{k_1}{1-k_4k_1}\overrightarrow{AA}$, but the denominator $1-k_4k_1 = 0$ so we take the formula for translation $\overrightarrow{AA'} = (1-k_4) \overrightarrow{AO}$, therefore the vector of translation is $(1-k_4)\overrightarrow{AA} + k_4 \overrightarrow{AA}$ which becomes $(1-k_4) \overrightarrow{AA} = (1+\frac{1}{2})\overrightarrow{AA} = \frac{3}{2} \overrightarrow{AA}$.

$f = h_2 \circ h_1$ is a homothety of ratio -

3. $\overrightarrow{B\Omega_f} = -\frac{1}{2}\overrightarrow{BA}$.

$g = h_2 \circ h_3$ is a translation with vector $\frac{3}{2}\overrightarrow{BC}$.

$h = h_4 \circ h_1$ is a translation. Since both centers are A, we have $\overrightarrow{AA'} = k_4 \overrightarrow{AA}$. But the ratio is 1 so this is a translation. However, since the center is the same, we have $\overrightarrow{AA'} = (1-k_4)\overrightarrow{AI}$ where I is the fixed point of $h_1$. In this case the vector of the translation is null. $\overrightarrow{AA} = (1-k_4)\overrightarrow{AI}$, so the vector of the translation is $\overrightarrow{AA'}=\frac{3}{2}\overrightarrow{AA}$. Since the fixed point of $h_1$ is A, $h$ is the identity transformation.

3. Final Answer

$f$ is a homothety with center $\Omega_f$ satisfying $\overrightarrow{B\Omega_f} = -\frac{1}{2}\overrightarrow{BA}$ and ratio -

3. $g$ is a translation with vector $\frac{3}{2}\overrightarrow{BC}$.

$h$ is the identity transformation.