We are given three non-collinear points $A$, $B$, and $C$. Point $C$ is the midpoint of segment $[BD]$. Let $f$ be an affine transformation defined by $f(A) = A$, $f(B) = D$, and $f(C) = B$. We need to: 1. Determine the set of invariant points under $f$.

1. Problem Description

We are given three non-collinear points $A$, $B$, and $C$. Point $C$ is the midpoint of segment $[BD]$.

Let $f$ be an affine transformation defined by $f(A) = A$, $f(B) = D$, and $f(C) = B$.

We need to:

1. Determine the set of invariant points under $f$.

2. Deduce that there exists a point $G$ on the line $(BC)$ that is invariant under $f$, and express $\vec{GB}$ in terms of $\vec{GC}$.

3. Prove that $f$ is an affinity and specify its characteristic elements.

2. Solution Steps

1. Invariant points:

Let $M$ be an invariant point. Then $f(M) = M$.

We can write $M$ as a barycentric combination of $A$, $B$, and $C$: $M = \alpha A + \beta B + \gamma C$, where $\alpha + \beta + \gamma = 1$.

Applying $f$ to $M$, we have:

$f(M) = \alpha f(A) + \beta f(B) + \gamma f(C) = \alpha A + \beta D + \gamma B$.

Since $f(M) = M$, we have:

$\alpha A + \beta D + \gamma B = \alpha A + \beta B + \gamma C$.

$\beta D + \gamma B = \beta B + \gamma C$.

Since $C$ is the midpoint of $[BD]$, we have $2C = B + D$, or $D = 2C - B$.

Substituting this into the equation, we get:

$\beta (2C - B) + \gamma B = \beta B + \gamma C$.

$2\beta C - \beta B + \gamma B = \beta B + \gamma C$.

$2\beta C - \beta B + \gamma B - \beta B - \gamma C = 0$.

$2\beta C - 2\beta B + \gamma B - \gamma C = 0$.

$(2\beta - \gamma)C + (\gamma - 2\beta)B = 0$.

Since $B$ and $C$ are linearly independent (as $A, B, C$ are non-collinear), we must have:

$2\beta - \gamma = 0$ and $\gamma - 2\beta = 0$.

Thus, $\gamma = 2\beta$.

Since $\alpha + \beta + \gamma = 1$, we have $\alpha + \beta + 2\beta = 1$, so $\alpha + 3\beta = 1$, or $\alpha = 1 - 3\beta$.

Therefore, the set of invariant points is the line $A + \lambda(B - \frac{1}{2}C)$, where $\lambda$ is a scalar.

2. Invariant point $G$ on line $(BC)$:

Let $G$ be a point on line $(BC)$ such that $f(G) = G$. We can write $G = \lambda B + (1 - \lambda)C$ for some scalar $\lambda$.

$f(G) = \lambda f(B) + (1 - \lambda) f(C) = \lambda D + (1 - \lambda) B$.

Since $C$ is the midpoint of $[BD]$, $D = 2C - B$.

$f(G) = \lambda (2C - B) + (1 - \lambda) B = 2\lambda C - \lambda B + B - \lambda B = (1 - 2\lambda) B + 2\lambda C$.

Since $f(G) = G$, we have $(1 - 2\lambda)B + 2\lambda C = \lambda B + (1 - \lambda)C$.

Equating coefficients, we get $1 - 2\lambda = \lambda$ and $2\lambda = 1 - \lambda$.

From either equation, we get $3\lambda = 1$, so $\lambda = \frac{1}{3}$.

Thus, $G = \frac{1}{3}B + \frac{2}{3}C$. This means that $G$ is the barycenter of $(B, 1)$ and $(C, 2)$.

Now, we need to express $\vec{GB}$ in terms of $\vec{GC}$.

$G = \frac{1}{3}B + \frac{2}{3}C$ implies $3G = B + 2C$.

Then, $\vec{GB} = \vec{G} - \vec{B} = \vec{G} - (3\vec{G} - 2\vec{C}) = 2\vec{C} - 2\vec{G} = -2(\vec{G} - \vec{C}) = -2\vec{CG} = 2\vec{GC}$.

So, $\vec{GB} = 2\vec{GC}$.

3. Affine transformation:

We have $f(A) = A$, $f(B) = D$, $f(C) = B$, and $D = 2C - B$. Since $f(G) = G$ and $G=\frac{1}{3}B+\frac{2}{3}C$, the line $(AG)$ is invariant, and $(AG)$ is the axis. Let's consider the direction $\vec{BC}$. $\vec{f(BC)} = f(\vec{BC}) = \vec{f(B)f(C)} = \vec{DB}$. Since $C$ is the midpoint of $BD$, we have $\vec{DB} = -2\vec{BC}$. Thus $f$ is an affinity with axis $(AG)$, direction $\vec{BC}$ and ratio -

2. Therefore, f is an affinity of axis (AG) and ratio -

2.

3. Final Answer

1. The set of invariant points is the line defined by $A + \lambda(B - \frac{1}{2}C)$, where $\lambda$ is a scalar.

2. $G = \frac{1}{3}B + \frac{2}{3}C$, and $\vec{GB} = 2\vec{GC}$.

3. $f$ is an affinity with axis $(AG)$ and ratio $-2$.