Let $A_{1} A_{2} A_{3}$ be a triangle. We construct the following points:

- $B_{1}, B_{2}$, and $B_{3}$ are the midpoints of $A_{1} A_{2}, A_{2} A_{3}$, and $A_{3} A_{1}$.

- $C_{1}, C_{2}$, and $C_{3}$ are the midpoints of $A_{1} B_{1}, A_{2} B_{2}$, and $A_{3} B_{3}$.

- $D_{1}$ is the intersection of $\left(A_{1} C_{2}\right)$ and $\left(B_{1} A_{3}\right)$, $D_{2}$ and $D_{3}$ are defined cyclically in the same way.
- $E_{1}$ is the intersection of $\left(A_{1} B_{2}\right)$ and $\left(C_{1} A_{3}\right)$, $E_{2}$ and $E_{3}$ are defined cyclically in the same way.

Calculate the quotient of the area of $\mathrm{D}_{1} \mathrm{D}_{2} \mathrm{D}_{3}$ by the area of $\mathrm{E}_{1} \mathrm{E}_{2} \mathrm{E}_{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_10_59677aa68548da326d08g-214.jpg?height=632&width=1371&top_left_y=1465&top_left_x=366)

## 2 Solutions
We will solve this exercise using barycentric coordinates. We notice that the two triangles of interest, $\mathrm{D}_{1} \mathrm{D}_{2} \mathrm{D}_{3}$ and $\mathrm{E}_{1} \mathrm{E}_{2} \mathrm{E}_{3}$, appear to be copies of the large triangle $A_{1} A_{2} A_{3}$, that is, they are the images of the large triangle by two homotheties. If we can find the ratios of these two homotheties, we can deduce the quotient of the areas. Let's first focus on the center of the homotheties: if $A_{1} \rightarrow D_{2}$, $A_{2} \rightarrow D_{3}$, and $A_{3} \rightarrow D_{1}$, the center must be at the intersection of the three lines $\left(A_{1} D_{2}\right)$, $\left(A_{2} D_{3}\right)$, and $\left(A_{3} D_{1}\right)$, which is the centroid $G$.

Now let's calculate the barycentric coordinates of all the points in the figure. Let's start with the obvious points:

$$
A_{1}:(1,0,0), A_{2}:(0,1,0), A_{3}:(0,0,0) \text { and } G:(1,1,1)
$$

Next, the $B_{i}$ are the midpoints of the segments, so

$$
B_{1}:(1,1,0), B_{2}:(0,1,1), B_{3}:(1,0,1)
$$

and the $C_{i}$ are at the quarter points of the segments, so

$$
C_{1}:(3,1,0), C_{2}:(0,3,1), C_{3}:(1,0,3)
$$

Now let's look at the $D_{i}$. The point $D_{1}$ is the intersection of $\left(A_{1} C_{2}\right)$ with $\left(A_{3} B_{1}\right)$. The course on barycentric coordinates tells us that the lines have equations of the form $u \alpha + \nu \beta + w \gamma = 0$, with $(u, v, w)$ unique up to a multiplicative factor. Let's look at the line $A_{1} C_{2}$: since $A_{1}:(1,0,0)$ is on the line, we deduce $u=0$, and since $C_{2}:(0,3,1)$ is also on the line, we deduce that $3 v + w = 0$. The equation of the line is therefore $\beta - 3 \gamma = 0$. Similarly, we find that the equation of $\left(A_{3} B_{1}\right)$ is $\alpha - \beta = 0$. The point $D_{1}$ is at the intersection of the two lines, so it satisfies both equations. Solving, we find that $D_{1}:(3,3,1)$. Now we want to find the ratio $\frac{G D_{1}}{G A_{3}}$. The point $G$ is on the line $A_{3} D_{1}$, so $G$ can be expressed as the barycenter of $\left(A_{3}, \lambda\right)$ and $\left(D_{1}, \mu\right)$, it suffices to find the appropriate weights $\lambda, \mu$. The property of barycentric coordinates of a barycenter says that:

$$
\mathrm{G}:\left(\frac{3 \mu}{7}, \frac{3 \mu}{7}, \frac{\mu}{7} + \lambda\right)
$$

we deduce that $\mu=7$ and $\lambda=2$, and

$$
\frac{\mathrm{GD}_{1}}{\mathrm{GA}_{3}}=\frac{-2}{7}
$$

The same proof gives the same result for the points $D_{2}$ and $D_{3}$, so the triangle $D_{1} D_{2} D_{3}$ is indeed the image of the triangle $A_{1} A_{2} A_{3}$ by a homothety of center $G$ and ratio $\frac{-2}{7}$. The ratio of the areas is therefore

$$
\frac{\left[D_{1} D_{2} D_{3}\right]}{\left[A_{1} A_{2} A_{3}\right]}=\left(\frac{-2}{7}\right)^2=\frac{4}{49}
$$

I leave the following results to you: $E_{1}:(3,1,1)$, and $\frac{\mathrm{GE}_{1}}{\mathrm{GE}_{3}}=\frac{2}{5}$ and

$$
\frac{\left[E_{1} E_{2} E_{3}\right]}{\left[A_{1} A_{2} A_{3}\right]}=\frac{4}{25}
$$

The sought quotient is therefore

$$
\frac{\left[\mathrm{D}_{1} \mathrm{D}_{2} \mathrm{D}_{3}\right]}{\left[\mathrm{E}_{1} \mathrm{E}_{2} \mathrm{E}_{3}\right]}=\frac{25}{49}
$$

## 4 Group D

## 1 Statements