Given triangle $AIJ$ inscribed in circle $(C)$ with center $O$. Points $B$ and $C$ are symmetric to $A$ with respect to $I$ and $J$ respectively. $M$ is the midpoint of $[IJ]$. $D$ is diametrically opposite to $A$ in circle $(C)$, and $N$ is the orthogonal projection of $D$ onto line $(BC)$. $h$ is a homothety with center $A$ and ratio $2$. 1) Determine the images of points $I, J, O$ under $h$. 2) What is the image of line $(IJ)$ and line $(OM)$ under $h$? 3) Prove that $A, M, N$ are collinear.

1. Problem Description

Given triangle $AIJ$ inscribed in circle $(C)$ with center $O$. Points $B$ and $C$ are symmetric to $A$ with respect to $I$ and $J$ respectively. $M$ is the midpoint of $[IJ]$. $D$ is diametrically opposite to $A$ in circle $(C)$, and $N$ is the orthogonal projection of $D$ onto line $(BC)$. $h$ is a homothety with center $A$ and ratio $2$.

1) Determine the images of points $I, J, O$ under $h$.

2) What is the image of line $(IJ)$ and line $(OM)$ under $h$?

3) Prove that $A, M, N$ are collinear.

2. Solution Steps

1) Let $I', J', O'$ be the images of $I, J, O$ under the homothety $h$ with center $A$ and ratio $2$. Then $\vec{AI'} = 2\vec{AI}$, $\vec{AJ'} = 2\vec{AJ}$, $\vec{AO'} = 2\vec{AO}$.

Since $B$ is symmetric to $A$ with respect to $I$, then $\vec{AI} = \vec{IB}$, thus $\vec{AB} = 2\vec{AI}$. Therefore $I' = B$.

Since $C$ is symmetric to $A$ with respect to $J$, then $\vec{AJ} = \vec{JC}$, thus $\vec{AC} = 2\vec{AJ}$. Therefore $J' = C$.

Since $D$ is diametrically opposite to $A$ in circle $(C)$, then $\vec{AD} = 2\vec{AO}$. Therefore $\vec{AO'} = \vec{AD}$, so $O'=D$.

Thus, the images of $I, J, O$ under $h$ are $B, C, D$ respectively.

2) Since $h$ is a homothety with ratio 2 and center $A$, the image of line $(IJ)$ under $h$ is a line parallel to $(IJ)$ passing through $B$ and $C$. Therefore, the image of line $(IJ)$ under $h$ is line $(BC)$.

Let $M'$ be the image of $M$ under $h$. Then $\vec{AM'} = 2\vec{AM}$. Since $M$ is the midpoint of $[IJ]$, $\vec{AM} = \frac{\vec{AI} + \vec{AJ}}{2}$. Therefore, $\vec{AM'} = 2\vec{AM} = \vec{AI} + \vec{AJ}$. Also $\vec{AI} = \frac{1}{2}\vec{AB}$ and $\vec{AJ} = \frac{1}{2}\vec{AC}$.

So $\vec{AM'} = \frac{1}{2}(\vec{AB} + \vec{AC})$.

Let $K$ be the midpoint of $BC$. Then $\vec{AK} = \frac{\vec{AB} + \vec{AC}}{2}$. Thus, $\vec{AM'} = 2\vec{AK}$ which is incorrect.

Since $M$ is the midpoint of $IJ$, its image $M'$ is the midpoint of $I'J'$, i.e., $M'$ is the midpoint of $BC$. Then $M'$ is $K$.

So we have that $O' = D$ and $M'=K$, where $K$ is the midpoint of $BC$. Thus, the image of line $(OM)$ is line $(DK)$.

We know that $N$ is the projection of $D$ onto $BC$, thus $DN \perp BC$. Let the circle $(C)$ have radius $R$.

Also, $M'$ is the midpoint of $BC$.

3) Let's prove that $A, M, N$ are collinear.

Since $N$ is the orthogonal projection of $D$ onto $BC$, $DN \perp BC$.

Since $O$ is the center of the circle and $M$ is the midpoint of $IJ$, $OM \perp IJ$. The image of line $(IJ)$ under $h$ is line $(BC)$, which is parallel to $(IJ)$. Also $\vec{AM'} = 2\vec{AM}$.

Since $DN \perp BC$ and $M'$ is the midpoint of $BC$, $BC$ is perpendicular to $DN$. Let $K$ be the midpoint of $BC$. $K=M'$.

We want to prove that $A, M, N$ are collinear. It is equivalent to prove that $\vec{AM} = \lambda \vec{AN}$ for some $\lambda$.

Since $D$ is diametrically opposite to $A$, $\angle AJA = \angle AIA = 90$. Thus $AJ\perp JC$ and $AI\perp IB$. Since $BC$ is the image of $IJ$ under the homothety, $(IJ)||(BC)$. Since $N$ is the projection of $D$ onto $(BC)$, we have $DN \perp BC$. Thus, $DN \perp (IJ)$.

Let the coordinates of the points be $A=(0,0), I=(x_i, y_i), J=(x_j, y_j)$.

$M = (\frac{x_i+x_j}{2}, \frac{y_i+y_j}{2})$. Thus, $\vec{AM} = (\frac{x_i+x_j}{2}, \frac{y_i+y_j}{2})$.

Since $B$ is the symmetric of $A$ with respect to $I$, we have $I$ is the midpoint of $AB$. Thus $x_i=\frac{x_b+0}{2}$, $y_i = \frac{y_b+0}{2}$, thus $x_b = 2x_i, y_b=2y_i$, so $B=(2x_i, 2y_i)$.

Similarly, since $C$ is the symmetric of $A$ with respect to $J$, we have $J$ is the midpoint of $AC$. Thus $x_j=\frac{x_c+0}{2}$, $y_j = \frac{y_c+0}{2}$, thus $x_c = 2x_j, y_c=2y_j$, so $C=(2x_j, 2y_j)$.

The line equation for $(BC)$ is $\frac{y-2y_i}{x-2x_i} = \frac{2y_j-2y_i}{2x_j-2x_i} \rightarrow \frac{y-2y_i}{x-2x_i} = \frac{y_j-y_i}{x_j-x_i}$.

Since $D$ is diametrically opposed to $A$ in the circle centered at $O$, then $\vec{OD} = -\vec{OA}$. This means $\vec{AD} = 2\vec{AO}$.

Final Answer:

1. The images of $I, J, O$ under $h$ are $B, C, D$ respectively.

2. The image of line $(IJ)$ under $h$ is line $(BC)$. The image of line $(OM)$ is line $(DK)$, where $K$ is the midpoint of $(BC)$.

3. $A, M, N$ are collinear. I am not sure how to prove it rigorously from the given information.